States Of Matter Gases And Liquids Long Answer Type Questions
Question 1. Determine the types of intermodular forces of attraction in the following instances. w-hexane, SO2,CO2, CHCI3, (CH3)2CO, (CH3)2O
Answer: London forces or instantaneous induced dipole instantaneous induced dipole attraction: n-hexane, CO2, Dipole-dipole attraction: SO2 CHCI3, (CH3)2O. However, the London force also acts in this case.
Question 2. Which type of intermodular forces of attraction act between O2 and water modules when O2 is dissolved in water?
Answer: O2 is a non-polar molecule, whereas the H2O molecule is polar. Hence, the force of attraction acting between O2 and H2O molecules is dipole-induced dipole attraction
Question 3. Which types of intermodular forces of attraction act between the modules in liquid HF?
Answer: HF is a polar covalent molecule. Hence, the forces of attraction between HF molecules in their liquid state are dipole-dipole attraction and H -H-H-bonding.
Question 4. What do you mean by the pressure of the gas?
Answer: Since E2>E1, T2>T1. This is because, with the increase in the absolute temperature of a gas, the average kinetic energy ofthe gas molecules increases
Question 5. At a constant temperature, the pressures of four gases A, B, C, and D are 0.2 atm, 250torr, 26.23k Pa, and 14.2 bar, respectively. Arrange them according to their increasing pressure.
Answer: \(A: 0.2 \mathrm{~atm}, B: 250 \mathrm{torr}=\frac{250}{760} \mathrm{~atm}=0.33 \mathrm{~atm}\)
⇒ \(A: 0.2 \mathrm{~atm}, B: 250 \mathrm{torr}=\frac{250}{760} \mathrm{~atm}=0.33 \mathrm{~atm}\)
⇒ \(D: 14.2 \mathrm{bar}=\frac{14.2}{1.013}=14.02 \mathrm{~atm}\)
∴ A<C<B<D
Question 6. Among the four quantities—mass, pressure, temperature, and volume, which are taken to be constant in the following gas laws?
- Boyle’slaw
- Charles’law
- Gay-Lussac’slaw
- Avogadro’s law.
Answer:
- Boyle’s law: Mass and temperature ofthe gas
- Charles’ law: Mass and pressure of the gas
- Gay-Lussac’s law: Mass and volume of the gas
- Avogadro’s law: Pressure and temperature of the gas
Question 7. Why does the volume of a given mass of gas increase by decreasing its pressure at constant temperature?
Answer: According to Boyle’s law, PV= constant for a fixed mass of gas at a constant temperature. That is, the product of P and V for a fixed mass of gas at a constant temperature is always constant. Suppose, P and V are the pressure and volume of a given mass of gas at a constant temperature.
Keeping the temperature constant, if the pressure is made to \(\frac{P}{x}\)(where x > 1 ), then the volume ofthe gas will be V x x because PV = constant
Question 8. Why does the volume of a gas increase by increasing its number of moles at a given temperature and pressure?
Answer: According to Avogadro’s law, at a given temperature and pressure the volume (V) of a gas is directly proportional to its number of moles (n), i.e., V ∝ n. Therefore, if the number of moles of a gas is increased at a constant temperature and pressure, its volume will increase
Question 9. Will the nature of the following graphical presentations for a given mass of gas be the same?
Answer:
P vs V ata constant temperature
For a fixed mass of a gas at a given temperature PV = constant. This relation expresses an equation of a rectangular hyperbola. Therefore, the P vs V plot for a fixed mass of gas at a given temperature will produce a curve of a rectangular hyperbola.
V vs T at constant pressure
For a fixed amount of a gas at a constant pressure, V = KxT (K = constant). This relation expresses an equation of a straight line passing through the origin. Therefore, the V vs T plot for a fixed mass of gas at constant pressure will give a straight line passing through the origin.
Question 10. When does the graph showing variation ofthe volume ofa given mass of gas with pressure at a constant temperature become linear?
Answer: For a given mass of gas at a fixed temperature, PV = K (constant). That is, \(P=\frac{K}{V}\) This relation expresses an equation of a straight line passing through the origin. Therefore, if P is plotted against \(\frac{1}{V}\) for a given mass of a gas at a fixed temperature, a straight line will be obtained
Question 11. N2 gas is present in a 1L desiccator at latm pressure. The pressure ofthe gas decreases to 78mmHg pressure when the desiccator is partially evacuated using a vacuum pump at a constant temperature. Find out the final volume of the gas.
Answer: Since the volume of the desiccator is fixed, the final volume of the gas will be 1L even after the desiccator is partially evacuated. In this process, the number of moles of the gas decreases but its volume and temperature remain the same. As the pressure ofthe gas is reduced
Question 12. Plot density vs pressure for a fixed mass of an ideal gas at a
constant temperature.
Answer: Since docs for a given mass of gas at a constant temperature.
Question 13. According to Boyle’s law, at a constant temperature, the volume ofa given mass of gas is inversely proportional to its pressure. But when a balloon is filled with air, both the volume and pressure ofthe gas inside it increase—Explain.
Answer: When the balloon is pumped, the quantity of air inside the balloon goes on increasing. As a result, the mass of air inside it does not remain constant. Moreover, pumping causes a rise in the temperature of the air inside the balloon. Thus, neither the temperature nor the mass of the air remains constant. So, Boyle’s law is not applicable in this case.
Question 14. Under different conditions, the following graph is obtained for an ideal gas. Mentioning A and B,- identify the conditions.
Answer: For a fixed amount of gas at a given temperature, PV is constant. This means the value of PV will always be constant for a fixed amount of a gas at a given temperature no matter what the pressure of the gas is. Hence PV vs P or PV vs V plot will give a straight line parallel to the P-axis or V-axis, respectively. Therefore, A will be equal to PV and B will be equal to P or V.
For a fixed amount of a gas at a given pressure V – K (constant) x T, i.e., V/T = K. This relation tells us that the value of V/ T will always be constant for a fixed amount of gas at a given pressure irrespective of the value of temperature of the gas. Hence, the V/T vs T plot will give a straight line parallel to the T axis. Hence, A will be equal to V/ T and B will be equal to T.
Question 14. What are the molar volumes of nitrogen and argon gases at 273.15K temperature and 1 atm pressure? [consider both the gases behave ideally]
Answer: At 273.15K and late pressure, both nitrogen and argon behave ideally. Hence, at this temperature and pressure, the molar volume of each of them will be 22.4L
Question 15. Comment on the validity of Boyle’s law for the following
reaction: N3O4(g) 2NO2(g)
Answer: The number of molecules of N2O4 and NO2 varies with the pressure at constant temperature. Hence, the mass ofthe gas mixture does not remain constant. Thus, Boyle’s law is not applicable here.
Question 16. If a substance were to be in a gaseous state at absolute zero temperature, what would be the theoretical value of its pressure?
Answer: \(V_t=V_0\left(1+\frac{t}{273}\right)\)
Pressure and mass of gas being constant] At absolute zero temperature, t = -273°C.
⇒ \(\text { Hence, } V_{-273^{\circ} \mathrm{C}}=V_0\left(1-\frac{273}{273}\right)=0 \text {. }\)
Since the volume of the gas is zero (0), the theoretical value of pressure will be zero (0).
Question 17. At a given pressure, the volume of a given amount of gas at 0°C is V0. Will the V vs t (celsius temperature) plot for this gas be linear? Will it be a straight line passing through the origin? If this straight line does not pass through the origin, then what will be its slope and intercept?
Answer: According to Charles’ law, Vt = V0 \(\left(1+\frac{t}{273}\right)\); where Vt and V0 are the volumes of a given mass of gas at a temperature t°C respectively, when the pressure of the gas is kept constant. Hence, the above equation can be rewritten as \(V_t=V_0+\frac{t}{273} V_0\)……………[1]
V0 is a fixed quantity for a given mass of gas at a constant pressure. Thus, the equation [1] represents a straight-line equation. Hence, the Vt vs t plot will give a straight line.
Equation [1] does not represent an equation of a straight line passing through the origin. Hence, the Vf vs t plot will not be a straight line passing through the origin
According to the equation [1], the straight line resulting from the plot of V( vs t has a slope of V0/273 and an intercept of V0.
Question 18. Under what conditions will the value of- always be the same irrespective ofthe value of T?
Answer: For a fixed mass of gas at constant volume, P ∝ T or, P = f Cx T (K = constant). Therefore, P/T = K. This relation indicates that the value of P/T is always constant for a fixed mass of a gas at constant volume irrespective of the value of T is
Question 19. Under what conditions will the value of PV always be the same irrespective of the value of P (or V)?
Answer: According to Boyle’s law, for a fixed mass of gas at a constant temperature, PV = constant. Therefore, the value of PV is always constant for a fixed mass of a gas at a constant temperature irrespective ofthe value of P.
Question 20. A certain amount of an ideal gas is enclosed in a cylinder fitted with a movable piston. What would be the changes in the volume ofthe gain in the following processes?
- The pressure of the gas is reduced by 25% at constant temperature.
- The temperature of the gas is increased by 50% at constant pressure.
Answer: According to Boyle’s law, P1V1 = P2V2 when the mass and temperature of a gas are constant. After the reduction of initial pressure (P1) by 25%, the final pressure (P2) becomes \(\left(P_1-P_1 \times \frac{25}{100}\right)=0.75 P_1\)
i.e., P2 = 0.75P1
∴ \(V_2=\frac{P_1 V_1}{P_2}=\frac{P_1 \times V_1}{0.75 P_1}=\frac{4}{3} V_1\)
Therefore, the change in volume \(=\frac{4}{3} V_1-V_1=\frac{V_1}{3}\)
According to Charles’ law, V1T2 = V2T1 when the mass and pressure of a gas remain constant. After increasing the initial temperature (T1) by 50%, the final temperature (T2) becomes \(\left(T_1+T_1 \times \frac{50}{100}\right)=1.5 T_1 \text {, i.e., } T_2=1.5 T_1\)
∴ \(V_1 \times 1.5 T_2=V_2 T_2 \quad \text { or, } V_2=1.5 \times V_1\)
∴ The change in volume = 1.5 V1– V1 = 0.5V1
Question 21. Determine the values of molar gas constant in the following units— mL torr K-1 mol-1; kPa-L-K-1-mol-1.
Answer: R =0.0821 L-atm -K-1. mol-1
= 0.0821 x 103 x 760 torr -mL -K-1.mol-1
[Since 1L = 103mL and latm = 760tor]
= 6.23 x 104 torr -mL -K-1 .mol-1
= 0.0821 L .atm -K-1 -mol-1
= 0.0821 X 1.013 x 1 02 kPa . L K-1. mol-1
= 8.31 kPa-L-K1-mol-1
[Since 1 atm = 1.013 x 105 Pa = 1.013 x 102kPa]
Question 22. For the same mass of two ideal gases X and Y at the same temperature and pressure, the volume of Y is found to be three times as large as that of X. Compare the values of their molar masses.
Answer: Px = Py, Tx = Ty, 3VX = Vy
According to the ideal gas equation, Pxvx = nXRTx and = PY VY = NY R TY
i.e., \(\frac{P_X}{P_Y} \times \frac{V_X}{V_Y}=\frac{n_X}{n_Y} \times \frac{T_X}{T_Y} \quad \text { or, } \frac{1}{3}=\frac{n_X}{n_Y}\)
Let, molar masses of X and Y be My and My respectively
∴ \(\frac{w / M_X}{w / M_Y}=\frac{1}{3} \quad \text { or, } \frac{M_X}{M_Y}=3\)
Question 23. At constant temperature and pressure volume of an ideal gas (molecular mass 28 g. mol-1) is 23.36 times greater than its mole number. Find out its density at the same temperature and pressure.
Answer: Let us suppose, the tire volume and density of n mol of an ideal gas are V L and d g.L-1 at P atm and T K.
Now, PV = nRT and d \(=\frac{P M}{R T}\)i.e., \(d=M \times \frac{n}{V}\)
Question 24. When a flask of fixed volume is filled with – mol of an ideal gas A at a constant temperature, the pressure ofthe gas becomes 2 atoms. Adding 2y mol of another ideal gas B to the flask at the same temperature causes the pressure of the system to increase to 4.0 atm.
Answer: For tyre gas. A: P = 2 atm and \(n=\frac{x}{2} \mathrm{~mol}\)
Hence, PV \(=n R T \text { or, } 2 \times V=\frac{x R T}{2}\)
For gas mixture: P=4atm \(\& n=\frac{x}{2}+2 y=\frac{1}{2}(x+4 y) \mathrm{mol}\)
∴ In case of gas mixture \(4 \times V=\frac{1}{2}(x+4 y) R T\)
Deriding [2] -r [1] we get \(\frac{4}{2}=\frac{x+4 y}{x} \text { or, } x=4 y\)
Question 25. Rank the gases N2, CO2, and CH5 in order of their increasing densities given temperature and pressure.
Answer: We know \(d=\frac{P M}{R T}\) Thus, docM at a certain temperature and pressure. Since, \(M_{\mathrm{CH}_4}<M_{\mathrm{N}_2}<M_{\mathrm{CO}_2}\) at a fixed temperature and pressure, \(d_{\mathrm{CH}_4}<d_{\mathrm{N}_2}<d_{\mathrm{CO}_2}\)
Question 26. Under which of the following conditions will the density of a, fixed mass of SO2 gas be higher?— STP 27°C and 3atm pressure.
Answer: \(d=\frac{P M}{R T}\) and M is constant for a particular gas. Hence ∝PIT. The value of PIT at 3atm and 27°C is greater than that at STP. Therefore, the density of SO2 at 3atm and 27°C will be greater than that at STP.
Question 27. Determine the SI unit of
Answer: \(\frac{P V^2 T^2}{n}=\frac{\mathrm{N} \times\left(\mathrm{m}^3\right)^2 \times \mathrm{K}^2}{\mathrm{~m}^2 \times \mathrm{mol}}=\mathrm{N} \cdot \mathrm{m}^4 \cdot \mathrm{K}^2 \cdot \mathrm{mol}^{-1}\)
Question 28. At 27°C and 1 atm pressure, the volume of a 5.0g mixture of He and Ar gases is 10dm3. Find the mass per cent of the two gases in the gas mixture.
Answer: Let, the amount of He and Ar in the mixture be ag and bg respectively. Hence, a + b = 5
⇒ \(\text { Now, } P V=\left(n_1+n_2\right) R T \text { or, } 1 \times 10=\left[\frac{a}{4}+\frac{b}{40}\right] \times 0.0821 \times 300\)
∴ 10 a +b = 16.24
By solving [1] and [2] we get, a = 1.25g and b = 3.75g
∴ \(\% \mathrm{He}=\frac{1.25}{5} \times 100=25 \text { and } \% \mathrm{Ar}=\frac{3.75}{5} \times 100=75\)
Question 29. Arrange O2, CO2, Ar, and SO2 gases to present a sample of air in order of their increasing pressures.
Answer: \(d_1=\frac{P M}{R T}, d_2=\frac{P M}{4 \times 8 R T}=\frac{1}{32} \frac{P M}{R T}=\frac{d_1}{32}\)
Question 30. A gas mixture consisting of O2 and N2 gases has a volume of 5 L at 25°C. In the mixture, if the mass of O2 gas is twice that of N2 gas, then which one of them will have a greater contribution to the total pressure ofthe mixture?
Answer: In the mixture \(n_{\mathrm{O}_2}=\frac{2 w}{32}=\frac{w}{16} \mathrm{~mol} \text { and } n_{\mathrm{N}_2}=\frac{w}{28} \mathrm{~mol} \text {. }\)
∴ Total number of moles (n) \(=\frac{w}{16}+\frac{w}{28}=\frac{11 w}{112} \mathrm{~mol}\)
⇒ \(\text { Hence, } x_{\mathrm{N}_2}=\frac{w}{28} \times \frac{112}{11 w}=\frac{4}{11} \text { and } x_{\mathrm{O}_2}=\frac{w}{16} \times \frac{112}{11 w}=\frac{7}{11}\)
Thus, xO2 > xN2. So, pO2> PN2
(Since pi = zip)
Question 31. In a gas mixture of H2 and He, the partial pressure of H2 is half that of He. Find the mole fractions of H2 and He in the mixture.
Answer: As given, p2 = pH2 or, pHe = 2 x PH2
or, xHe x P = 2 x XH2 x P or, xHe = 2x xe = 2x xH2
Hence, 2 x xH2+ xH2 = 1 or, \(x_{\mathrm{H}_2}=\frac{1}{3} \approx 0.34\)
∴ xHe = 2 X XH2 = 0.68
Question 32. A closed vessel contains an equal mass of O2 and CH4 gases at 25°C. What fraction of the total pressure is contributed by CH4 gas?
Answer: \(n_{\mathrm{CH}_4}=\frac{w}{16}, n_{\mathrm{O}_2}=\frac{w}{32}\)
∴ Total number of moles = \(\frac{w}{16}+\frac{w}{32}=\frac{3 w}{32}\)
∴ \(x_{\mathrm{CH}_4}=\frac{w}{16} \times \frac{32}{3 w}=\frac{2}{3}\)
⇒ \(p_{\mathrm{CH}_4}=\frac{2}{3}\); Thus Memthane Contributes \(\frac{2}{3}\) rd of the total pressure.
Question 33. A mixture of O2 and H2 gases contains 20% of H2 gas. At a certain temperature, the total pressure of the mixture is found to be 1 bar. What is the partial pressure of O2 (in bar) in the mixture?
Answer: In the mixture, 20% H2 is present. Hence extent of oxygen = 80%
∴ \(n_{\mathrm{H}_2}=\frac{20}{2}=10 \text { and } n_{\mathrm{O}_2}=\frac{80}{32}=\frac{5}{2}\)
∴ Total number of moles \(=\left(10+\frac{5}{2}\right)=\frac{25}{2}\)
∴ \(p_{\mathrm{O}_2}=\frac{5}{2} \times \frac{2}{25} \times 1 \mathrm{bar}=0.2 \mathrm{bar}\)
Question 34. At a constant temperature, gas A (volume VA and pressure PA) is mixed with gas B (volume VR and pressure PB). What will the total pressure ofthe gas mixture be?
Answer: Total pressure ofthe gas mixture \(P=\frac{P_A V_A+P_B V_B}{V_A+V_B}\)
Question 35. For which of the following gas mixtures is Dalton’s law
of partial pressures applicable? NO + O2, CO2 + CO , CO + O2 ,CH4 + C2H6,CO + H2
Answer: Dalton’s law is applicable in case NO reacts with O2 to form NO2. CO reacts with O2 to form CO2. Hence, Dalton’s law does not apply to 1 and 3.
Question 36. In a mixture of A, B, and Cgases, the mole fractions of A are 0.25 and 0.45, respectively. If the total pressure of the mixture is P, then find the partial pressure of B in the mixture.
Answer: In the gas mixture, the mole of B =1- (0.25 + 0.45) = 0.3.
∴ The partial pressure of B in the mixture = 0.3 p.
Question 36. A gas mixture consists of three gases A, B, and C with the number of moles 1, 2, and 4, respectively. Which of these gases will have a maximum partial pressure if the total pressure ofthe mixture is Pata given temperature t what temperature will the average velocity of O2 molecules be equal to that of U2 molecules at 20K?
Answer: In the mixture, the total number of moles of the constituent gases =(1 + 2 + 4) = 7mol. The mole fractions of A, B, and C \(\frac{1}{7}, \frac{2}{7} \text { and } \frac{4}{7} \text {, }\) respectively. Since the mole fraction of C is the highest, its partial pressure will be the highest in the mixture.
Question 37. Rank Cl2, SO2, CO2, and CH4 gases in Increasing order of their rates of diffusion under identical set of conditions.
Answer: The molar mass of the gases follows the order: \(M_{\mathrm{CH}_4}<M_{\mathrm{CO}_2}<M_{\mathrm{SO}_2}<M_{\mathrm{Cl}_2}\)
Hence, at constant temperature and pressure, the rates of diffusion will be the order: \(r_{\mathrm{Cl}_2}<r_{\mathrm{SO}_2}<r_{\mathrm{CO}_2}<r_{\mathrm{CH}_4}\)
Question 38. Why are the rates of diffusion of N2O and CO2 gases the same under identical set of conditions?
Answer: N2O and CO2 have the same molar mass. So, their rates of diffusion will be equal at a certain temperature and pressure.
Question 39. At constant temperature and pressure, the rate of diffusion of H2 gas is A/15 times. Find the value of n.
Answer: \(\frac{r_{\mathrm{H}_2}}{r_{\mathrm{C}_n \mathrm{H}_{4 n-2}}}=\sqrt{\frac{16 n-2}{2}}=\sqrt{8 n-1}=\sqrt{15}\)
Question 40. Under the same conditions, a gas diffuses times as fast as an SO2 gas. Find the molecular mass ofthe gas.
Answer: \(\frac{r_{\text {gas }}}{r_{\mathrm{SO}_2}}=\sqrt{\frac{M_{\mathrm{SO}_2}}{M_{\text {gas }}}}=\sqrt{2}\) \(\text { or, } \frac{M_{\mathrm{SO}_2}}{M_{\text {gas }}}=2 \text { or, } M_{\mathrm{gas}}=32 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)
or, 8n-1 = 15 or, n = 2
Question 41. Besides the lower layer, CO2 is also found in the upper layer ofthe atmosphere although it is heavier than O2 or N2—Explain.
Answer: The rate of diffusion of a gas is not influenced by the gravitational force. Hence, CO2 diffuses throughout the atmosphere instead of residing only at the lower layer of the atmosphere.
Question 42. The molecular masses of A, B, and C are 2, 4, and 28, respectively. Arrange them according to their increasing rates of diffusion.
Answer: Molar masses of the three gases follow the order: MA < MB < Mc So, at a certain temperature and pressure, the rates of diffusion will be in the order: rC<rB<rA.
Question 43. A closed vessel holds a gas mixture consisting of C2H6, C2H4, and CH4, each with an amount of 2.5 mol. However, due to a pinhole in the vessel, the gas mixture undergoes effusion. What will be the order of partial pressures of the gases in the vessel after some time?
Answer: Molar masses of C2H6, C2H4 and CH4 follow the order: \(M_{\mathrm{CH}_4}<M_{\mathrm{C}_2 \mathrm{H}_4}<M_{\mathrm{C}_2 \mathrm{H}_6}\) Thus, at a certain temperature and pressure, their rates of effusion will be in the order \(r_{\mathrm{C}_2 \mathrm{H}_6}<r_{\mathrm{C}_2 \mathrm{H}_4}<r_{\mathrm{CH}_4}\) Therefore, the order of partial pressure after some time will be: \(p_{\mathrm{CH}_4}<p_{\mathrm{C}_2 \mathrm{H}_4}<p_{\mathrm{C}_2 \mathrm{H}_6}\)
Question 44. Under similar conditions of temperature and pressure, the times it takes for the effusion of the same volume of H2, N2, and O2 gases through the same porous wall are t1 t2 and t3, respectively. Arrange t1, t2, and t3 in order of their increasing values.
Answer: Let the volume of effused gas be the rates of effusion ofthe three gases will be as follows—
⇒ \(\text { For } \mathrm{H}_2: \frac{V}{t_1} \propto \frac{1}{\sqrt{M_{\mathrm{H}_2}}} \cdots[1] ; \text { For } \mathrm{N}_2: \frac{V}{t_2} \propto \frac{1}{\sqrt{M_{\mathrm{N}_2}}}\)
For \(\mathrm{O}_2: \frac{V}{t_3} \propto \frac{1}{\sqrt{M_{\mathrm{O}_2}}}\)
From [1] and [2] we get, t2/ty = \(\sqrt{M_{\mathrm{N}_2} / M_{\mathrm{H}_2}}\) and form and we get \(t_3 / t_2=\sqrt{M_{\mathrm{O}_2} / M_{\mathrm{N}_2}}.\)
Question 45. What will happen if the collisions ofthe gas molecules with each other are not perfectly elastic?
Answer: In the case of inelastic collisions, the total kinetic energy of gas molecules decreases, leading to a decrease in molecular speeds. As a result, the gas molecules will gradually settle down at the bottom of the container, thereby causing the pressure ofthe gas to go on decreasing gradually. A time will come when the pressure ofthe gas will come duce to zero.
Question 46. At what temperatures rms velocity, average velocity & most probable velocity of O2 molecules will be 1500 m-s-1?
Answer: Molar mass (M) of O2 – 32g-mol-1 =0.032kg. mol 1
In case of rms velocity: \(\frac{3 R T_1}{M}=(1500)^2\)
⇒ \(\text { or, } \frac{3 \times 8.314 \times T_1}{0.032}=(1500)^2 \text { or, } T_1=2886.697 \mathrm{~K} \text {. }\)
In case of average velocity: \(\frac{8 R T_2}{\pi M}=(1500)^2\)
⇒ \(\text { or, } \frac{8 \times 8.314 \times T_2}{0.032 \times 3.14}=(1500)^2 \text { or, } T_2=3399.085 \mathrm{~K} \text {. }\)
In case of most probable velocity: \(\frac{2 R T_3}{M}=(1500)^2\)
⇒ \(\text { or, } \frac{2 \times 8.314 \times T_3}{0.032}=(1500)^2 \text { or, } T_3=4330.045 \mathrm{~K} \text {. }\)
Question 47. For which type of gas molecules are the total kinetic energy and translational kinetic energy equal?
Answer: For monoatomic gases (He, Ne, etc.) the total kinetic energy and translational kinetic energy are equal.
Question 48. Of the following types of velocity, which one has the highest value and which one has the lowest value at a given temperature?
Answer: At a given temperature; the average velocity, the root mean square velocity, and the most probable velocity of the molecules of gas with molar mass M are given by average velocity \((\bar{c})=\sqrt{\frac{8 R T}{\pi M}}\) and most Probable velocity \(\left(c_m\right)=\sqrt{\frac{2 R T}{M}}\)
These expressions indicate that at a given temperature a given gas has the highest value and cm has the lowest value.
Question 49. Which type of velocity does a gas molecule with average kinetic energy possess?
Answer: The average kinetic energy of a gas molecule \(=\frac{1}{2} m c_{r m s}^2.\). So gas molecule with average kinetic energy has the root mean square velocity.
Question 50. How does the average velocity or the root mean square velocity of gas molecules depend on temperature and pressure?
Answer: The average kinetic energy (c) and the root mean square velocity (c2) of gas molecules are given by \(\bar{c}=\sqrt{\frac{8 R T}{M}}\) and \(c_{r m s}=\sqrt{\frac{3 R T}{M}}\)
Question 51. At a given temperature, the root mean square velocities of the molecules of gases A, and B are x and ycm. s-1, respectively. If x is greater than y, then which gas has a larger molar mass?
Answer: \(\text { For gas } A: c_{r m s}=\sqrt{\frac{3 R T}{M_A}}=x \mathrm{~cm} \cdot \mathrm{s}^{-1}\)
⇒ \(\text { For gas } B: c_{r m s}=\sqrt{\frac{3 R T}{M_B}}=y \mathrm{~cm} \cdot \mathrm{s}^{-1}\)
∴ \(\frac{x}{y}=\sqrt{\frac{M_B}{M_A}} ;\) Since x> y, MB will be greater than MA.
Question 52. According to the kinetic theory of gases, the average kinetic energies of O2 and N2 molecules are the same at a particular temperature. State whether the velocities of the molecules ofthe two gases at a given temperature will be the same or not.
Answer: The average kinetic energy of the molecules of a gas depends only on the absolute temperature of the gas. It does not depend on the mass of gas molecules. At a given temperature, a lighter gas molecule has the same average kinetic energy as that of a heavier gas molecule.
On the contrary, the root mean square velocity of the molecules of a gas at a given temperature is inversely proportional to the molar mass ofthe gas. Hence, at a given temperature, the root mean square velocities of the molecules of N2 and O2 gas will not be the same.
Question 53. 1 mol of N2 & 3 mol of O2 are kept In two different containers with a volume of V at a fixed temperature. Compare—(1) the average kinetic energy and (11) the total kinetic energy ofthe molecules.
Answer: The average kinetic energy of gas molecules depends only on the absolute temperature ofthe gas. Since both gases are at the same temperature, they will have equal average kinetic energy.
At T K, the total kinetic energy of nmol gas molecules \(=n \times \frac{3}{2} R T\) Hence, at T K, the total kinetic energy of 3mol O2 molecules is 3 times that of 1 mol N2 molecules.
Question 54. On what factors does the total kinetic energy of the molecules in a gas depend?
Answer: The total kinetic energy ofthe molecules in a gas depends on the absolute temperature as well as the amount of the gas.
Question 55. At a given temperature, the most probable velocity of the molecules of gas A is the same as the average velocity of the molecules of gas B. Which has a larger molar mass
Answer: Suppose, the molar masses of gases A and B are MA And MB, and the temperature of body gases is 7′ K. Therefore, at this temperature, the most probable velocity of the molecules of gas A \(c_m=\sqrt{\frac{2 R T}{M_A}}\)
and the average velocity ofthe molecules of gas B, c \(\bar{c}=\sqrt{\frac{8 R T}{\pi M_{\mathrm{B}}}}\) As given, cm = c.
∴ \(\sqrt{\frac{2 R T}{M_A}}=\sqrt{\frac{8 R T}{\pi M_{\mathrm{B}}}} \quad \text { or, } \frac{1}{M_A}=\frac{4}{\pi M_{\mathrm{B}}}\)
∴ \(M_B=1.27 M_A\)
Therefore, gas B has a higher molar mass
Question 56. Between H2 and CO2 gas, which one has the value of compressibility factor greater than 1 at ordinary temperature and pressure?
Answer: At ordinary temperature and pressure, the compressibility factor of H2 is greater than 1.
Question 57. For a real gas, the van der Waals constant ‘a’ is zero. Can the gas be liquefied? Explain.
Answer: The van der Waals constant ‘a’ measures the magnitude of intermolecular forces of attraction in a gas. Hence, a real gas with ‘a’= 0 signifies that there are no intermolecular attractive forces in the gas. Consequently, such a gas cannot be liquefied.
Question 58. Why are the deviations from the ideal behaviour of CO2 and CH4 greater than those of H2 and He?
Answer: The Molar masses of CO2 and CH4 are greater than those of H2 and He. Hence, the intermolecular attractive forces in CO2 and CH4 are also greater in magnitude than those in H2 and He. This results in a greater deviation from ideal behaviour for CO2 and CH4 than that for H2 and He.
Question 59. At a given temperature and pressure, 1 mol of an ideal gas occupies a volume of 20.8 L. For mol of a real gas at the same temperature and pressure—
- Z will be equal to 1 if the volume ofthe real gas is…
- Z will be greater than 1 if the volume ofthe real gas is…
- Z will be less than 1 if the volume ofthe real gas is…
Answer: \(Z=\frac{V}{V_i}\); the volume of a certain amount of ideal gas at a given temperature and pressure and V = the volume of the same amount of a real gas at the same temperature and pressure.
As given, V- = 20.8L. Therefore, if
- V = 20.8L, then Z will be equal to1
- V> 20.8L, then Z will be greater than1
- V< 20.8L, then Z will be less than1
Question 60. A real gas follows the equation P(V- nb) = nRT under all conditions of temperature and pressure. Show that the compressibility factor of this gas is always greater than one.
Answer: The equation of state for the gas is: P(V-nb) = nRT The compressibility factor of the gas can be expressed as
⇒ \(\begin{aligned}
& P V-P n b=n R T^{\prime} \text { or, } P V=n R T+P n b \\
& \text { or, } \frac{P V}{n R T}=1+\frac{P b}{R T} \text { or, } Z=1+\frac{P b}{R T}
\end{aligned}\)
As \(\frac{P b}{R T}\) is always positive the value of z will be greater than 1.
Question 61. The van dar Waab constant ‘a’ for CO2 and CH4 gases are 3.6 and 2.3 L2-atm-mol-2. Which one of these two gases can easily be liquefied?
Answer: The van der Waals constant for a real gas is a measure ofthe strength of intermolecular forces of attraction in the gas. The stronger the intermolecular forces of attraction in a gas, the greater the value. Again, a gas with stronger intermolecular forces of attraction can easily be liquefied. As the value of CO2 is greater than that of CH4, it will be easier to liquefy CO2 gas.
Question 62. Why is it not possible to liquefy an ideal gas?
Answer: Because of the absence of intermolecular forces of attraction in an ideal gas, such a gas cannot be liquefied.
Question 63. For H2 gas: a = = 0.024 L2 atm.mol-2, b = 0.026 Lmol-1 a = 2.28 L2-atm .mol-2, b = 0.042 L-mol-1. and for CH4 gas: (1) At ordinary temperature and pressure, which one ofthe two gases will behave more like an ideal gas? (U) Which one ofthe two gases has a larger molecular size?
Answer: At ordinary temperature and pressure, a real gas behaves more like an ideal gas if the values of ‘a’ and ‘b’ are very small. The values of ‘a’ and ‘b ‘ for H2 gas are smaller than those for CH4 gas. Obviously, at ordinary temperature and pressure, H2 gas will behave more like an ideal gas.
The value of ‘b’ for a real gas reflects the sizes of molecules of the gas. A gas whose molecules are large has a high value of ‘b’. As the value of CH4 gas is greater than that of H2 gas, the size of the CH4 molecule will be larger than that of the H2 molecule.
Question 64. When does the effect of molecular volume dominate over
Answer: The compressibility factor of a real gas becomes Z>1 when the effect of molecular volume dominates over the effect of intermolecular forces of attraction. Again, Z will be greater than one for a real gas if the pressure ofthe gas is very high. Therefore, the effect of molecular volume becomes greater than the effect of intermolecular forces of attraction if the pressure ofthe gas is very high.
Question 65. At ordinary temperature, why can CO2 but not O2 gas be liquefied by applying pressure? Give reason.
Answer: A gas can be liquefied by applying the necessary pressure if its temperature is equal to or below its critical temperature. The critical temperature of CO2 is above the ordinary temperature (usually 25°C), while that of O2 gas is well below one ordinary temperature. Hence, CO2 can be liquefied by applying pressure at ordinary temperature.
Question 66. The critical temperature and the critical pressure of gas are Tc and Pc, respectively. If the gas exists at a temperature of T and a pressure of P, then under which of the following conditions will the gas not be liquefied?
- T> Tc; P>PC
- r=rc; P>PC
- T = Tc; P<PC
- T<TC-P = PC
Answer: Under the conditions of T> Tc and P> Pc, the gas cannot be liquefied because its temperature is above under the conditions of T = Tc and P>PC, it is possible to liquefy the gas. Because the gas Is at its critical temperature and its pressure is above critical pressure.
Under the conditions of T = Tc and P<PC, it is not possible to liquefy the gas as the minimum pressure needed to liquefy a gas at its critical temperature must be equal to Pc or greater than Pc.
Under the conditions of T< Tc and P = Pc, the gas can be liquefied. Because the gas is below its critical temperature and the pressure of the gas is equal to its critical pressure, the minimum pressure required to liquefy a gas at its critical temperature.
Question 67. The critical temperatures of H2, NH3, and CO2 gases are 5K, 405K, and 304K, respectively. Arrange them in the increasing order of their intermolecular forces of attraction.
Answer: A gas with high critical temperature possesses strong intermolecular forces of attraction. The order of critical temperatures of the gases H2, NH3, and CO2 is NH3 > CO2 > H2. So, the increasing order of their strength of intermolecular forces will be H2<CO2<NH3.
Question 68. The critical temperatures of NH3 and SO2 gases are 405.0K and 430.3K, respectively. For which gas is the value of a der Waals constant greater, and why?
Answer: The higher the critical temperature of a gas, the stronger its intermolecular forces of attraction, and hence the larger the value of the van der Waals constant the gas has. Thus, between NH3 and SO2, the value of a will be larger for SO2 because its critical temperature is higher than that of NH3.
Question 69. The critical temperatures of NH3, CO2, and O2 gases are 405.6K, 304.1K, and 154.2K, respectively. If the gases are cooled from 500K to their respective critical temperatures, then which gas will be liquefied first?
Answer: If the given gases are cooled from 500K, NH3 gets liquefied first (critical temperature 405.6K). The reason is that the critical temperatures of CO2 and O2 are lower than that of NH3. As a result, the liquefaction of either CO2 or O2 is not possible at the critical temperature of NH3.
Question 70. The values of van der Waals constants ‘a’ and ‘b’ for X, Y, and Z gases are 6, 6, 20, and 0.025, 0.15, and 0.11, respectively. Which one has the highest critical temperature?
Answer: The greater the magnitude of intermolecular forces of attraction of a gas, the higher the critical temperature it will have. Among the given gases, the value of a is maximum for gas Z. Thus, the magnitude of effective intermolecular forces of attraction is also maximum for Z, resulting in its higher value of critical temperature.
Question 71. At 20°C, the surface tension of water is three times that of CCI4 —give reason.
Answer: The surface tension of a liquid depends on the magnitude of intermolecular forces of attraction. It decreases or increases, respectively, with a decrease or increase in the magnitude of intermolecular forces of attraction. The only attractive forces that exist in carbon tetrachloride are to London fores exists. As the H-bond is stronger than the London force, the surface tension of water is 3 times that of CC14 at 20°C.
Question 72. At tC and t2°C, the values of viscosity coefficients of a liquid are x poise, and y poise respectively. If x>y, then which one is higher, t1 or t2?
Answer: With the increase in temperature, the viscosity of a liquid decreases. Now, viscosity directly varies with the value of viscosity coefficients As the viscosity coefficient of the liquid at t2°C is smaller than that at t1°C, t2 > t1.
Question 73. Why is the nib of the fountain pen split?
Answer: The split part of the nib of a fountain pen acts like a capillary tube. The ink moves towards the tip of the die nib by capillary action against gravitation through the split par.
Question 74. At 20°C the increasing order of viscosity of acetic acid, acetone, and methanol is: acetone < methanol < acetic acid. Arrange the liquids according to their increasing intermolecular attractive forces.
Answer: With the increase or decrease in the intermolecular force of attraction of a liquid, the value of viscosity confidence of the liquid increases or decreases. At 20°C, the increasing order of the values of the viscosity coefficient of the given liquids is acetone < methanol < acetic add. Thus, the order of increasing the intermolecular attractive force of these liquids is acetone < methanol < acetic add.
Question 75. Why does the surface energy increase on the dispersion of a liquid? large water drop into smaller droplets?
Answer: When a large water drop disperses into smaller water droplets, the total surface area ofthe small water droplets becomes greater than the surface area of the large water drop. As the surface energy increases with an increase in surface area, the dispersion of a large water drop into smaller droplets leads to an increase in surface.
Question 76. Why does oil spread over water when it is poured over water
Answer: The surface tension of water is greater than that of oil. Also, the density of oil is less than that of water. When oil is poured over water, the higher surface tension of water causes oil to spread over water.
Question 78. \(c_{r m s}=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 P V}{M}} .\) According to this equation, with the increase in pressure or volume, the value of arms increases. Justify this statement.
Answer: The rms velocity of the molecules of a gas depends only on the temperature and molar mass of the gas. Its value does not depend on the pressure or volume of the gas because, at a constant temperature, PV for a given mass of gas always remains constant, irrespective of the values of pressure or volume.
Question 79. The volumes, energy numbers of the two ideal molecules gases A and and B average the same. What is the relationship between the pressures of these two gases?
Answer: The average kinetic energy of gas molecules is dependent only on the temperature ofthe gas. As the molecules of A and B gases have the same values of average kinetic energy, both the gases have die same absolute temperature.
Since the volume, the number of molecules and the temperature of both gases are identical, the pressure will be the same for both gases.
Question 80. At constant pressure, the value of V/T for different quantities of an ideal gas will be different. Is this statement true or false?
Answer: The equation of state for n moles of ideal gas is: PV=nRT
Therefore \(\frac{V}{T}=\frac{n R}{P}\)
When the pressure remains constant, V/ Toe n [since R= constant]
Thus, at constant pressure, V/T for different quantities of an ideal gas will be different.
Question 81. Two ideal gases A and B are mixed at temperature T and pressure P. Show that \(d=\left(X_A M_A+X_B M_B\right) \frac{P}{R T}\) Id = density of the mixture, X2 = mole fraction of A, XD = mole fraction of B, M A = Molar mass of A, Mlt = Molar mass of B]
Answer: Suppose, the total volume of the gas mixture is V, and nA and nB are the respective number of moles of A and B in the mixture. If W A and be the masses of A and B respectively, in the mixture, then \(n_A=\frac{w_A}{M_A} \text { and } n_B=\frac{W_B}{M_B}\) Total mass ofA & Bin mixture = \(=W_A+W_B=n_A M_A+n_B M_B\)
∴ Density Of The Mixture \(=\frac{W_A+W_B}{V}=\frac{n_A M_A+n_B M_B}{V}\) for the gas mixture
⇒ \(P V=\left(n_A+n_B\right) R T ;\)
∴ \(V=\left(n_A+n_B\right) \frac{R T}{P}\)
⇒ \(\text { So, } d=\frac{n_A M_A+n_B M_B}{n_A+n_B} \times \frac{P}{R T}=\left(\frac{n_A M_A}{n_A+n_B}+\frac{n_B M_B}{n_A+n_B}\right) \frac{P}{R T}\)
∴ \(\left.d=X_A M_A+X_B M_B\right) \frac{P}{R T}\left[X_A=\frac{n_A}{n_A+n_B} ; X_B=\frac{n_B}{n_A+n_B}\right]\)
Question 82. A closed container holds a mixture of H2, SO2 and CH4 gases, each with an amount of 0.5 mol. If these gases effuse through a fine orifice in the container, arrange them in the increasing order of their partial pressures once the effusion begins.
Answer: According to Graham’s law of diffusion, at constant temperature and pressure, the rate of effusion of a gas is inversely proportional to the square root of its molar mass, i.e \(r \propto \frac{1}{\sqrt{M}}.\)
The increasing order of molar masses ofthe given gases is H2 < MCH4 < MSO2– SO2, and the order of the rate of effusion of these gases at a particular temperature and pressure will be rH2>rCH4>rSO2– Once effusion begins, the order of their number of moles will be nu2 < raCH4< WSO4 Therefore, the order oftheirpartialpressures will be pH <pCH <PSO2.
Question 83. “The total kinetic energy of the molecules in an ideal gas with a volume V at pressure P and temperature T is equal to the total kinetic energy of the molecules present in the same volume of another ideal gas at the same pressure and temperature 2T”—Justify the statement.
Answer: Suppose, the number of molecules present in volume V of the first gas =. If the root mean square velocity of the molecules in the first gas is cl, and the mass of each molecule is mx, then from the kinetic gas equation we have,
⇒ \(P V=\frac{1}{3} m_1 n_1 c_1^2 \quad \text { or, } P V=\frac{2}{3} n_1 \times \frac{1}{2} m_1 c_1^2\) or \(P V=\frac{2}{3} E_1 \text { [where } E_1=n_1 \times \frac{1}{2} m_1 c_1^2\) energy ofthe molecules.] Similarly, if the number of molecules in volume V of the second gas = n2, the mass of each molecule = m2 and the root mean square velocity of molecules= c2, then
⇒ \(P V=\frac{1}{3} m_2 n_2 c_2^2=\frac{2}{3} n_2 \times \frac{1}{2} m_2 c_2^2 \quad \text { or, } P V=\frac{2}{3} E_2\)
[where E2 = total kinetic energy ofthe molecules of second gas] As P and V of both the gases are equal, so E1 = E2.
Question 84. Prove that at a certain pressure, the rate of diffusion of a gas is proportional to the square root of the I absolute temperature of the gas
Answer: The rate of diffusion of a gas at constant pressure (r) oc root mean square velocity ofthe gas (Crms)
∴ \(r \propto c_{r m s} \text { or, } r \propto \sqrt{\frac{3 R T}{M}}\)
Since \(c_{r m s}=\sqrt{\frac{3 R T}{M}}\)
At constant pressure, if and r2 are the rates of diffusion of a particular gas at temperatures and T2, respectively, then \(r_1 \propto \sqrt{\frac{3 R T_1}{M}} \text { and } r_2 \propto \sqrt{\frac{3 R T_2}{M}}\)
∴ \(\frac{r_1}{r_2}=\sqrt{\frac{T_1}{T_2}} \text { or, } r \propto \sqrt{T}\)
So, at constant pressure, the rate of diffusion of a gas is proportional to the square root of the absolute temperature of the gas.
Question 85. The value of the compressibility factor (Z) for a gas at STP is less than 1. What is the molar volume of this gas at STP?
Answer: We Know, \(Z=\frac{V}{V_i}\) = molar volume of a real gas at a certain temperature arid pressure, V2 – molar volume of an ideal gas at the same temperature and pressure.]
According to the question, Z <1
∴ \(\frac{V}{V_i}<1 \text { or, } V<V_i \text { or, } V<22.4 \mathrm{~L}\)
Since the molar volume of an ideal gas at STP = 22.4 litres] The volume ofthe real gas at STP will be less than 22.4 litres.
Question 86. At 0°C, plots of PV vs P for three real gases A, B and C are given below. O Which gas is present above its Boyle temperature? 0 Which gas can be liquefied more easily?
Answer: The PV vs P plot for gas above its critical temperature does not possess any minimum, and the value of PV increases continuously with pressure from the beginning. So, gas A is present above its Boyle temperature.
According to the given figure, the depth of minimum in the PV vs P curve is maximum for the gas C. Hence, the compressibility of gas C is greater than that of either A or B gas. This implies that the forces of attraction between the molecules are stronger in gas C than other two gases. Again, the stronger the intermolecular forces of attraction in a gas, the easier it is to liquefy the gas. So, gas C can be liquefied more easily.
Question 87. At constant temperature and pressure, the compressibility factor (Z) for one mole of a van der Waals gas is 0.5. If the volumes of the gas molecules are considered to be negligible, then show that \(a=\frac{1}{2} V_m\) where Vm and Tare the molar volume and temperature of the gas respectively.
Answer: We known \(Z=\frac{P V}{n R T}\); Given Z= 0.5 and n=1
∴ \(P V=0.5 R T=\frac{1}{2} R T\)
The equation of state for mol of a van der Waals gas is, \(\left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=R T\) [Vm= molar volume]
If volumes of the molecules are considered to be negligible, as per the given condition, then \(V_m-b \approx V_m\)
∴ \(\left(P+\frac{a}{V_m^2}\right) V_m=R T \quad \text { or, } P V_m+\frac{a}{V_m}=R T\)
Again, from equation [1] we get, \(P V=\frac{1}{2} R T\)
For 1 mol (n = 1) ofthe gas, V = Vm (molar volume)
Therefore \(P V_m=\frac{1}{2} R T\)
From equations [2] and [3] we have,
⇒ \(\frac{1}{2} R T+\frac{a}{V_m}=R T \quad \text { or, } a=\frac{1}{2} V_m R T\)
Question 88. Rubber balloon filled with H2 gas gets deflated after some time—explain why.
Answer: As rubber is a porous substance, a rubber balloon contains many invisible pores on its surface. The balloon contains H2 gas at high pressure. But the pressure of air outside the balloon is comparatively lower.
As a result, H2 gas escapes from the balloon by effusion. This is why the pressure inside the balloon gradually falls and after some time the balloon gets deflated.
Question 89. At a given temperature and pressure, the volume fraction of an ideal gas is equal to its mole fraction in a mixture of ideal gases—is it true or false?
Answer: Suppose, at a given temperature (T) and pressure (P), the volumes of two ideal gases are VA and Vg, respectively, and nA and nB are their respective number of moles. Let at the same temperature and pressure, the total volume ofthe mixture of these two gases is V.
∴ According to Amagat’s law of partial volume, \(V=V_A+V_B\)
So, the partial volume of A in the mixture \(=\frac{V_A}{V}\) and its mole fraction \(=\frac{n_A}{n_A+n_B}\)
Now, applying the ideal gas equation to each component gas as well as the gas mixture, we get
⇒ \(\begin{array}{llll}
P V_A=n_A R T & \cdots[1] ; & P V_B=n_B R T & \cdots[2] \\
P V=\left(n_A+n_B\right) R T & & & \cdots[3]
\end{array}\)
Dividing equation no. [1] by equation no. [3], we have
⇒ \(\frac{P V_A}{P V}=\frac{n_A}{n_A+n_B}\)
∴ \(\frac{V_A}{V}=\frac{n_A}{n_A+n_B}\)
That, the volume fraction of A = the mole fraction of A
Similar \(\frac{V_B}{V}=\frac{n_B}{n_A+n_B}\)
[Dividing equation no. [2] by equation no. [3] That is the volume fraction of B = the mole fraction of B. Therefore, at a given temperature and pressure, the volume fraction of an ideal gas in a mixture of ideal gases is equal to its mole fraction. Hence, the given statement is true.
Question 90. For a real gas which obeys the van der Waals equation, a graph is obtained by plotting the values of P Vm along the y-axis and the values of P along the x-axis. What is the value of the intercept on the y-axis of the graph?
Answer: The plot of JPVM against P may give a graph like A or B. In both cases, the graphs intersect the y-axis at P = 0. Now at very low pressure (i.e., P), the van der Waals equation reduces to the ideal gas equation.
∴ PV = nRT
or \(P\left(\frac{V}{n}\right)=R T\)
or, PVm= RT, [Vm= molar volume]
So, the value intercept on the y-axis is RT.
Question 91. A 15.0 L vessel containing 5.6 g of N2 is connected to a 5.0 L vessel containing 8.0 g of 02 using a valve. After the valve is opened and the gases are allowed to mix, what will be the partial pressure of each gas in the mixture at 27°C?
Answer: \(5.6 \mathrm{~g} \mathrm{~N}_2=\frac{5.6}{28}=0.2 \mathrm{~mol} \mathrm{~N}_2, 8.0 \mathrm{~g} \mathrm{O}_2=\frac{8}{32}=0.25 \mathrm{~mol} \mathrm{O}_2\) and T = (273 + 27)K = 300 K.
After the opening of the valve, the total volume ofthe gas mixture, V = (15 + 5)L = 20 L
If the partial pressure of N2 gas in the mixture is pN, then pN x V = nN RT or,
PN2 x 20=0.2 x0.0821 x 300
∴ PN2 = 0.2463 atm
If the partial pressure of O2 gas in the mixture is pO2
then XV= NO2RT
Or, PO2 x 20=0.25 x 0.00821 x 300
∴ PO2= 0.30+748 atm
Question 92. The molecular speeds of gas molecules are analogous to the speeds of rifle bullets. Why is the odour of a gas not detected so fast?
Answer: Gas molecules move almost at the same speed as rifle bullets, but the molecules do not follow a straight line path. Since the molecules collide with each other at a very fast rate, the path becomes zig-zag. Hence, the odour of a gas can not be detected as fast as its molecules move.
Question 93. Why is the quantity of air required to inflate the tyre of a car in summer less than that required in winter?
Answer: According to the kinetic theory of gases, the pressure of a gas originates due to the bombardment of the gas molecules on the walls ofthe container. At higher temperatures, the molecules in a gas have higher velocities or average kinetic energy, and they collide with the walls of their container more frequently and with greater force. As a result, the pressure of a gas increases with an increase in temperature if the volume of the gas remains fixed.
The summer temperature is higher than the winter temperature, and the average kinetic energy of the air molecules in summer is comparatively more than that in the winter. As a result, the air molecules in the summer will exert a greater amount of force on the walls than that exerted by the same number of molecules in the winter.
Therefore, if the volume remains fixed, the pressure of a certain amount of air in summer will be more than that in winter.
Question 94. Two flasks of equal volume are connected by a narrow tube of negligible volume and are filled with N2 gas. When both the flasks are immersed in boiling water the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in ice water while the other flask is in boiling water
Answer: Temperature of the gas when flasks are in boiling water = 100 + 273 = 373 K and pressure = 0.5 atm
The average temperature of the gas when one flask is in ice and the other in boiling water
⇒ \(=\frac{0+100}{2}=50^{\circ} \mathrm{C}=50+273=323 \mathrm{~K}\)
∴ Temperature of the gas when flasks are in boiling water = 100 + 273 = 373 K and pressure = 0.5 atm The average temperature of the gas when one flask is in ice and other boiling water \(=\frac{0+100}{2}=50^{\circ} \mathrm{C}=50+273=323 \mathrm{~K}\)
Question 95. Assuming the same pressure in each case, calculate the mass of hydrogen required to inflate a balloon to a certain volume at 100°C if 3.5g He is required to inflate the balloon to half the volume at 25°C.
Answer: Volume of 3.5gHe at 25°C and pressure P is \(V=\frac{n R T}{P}=\frac{3.5}{4} \times \frac{R T}{P}=\frac{3.5 \times 298 R}{4 P}\)
To fill the balloon with H2 at 373 K, the volume of Hg gas required, VH = 2V. Hence \(2 V=\frac{n R T}{P}=\frac{w}{2} \times \frac{R \times 373}{P}\)
Dividing [2] by [1] gives \(2=\frac{w \times 373}{2} \times \frac{4}{3.5 \times 298} \quad \text { or, } w=2.796 \mathrm{~g}\)
Question 96. The given figure indicates the plot of vapour pressure vs. temperature for the three liquids, A, B & C. Arrange them in the increasing order of their intermolecular forces of attraction and normal boiling points.
Answer: According to the given plot, at a particular temperature, the vapour pressure of A is higher than that of B, which in turn is higher than C. Now, a liquid with weak intermolecular forces of attraction has high vapour pressure. Therefore, the order of intermolecular forces of attraction of the given liquids will be A < B < C.
The lower the vapour pressure of a liquid, the higher its normal boiling point. Alternatively, the higher the vapour pressure of a liquid, the lower its normal boiling point. Therefore, the order of normal boiling points of the given liquids will be: A < B < C
Question 97. Water spreads on a glass surface but It forms beads on a glass surface polished by paraffin—why?
Answer: Adhesive forces between the molecules of glass and water are stronger than the cohesive forces between the water molecules. For this reason, water can spread on the glass surface.
On the other hand, adhesive forces between the molecules of paraffin (non-polar) and water are weaker than cohesive forces between the water molecules. For this reason, water forms small beads on a glass surface of polished paraffin.
States Of Matter Gases And Liquids Short Answer Type Questions
Question 1. When a football is pumped, both the volume and pressure of the gas inside it increase. Does this observation contradict Boyle’s law?
Answer: According to Boyle’s law, at a constant temperature, the volume of a given mass of a gas is inversely proportional to its pressure. When the football is pumped, the quantity of air inside the football goes on increasing. As a result, the mass of air inside the football does not remain constant. Moreover, 2 pumping causes a rise in the temperature of air inside the football. Thus, neither the temperature nor the mass of the air remains constant. So, Boyle’s law is not applicable in this case.
Question 2. Equal mass of two gases A and B are kept in two separate containers under the same conditions of temperature and pressure. If the ratio of molar masses of A and JB is 2:3, then what will be the ratio of volumes of the two containers?
Answer: Let the volume of the container holding gas A = VA and that of the container holding gas B = VB. Suppose, the molar masses of A and B are MA and MB, respectively. Thus, for equal mass (say ‘m’) of each gas, the number of moles of
⇒ \(A\left(n_A\right)=\frac{m}{M_A} \text { and that of } B\left(n_B\right)=\frac{m}{M_B}\)
∴ \(\text { For } A: P V_A=\frac{m}{M_A} R T \text { and for } B: P V_B=\frac{m}{M_B} R T\)
P V=\frac{2}{3} \times n \times \frac{1}{2} m c^2 \quad \text { or, } P=\frac{2}{3} \times\left(\frac{n}{V}\right) \times \frac{1}{2} m c^2 \(\frac{V_A}{V_B}=\frac{M_B}{M_A}=\frac{3}{2}\)
Hence, the ratio of volumes ofthe containers = 3:2
Question 3. 4 gas-mixture consists of two gases, A and B, each with equal mass. The molar mass of B is greater than that of 4. Which one of the two gases will contribute more to the total pressure of the gas mixture?
Answer: In a gas mixture, the component gas with higher partial pressure will have a greater contribution to the total pressure of the mixture. Masses of A and B in the mixture are the same but the molar mass of B is greater than that of A. Hence, in the mixture, the number of moles or the mole fraction (xA) of A will be greater than that (xB) of B. Suppose, in the mixture, the partial pressures of A and B are pA and pB respectively and the total pressure is P. According to Dalton’s law of partial pressures, pA = xAP and pB = xgP.
As xA>xB, pA will be greater than pB. Hence, the contribution of gas A to the total pressure of the mixture will be more than that of gas B.
Question 4. Under the same conditions of temperature and pressure, the rate of diffusion of hydrogen gas is four times that of oxygen gas—explain
Answer: At constant temperature and pressure, rates of diffusion (r) of different gases are inversely proportional to the square roots of their molecular masses (M) \(r \propto \frac{1}{\sqrt{M}}\)
⇒ \(\frac{r_{\mathrm{H}_2}}{r_{\mathrm{O}_2}}=\frac{\sqrt{M_{\mathrm{O}_2}}}{\sqrt{M_{\mathrm{H}_2}}}=\frac{\sqrt{32}}{\sqrt{2}}=4 \text { or, } r_{\mathrm{H}_2}=4 \times r_{\mathrm{O}_2}\)
So, under the same conditions of temperature and pressure, the rate of diffusion of H2 gas is four times that of O2 gas.
Question 5. Four tyres of a motor car were filled with nitrogen, hydrogen, helium and air. In which order are these tyres to be filled with the respective gases again
Answer: According to Graham’s law of diffusion, under the conditions of temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. The order of molar masses of the given gases is—
⇒ \(M_{\mathrm{H}_2}<M_{\mathrm{He}}<M_{\mathrm{N}_2}<M_{\text {air }}\)
At the same temperature and pressure, the order of rates of diffusion of these gases would be—rH > rHe > rN > air After a certain time, the order of decrease in pressure in these three tyres tyre (air) < tyre (N2) < tyre (He) < tyre (H2). Hence, tyres are to be filled again with the respective gases in the following order, tyre(H2) tyre(He), tyre(N2), and tyre(air).
Question 6. At constant pressure for a given amount of gas, will the graphs obtained by plotting V vs t°C and V vs TK be different?
Answer: According to Charles’ law, V = KT – [1] [at constant pressure for a given mass of a gas]
But, T = 273 + t, hence, V = K(273 + t) -[2] Both the equations [1] & [2] express equations for straight lines. Equation no. [1] represents a straight line passing through the origin. Equation no. [2] does not represent a straight line passing through the origin. The plot of V vs t results in a straight line that cuts the V-axis at 0°C. If this straight line is extrapolated backwards, it meets the y-axis at -273°C.
From the, above two graphs it is evident that there are no actual differences between the two graphs, because -273°C and OK express the same temperature.
Question 7. Under similar conditions of temperature and pressure, if the time taken for effusion of the same volume of H2, N2 and CO2 gas through the same porous wall are t1, t2 and t2 respectively, then arrange t1, t2 and f3 in their increasing order.
Answer: At constant temperature and pressure, if V volume of H2 gas effuses in time t j, then according to Graham’s law, \(\frac{V}{t_1} \propto \frac{1}{\sqrt{M_{\mathrm{H}_2}}}\)
For N2 gas and CO2 gas, the equations are respectively,
⇒ \(\frac{V}{t_2} \propto \frac{1}{\sqrt{M_{\mathrm{N}_2}}}\)
⇒ \(\frac{V}{t_3} \propto \frac{1}{\sqrt{M_{\mathrm{CO}_2}}}\)
Volume of gas effused in each case is the same From equations [1] & [2]; from equations [2] & [3] we get \(\frac{t_2}{t_1}=\sqrt{\frac{M_{\mathrm{N}_2}}{M_{\mathrm{H}_2}}}; \frac{t_3}{t_2}=\sqrt{\frac{M_{\mathrm{CO}_2}}{M_{\mathrm{N}_2}}}\)
Since, MH2 < MN2 < Mc2, therefore, t2 > and t2 > t2 Hence, t1,t2 and t2 will be in order: t1<t2<t2
Question 8. The molar mass of UF6 is 176 times as high as that of H2, yet at a particular temperature, the average kinetic energy of both is found to be the same—why?
Answer: According to the kinetic theory of gas, the average kinetic energy of the molecules in a gas is directly proportional to the absolute temperature of the gas, and it does not depend upon the molar mass of the gas.
The average kinetic energy of a molecule of a gas at temperature TK is given by \(\frac{3}{2} k T\) is Boltzmann constant]. Since the value of is constant at a given temperature and is independent of the mass of the gas molecule, \(\frac{3}{2} k T\) the average kinetic energy of the molecule of a heavier gas will be the same as that of the molecule of a lighter gas. Thus, at a given temperature, the average kinetic energy of a UF6 molecule will be the same as that of a H2 molecule.
Question 9. If, at a given temperature, the total kinetic energy of the molecules in a unit volume of an ideal gas is E, show that the pressure of the gas, P = 2/3E.
Answer: According to the kinetic gas equation, \(\mathrm{PV}=\frac{1}{3} \mathrm{mnc}^2\)
Suppose, m = mass of each molecule of the gas and n = number of molecules present in V volume of the gas, c = root mean square velocity of gas molecule
∴ \(P V=\frac{2}{3} \times n \times \frac{1}{2} m c^2 \quad \text { or, } P=\frac{2}{3} \times\left(\frac{n}{V}\right) \times \frac{1}{2} m c^2\)
where \(\frac{n}{V}\) the number of molecules per unit volume and \(\frac{1}{2} m c^2\) = the average kinetic energy of each molecule
∴ \(\left(\frac{n}{V}\right) \times \frac{1}{2} m c^2\) hre total kinetic energy of the molecules present per unit volume = E
∴ P = 2/3E (Proved)
Question 10. For the molecules of a given gas at a constant temperature, arrange the most probable velocity (cm), root mean square velocity (CRM) and average velocity (CFL) in the order of their increasing values. With the increase in temperature, will the ratio of these velocities increase, decrease or remain constant? What will the effect of increasing temperature be on the value of (crnls- cm) for a given gas?
Answer: First Part: At temperature T, the most probable velocity (cm) ofthe gas molecules = \(\sqrt{\frac{2 R T}{M}}\) [M – molar mass ofthe gas] rms velocity \(\left(c_{r m s}\right)=\sqrt{\frac{3 R T}{M}}\) average velocity \(\left(c_a\right)=\sqrt{\frac{8 R T}{\pi M}}\)
∴ \(c_{r m s}>c_a>c_m\)
Second Part: Since the value of each of the velocities, crms’ ca and cm’ proportional to Jf, their ratio remains unaltered with temperature rise.
Third: Suppose, at temperature, riK, the root mean square velocity (arms) and most probable velocity (cm) of the molecules of a gas are (arms)j and (cm) respectively. If the difference between the values of these velocities is Axx, then
⇒ \(\Delta x_1=\sqrt{\frac{3 R T_1}{M}}-\sqrt{\frac{2 R T_1}{M}}=\sqrt{\frac{R T_1}{M}}(\sqrt{3}-\sqrt{2})\)
Let the temperature of the gas be raised to T2K and at this temperature, the difference between the values of these two velocities is Ax2, then,
⇒ \(\Delta x_2=\left(c_{r m s}\right)_2-\left(c_m\right)_2=\sqrt{\frac{R T_2}{M}}(\sqrt{3}-\sqrt{2})\)
Since T2 > , Ax2 > Ax1. Hence, the difference between the values of these velocities increases with the temperature rise.
Question 11. A and B are closed flasks having the same volume. In flask A, O2 gas is present at TK and 1 atm pressure. In flask B, H2 gas is present at 1 atm pressure. If these gases behave ideally, then compare their O total kinetic energies, total number of molecules, and root mean square velocities.
Answer: Suppose, M1 and n2 are the number ofmoles ofthe gases present in flask A and flask B, respectively. Applying the ideal gas equation to the gases O2 and H2, we obtain
⇒ \(P V=n_1 R T \text { and } P V=n_2 R \frac{T}{2}\)
∴ \(\frac{n_1}{n_2}=\frac{1}{2}\)
The total kinetic energy of the molecules of mol O2 in flask A, \(E_1=n_1 \times \frac{3}{2} R T\) and the total kinetic energy of the molecules of n2 mol H2 in flask B, \(E_2=n_2 \times \frac{3}{2} R \times \frac{T}{2}\)
[ since Total kinetic energy of the molecules of1mol gas =[Rx absolute temperature]
∴ E1=E2
So, the total kinetic energy of the molecules of O2 gas = the total kinetic energy of the molecules of H2 gas
If n’1 and n’2 be the number of molecules of O2 gas and H2 gas respectively then n1 = nxN and n2 = n2xN [AT= Avogadro’s number]
∴ \(\frac{n_1^{\prime}}{n_2^{\prime}}=\frac{n_1 \times N}{n_2 \times N}=\frac{1}{2}\)
∴ Number of molecules in H2 gas = 2 x number of molecules in oxygen gas
ms velocity of O2, Crms \(=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 R T}{32}}\)
rms velocity of H2, Cms \(=\sqrt{\frac{3 R \frac{T}{2}}{2}}=\sqrt{\frac{3 R T}{4}}\)
∴ \(\frac{c_{r m s}\left(\mathrm{O}_2\right)}{c_{r m s}\left(\mathrm{H}_2\right)}=\sqrt{\frac{4}{32}}=\sqrt{\frac{1}{8}}=\frac{1}{2 \sqrt{2}}\)
Question 12. Which one of the gases, under the given conditions, exhibits real gas behaviour? Q) 0.25 mol CO2, T = 1200K, 1. = 24.63 atm, V=1 L 2. 1.0 mol SO2, T = 300 K, P = 50 atm, V = 0.35 L
Answer: At a certain temperature and pressure, if the compressibility factor (Z) of a gas is less than or more than 1, then at that temperature and pressure the gas behaves as a real gas. If Z = 1, then the gas exhibits ideal behaviour
For CO2 gas \(Z=\frac{P V}{n R T}=\frac{24.63 \times 1}{0.25 \times 0.0821 \times 1200}=1\)
For SO2 gas ,Z \(=\frac{P V}{n R T}=\frac{50 \times 0.35}{1 \times 0.0821 \times 300}=0.71\)
Under the given conditions, for CO2 gas, Z = 1 and hence it shows ideal behaviour. In the case of SO2, Z < 1. So, under the given conditions, S02 behaves as a real gas.
Question 13. Under what conditions can a gas be liquefied?
- T = Tc and P < Pc
- T<TC and P – Pc
Answer: It is possible to liquefy a gas by the application of pressure, provided that the temperature of the gas is equal to or less than its critical temperature. If the temperature (T) of a gas is equal to its critical temperature ( Tc), then it is possible to liquefy the gas if the pressure (P) of the gas is equal to or above its critical pressure (Pc).
So, under the condition, the gas cannot be condensed into a liquid. On the other hand, if the temperature (T) of a gas is below its critical temperature (Tc), then the gas can be transformed into liquid if the applied pressure on the gas is greater than or less than or equal to its critical pressure. So, under certain conditions, the gas can be liquefied.
Question 14. Two gases, obeying the van der Waals equation, have identical values of ‘b’ but different values of ‘a’. Which one of the two gases will occupy less volume under identical conditions? If the values of ‘a’ for the two gases are the same but the values of‘V are different, then under identical conditions which gas will be more compressible?
Answer: The larger the value of ’a’ of a gas, the stronger the intermolecular forces of attraction in the gas. So, under identical conditions, a gas with a larger value of ‘a’ will be more compressible than that with a smaller value. Hence, between the two gases, the one with a higher value of ‘a’ will occupy less volume under identical conditions.
If the value of ‘a’ for two gases is the same the values of ‘b’ differ, then the gas with a smaller value of ‘b’ will be more compressible because a small value of ‘b’ for a gas signifies that the volume occupied by the molecules of the gas is small. So, this gas can be compressed to a greater extent.
Question 15. A real gas follows van der Waals equation. Find the compressibility for 1 mol of the gas at its critical temperature
Answer: At the critical point
⇒ \(V=V_c=3 b ; P=P_c=\frac{a}{27 b^2} \text { and } T=T_c=\frac{8 a}{27 R b}\)
∴ \(Z=\frac{P V}{R T}=\frac{\frac{a}{27 b^2} \times 3 b}{R \times \frac{8 a}{27 R b}}=\frac{3}{8}\)
Question 16. Derive the van der Waals equation for ‘n’ mol of a real gas from the equation for 1 mol of the gas.
Answer: Van der Waals equation for1 mol of real gas is given by—
⇒ \(\left(P+\frac{a}{v^2}\right)(\nu-b)=R T\)
If at a pressure P and a temperature T, the volume of ‘n’ moles of this gas be V, then \(v=\frac{V}{n}\) Putting v =\(v=\frac{V}{n}\) in equation n n [1], we have
⇒ \(\left(P+\frac{a}{\left(\frac{V}{n}\right)^2}\right)\left(\frac{V}{n}-b\right)=R T \text { or }\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T\)
Question 17. Write the van der Waals equation for a real gas containing n molecules.
Answer: If v is the volume of mol of a real gas at a pressure P and temperature T then the van der Waals equation for the gas is
⇒ \(\left(P+\frac{a}{v^2}\right)(v-b)=R T\)
Suppose, the volume for a real gas containing n molecules = V. So, volume for n/N moles of real gas = V [N= Avogadro’s number]
Therefore, the volume for 1 mol of real gas \(=\frac{V \times N}{n}=v\)
Substituting the value of v in equation [1], we obtain \(\left(P+\frac{n^2 a}{V^2 N^2}\right)(V N-n b)=n R T\)
This is the van der Waals equation for a real gas containing n molecules
Question 18. What will the nature of the PV vs P graph be for a real gas at Boyle temperature?
Answer: AOS, At Boyle temperature the value of PV, particularly in the low-pressure region, becomes constant. Thus, the gas shows ideal behaviour. So, at this temperature, the plot pv of PV against P will give a straight line parallel to the P-axis
Question 19. What will the value of compressibility factor (Z) be for a gas if the pressure correction term in the van der Waals equation for the gas is neglected?
Answer: For 1 mol of a real gas \(\left(P+\frac{a}{V^2}\right)(V-b)=R T\)
If the pressure correction term is neglected, then \(P+\frac{a}{V^2} \approx P\)
∴ P(V-b) = RT or, PV -Pb = RT
Or, \(P V=R T+P b \quad \text { or, } \frac{P V}{R T}=1+\frac{P b}{R T}\)
∴ \(Z=1+\frac{P b}{R T}\)
Question 20. The value of van der Waals constant ‘a’ for nitrogen gas is 1.37 L2-atmmol-2, but that for ammonia gas is 4.30 L2-atm-mol-2. What is the reason for this large difference? Which one of these two gases would you expect to have a higher critical temperature?
Answer: Since N2 is a non-polar molecule, the only attractive forces that operate between the molecules in N2 gas are weak London forces. On the otherhand, NH3 is apolarmolecule. So, the molecules in NH3 gas experience dipole-dipole attractive forces in addition to London forces.
Hence, the intermolecular forces of attraction in NH3 gas are much stronger than those in N2 gas, and this makes the value of ‘a’ for NH3 gas higher. The critical temperature of a gas depends upon the strength of intermolecular forces of attraction in the gas.
The stronger the intermolecular forces of attraction in a gas, the higher the critical temperature of the gas. Since the intermolecular forces of attraction are stronger in NH3 gas than those in H2 gas, the critical temperature of NH3 will be higher than that of H2.
Question 21. The values of ‘a’ and ‘b’ for three A real gases B A, Band C Care—
⇒ \(\begin{array}{|c|c|c|c|}
\hline & A & B & C \\
\hline a\left(\mathrm{~L}^2 \cdot \mathrm{atm} \cdot \mathrm{mol}^{-2}\right) & 6 & 8 & 20 \\
\hline b\left(\mathrm{~L} \cdot \mathrm{mol}^{-1}\right) & 0.025 & 0.12 & 0.10 \\
\hline
\end{array}\)
- Which one of these gases has the largest molecular size?
- Which one of these will behave most like an ideal gas at STP?
Answer: The volume of a molecule depends on its radius. Now we know b ∝ r3. Hence, the molecular size will be the largest for the gas which has the highest value of ‘b’. From die given values, it is found that gas B has the highest value of ‘b’ So, so the molecule ofthe gas will be the largest.
A gas with small values of ‘a’ and ‘b‘ behaves close to an ideal gas. For gas A, these two quantities have the smallest values. Hence at STP gas A will show the most ideal behaviour.
Question 22. Between methanol (CH3OH) and water (H2O) whose surface tension is greater, and why?
Answer: Molecules in methanol and water both are capable of forming hydrogen bonds. However, the number of hydrogen bonds per molecule in water is greater than that in methanol. So, the intermolecular forces of attraction are stronger in water than those in methanol. Again, the stronger the intermolecular forces of attraction in a liquid, the greater the surface tension of the liquid. So, the surface tension of water is greater than that of methanol.
Question 23. A liquid has a high normal boiling point. Will its viscosity and surface tension values be high or low?
Answer: The high normal boiling point of a liquid indicates that the liquid possesses strong intermolecular forces of attraction. The values of viscosity and surface tension of a liquid depend upon the strength of intermolecular attractive forces in the liquid. The stronger the intermolecular attractive forces, the higher the values of viscosity and surface tension of the liquid. As the intermolecular forces of attraction are strong for the concerned liquid, the values of both surface tension and viscosity of the liquid will be high.
Question 24. How is the vapour pressure of a liquid affected if the surface area of the liquid is increased at a given temperature?
Answer: At a given temperature, the vapour pressure of a liquid does not depend upon the surface area of the liquid. If the surface area of a liquid is increased, keeping the temperature constant, then its rate of evaporation increases because at a particular time, more molecules are now able to leave the liquid surface and go to the vapour phase.
The increased surface area of the liquid also causes the increase in the die rate of condensation because, at a particular time, more molecules are now able to reenter from vapour to the liquid phase.
Therefore, the rates of both evaporation and condensation are increased to the same extent So, the vapour pressure remains the same at a constant temperature and is not affected by the surface area of the liquid
Question 25 Compare the viscosity coefficients of the following liquids at a particular temperature: Propanol (CH3CH2CH2OH), ethylene glycol (HOCH2—CH2OH) and glycerol (HOCH2—CHOH—CH2OH).
Answer: The number of OH groups in the molecules of CH3CH2CH2OH, HOCH,—CH2OH HOCH2—CHOH—CH2OH are 1, 2 and 3, respectively. So, the number of hydrogen bonds formed per molecule of CH3CH2CH2OH HOCH2—CH2OH and HOCH2 —CHOH —CH2OH are 1, 2 and 3, respectively. Thus, the order of the intermolecular forces of attraction of the given liquids will be propanol < ethylene glycol < glycerol.
Again, the stronger the intermolecular forces of attraction of a liquid, the higher the value of its viscosity coefficient. Thus, the order of viscosity is propanol < ethylene glycol < glycerol.
Question 26. Find out the minimum pressure required to compress 2 X M 500 dm3 of air at bar to 200 dm1 at 30nC
Answer: In this process, the mass and temperature of the gas remain constant but the volume and pressure of the gas change.
Given P1=1 bar, V1= 500 dm³, v2= 200 dm³.
Appyling Boyle’s law , pe \(P_2=\frac{P_1 V_1}{V_2}=\frac{1 \times 500}{200}=2.5 \text { bar }\)
So, the minimum pressure needed to V2 compress 200 the gas is 2.5 bar.
Question 27. A vessel of 120 mi. capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 1 HO ml. at 35°C. What would be Its pressure?
Answer: In this process, the pressure and volume ofthe gas change but its mass and temperature remain constant.
Applying Boyle’s law to the process, we have \(P_2=\frac{P_1 V_1}{V_2}\)
Given: p1= 1.2 bar, V1 = 120ml and V2 = 180 ml
∴ \(p_2=\frac{1.2 \times 120}{180}=0.8 \text { bar }\)
Question 28. Using the equation of state PV = nBT, show that at a given temperature, the density of a gas is proportional to the gas pressure P.
Answer: \(P V=n R T \text { or, } P V=\frac{g}{M} R T\) M= Molar mass of gas in g.mol-1, g= mass of gas in g] or \(P M=\frac{g}{V} R T=d R T\) [d= g/v= density of the gas]
Since, B is constant dec P, when T is constant.
Question 29. At 0°C, the density of a certain oxide of a gas at 2 bar is the same as that of dinitrogen at 5 bar. What is the the molecular mass of the oxide?
Answer: The equation expressing the relation between density (d), pressure (P), absolute temperature (T) and molar mass (M) of a gas is d \(=\frac{P M}{R T}\)
Density ofthe oxide of the gas, \(d_1=\frac{2 \times M}{R \times 273}\)
Density of N2 gas, d2 \(=\frac{5 \times 28}{R \times 273}\)
It is mentioned that under the given conditions, d1 = d2
∴ \(\frac{2 \times M}{n \times 273}=\frac{5 \times 28}{R \times 273} \quad \text { or, } M=70\)
Therefore, the molecular mass ofthe oxide is 70 g-mol-1
Question 30. The pressure of Ig of an Ideal gas A at 27°C Is found to be 2 bar. When 2 g of another ideal gas B Is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses
Answer: Suppose, the molar masses of gases A and B are MA and MB g-mol-1 respectively.
⇒ \(\lg \text { of } A=\frac{1}{M} \mathrm{~mol} \text { of } A \text { and } 2 \mathrm{~g} \text { of } B=\frac{2}{M} \mathrm{~mol} \text { of } B \text {. }\)
Using the ideal gas equation (PV = NRT) for the gas A and the gas mixture of A and B, we have—
⇒ \(2 \times V=\frac{1}{M_A} R T \cdots[1] \text { and } 3 \times V=\left(\frac{1}{M_A}+\frac{2}{M_B}\right) R T \cdots[2]\)
Dividing Equation [2] by[1], we have
⇒ \(1.5=\frac{M_B+2 M_A}{M_B} \quad \text { or, } 2 M_A=0.5 M_B \quad \text { or, } \frac{M_A}{M_B}=\frac{1}{4}\)
Question 31. The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and lbar will be released when 0.15g aluminium reacts?
Answer: The following reaction involving aluminium (Al) and caustic soda (NaOH) produces H2 gas
⇒ \(\begin{array}{ll}
2 \mathrm{Al}(s)+2 \mathrm{NaOH}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{NaAlO}_2(a q)+ & 3 \mathrm{H}_2(g) \\
2 \times 27 \mathrm{~g} & 3 \times 1 \mathrm{~mol}
\end{array}\)
According to this reaction 54g of Al = 3 mol of H2
∴ \(0.15 \mathrm{~g} \text { of } \mathrm{Al} \equiv \frac{3}{54} \times 0.15 \equiv 8.33 \times 10^{-3} \mathrm{~mol} \mathrm{of}_2\)
To calculate the volume ofthe liberated H2 gas, we apply the ideal gas equation, PV = nRT. Given: P = bar and T = (273 + 20)K =293K The number of moles of liberated H2 gas (n) =8.33 x 10-3 mol
∴ \(V=\frac{n R T}{P}=\frac{8.33 \times 10^{-3} \times 0.0821 \times 293}{0.987} \mathrm{~L}=0.203 \mathrm{~L}=203 \mathrm{~mL}\)
Question 32. What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9dm3 flask at 27°C?
Answer: Total number of moles (n) in the mixture \(=\left(\frac{3.2}{16}+\frac{4.4}{44}\right)\) 0.3 mol [Molar mass: \(\mathrm{CH}_4 \Rightarrow 16, \mathrm{CO}_2 \Rightarrow 44\) As given, V = 9 dm3, T = (273 + 27)K = 300K, P = ? So, \(p=\frac{n R T}{V}=\frac{0.3 \times 0.0821 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 300 \mathrm{~K}}{9 \mathrm{dm}^3}\)
= 0.821 aztm [ 1l=1dm3]
= 8.316x104pa[1atm=1.013x105pa]
Question 33. What will be the pressure of the gaseous mixture when 0.5L of H2 at 0.8 bar and 2.0L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?
Answer: To calculate the number of moles of unmixed H2 and O2 m gases, we apply the ideal gas equation, PV = nRT
In case of H2, P = 0.8bar, V = 0.5L and T = (273 + 27)K = 300K and in case of O2, P = 0.7bar, V = 2.0L and T = (273 + 27)K = 300K
Therefore \(n_{\mathrm{H}_2}=\frac{P V}{R T}=\frac{0.8 \times 0.5}{R \times 300}=\frac{0.4}{300 R} \mathrm{~L} \cdot \text { bar }\) and \(n_{\mathrm{O}_2}=\frac{P V}{R T}=\frac{2 \times 0.7}{R \times 300}=\frac{1.4}{300 R} \mathrm{~L} \cdot \text { bar }\) In the mixture of H2 and 02, total number of mol \(=n_{\mathrm{H}_2}+n_{\mathrm{O}_2}=\frac{1}{300 R}(0.4+1.4) \mathrm{L} \text { bar }=\frac{1.8}{300 R} \mathrm{~L} \cdot \text { bar }\) For this mixture V = 1L and T = (273 + 27)K = 300K
If the pressure of the mixture is P, then \(P=\frac{n R T}{V}=\frac{1.8}{300 R} \times \frac{R \times 300}{1} \mathrm{bar}=1.8 \mathrm{bar}\)
Question 34. The density of a gas is found to be 5.46 g/dm3 at 27°C and 2 bar pressure. What will be its density at STP?
Answer: Density (d), pressure (P), absolute temperature (T) and molar mass of a gas (M) are related by, d = PM/RT Under the conditions of T = (273 + 27)K = 300K and P = 2 bar, the density ofthe gas (d) = 5.46 g.dm-3
Therefore \(5.46 \mathrm{~g} \cdot \mathrm{dm}^{-3}=\frac{(2 \times 1.013) \mathrm{atm} \times M}{0.0821 \mathrm{~atm} \cdot \mathrm{dm}^3 \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 300 \mathrm{~K}}\)
Or M = 66.37 g.mol-1
At STP, ifthe density of the gas be d1, then \(\begin{aligned}
d_1=\frac{P M}{R T} & =\frac{1 \times 66.37}{0.0821 \times 273}[\text { AtSTP, } T=273 \mathrm{~K} P=1 \mathrm{~atm}] \\
& =2.96 \mathrm{~g} \cdot \mathrm{L}^{-1}=2.96 \mathrm{~g} \cdot \mathrm{dm}^{-3}
\end{aligned}\)
Question 35. 4.05mL of phosphorus vapour weighs 0.0625g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?
Answer: Suppose, the molar mass of phosphorus =M g-mol-1
So, 0.0625g of phosphorus \(=\frac{0.0625}{\mathrm{M}} \mathrm{mol}\) of phosphorus As given, P = 0.1 bar, T = (273 + 546)K = 819K, V = 34.05 mL =34.05 X 10-3L Therefore, 0.1 bar x 34.05 x 10-3L.
⇒\(=\frac{0.0625}{M} \mathrm{~mol} \times 0.082 \mathrm{~L} \cdot \mathrm{bar} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 819 \mathrm{~K}\)
or, M = 1232.71, Hence, Molar mass = 1232.71g.mol-1.
Question 36. A student forgot to add the reaction mixture to the round-bottomed flask at 27°C but instead, he/she placed the flask on the flame. After a lapse of time, he realised his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Answer: Suppose the number of mol of air inside the flask at 27°C is n1 and that at 477°C is n2. Since the flask is opened to air, the pressure of air inside the flask at 27°C and 477°C is the same as that of the atmospheric pressure. Let this pressure be P. Again on heating the volume of a round bottom flask remains the same. Hence, the volume ofthe flasks the some at both 27°C and 477°C. Let this volume be V.
Applying the ideal gas equation we have, PV= n1R(273 + 27)
= HJ X 300R n2PV = n2 (273 + 477) = n2 x 750R
Hence, n2 x 300R = n2 x 750R or, \(\text { or, } n_2=\frac{2}{5} n_1\)
∴ Fraction of air that would have been expelled out = \(=\frac{n_1-\frac{2}{5} n_1}{n_1}=\frac{3}{5}\)
Question 37. Calculate the temperature of 4.0 mol of gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar. dm3- K-1. mol-1)
Answer: To calculate the temperature ofthe gas, we apply the gas equation, PV = nRT. As given: P = 3.32 bar, V = 5 dm3 and n = 4 mol Putting these values into the equation, PV = nRT, we have \(T=\frac{P V}{n R}=\frac{3.32 \times 5}{4 \times 0.083} \mathrm{~K}=50 \mathrm{~K}\)
Question 38. Calculate the total number of electrons present In 1.4 g of dinitrogen gas.
Answer: \(1.4 \mathrm{~g} \text { of } \mathrm{N}_2=\frac{1.4}{28}=0.05 \mathrm{~mol} \text { of } \mathrm{N}_2\)
1 molecule of N2 contains 14 electrons, Hence, 0.05 mol of N2, i.e., 0.05 X 0.022 X 1023 molecules of = 4,2154 x 1023 electrons. N2 contain 14 x 0,05 X 0,023 X 1023
Question 39. How much time would It take to distribute one Avogadro number of wheat grains If 1010 grains are distributed each second?
Answer: 1010 grains are distributed in 1 sec. So, time required to distribute 6.022 x IO23 grains would be \(\begin{aligned}
\frac{1 \times 6.022 \times 10^{23}}{10^{10}}=6.022 \times 10^{13} \sec & =696.990 \times 10^6 \text { days } \\
& =1.909563 \times 10^6 \text { years }
\end{aligned}\)
Question 40. Calculate the total pressure In a mixture of 8 g of dioxygen and 4 g of dihydrogen confined In a vessel of 1 dm3 at 27°C, R = 0.083 bar -dm3- Kr1- mol-1
Answer: \(\left(\frac{8}{32}+\frac{4}{2}\right)=2.25 \mathrm{~mol}\) Molar mass of O2 and H2 are 32 and 2 g mol-1 respectively]
As given: V = 1dm3 and T = (273 + 27)K = 300K
∴ \(P=\frac{n R T}{V}=\frac{2.25 \times 0.083 \times 300}{1}=56.025 \text { bar. }\)
Question 41. Payload Is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload when a balloon of radius 10 m, mass 100 kgs filled with helium at 1.66 bar at 27°C. (R = 0.083 bar. dm3. K-). mol-1 and density of air = 1.2kg.m-3
Answer: Volume (V) of the balloon \(=\frac{4}{3} \pi r^3=\frac{4}{3} \pi \times(10)^3 \mathrm{~m}^3\) = 4186.66m3 =4186.66 X 103 dm3
To calculate the number of moles of He gas enclosed in the balloon, we apply the ideal gas equation, PV = nRT, Given: P = 1.66 bar, T = (273 + 27)K = 300K So, according to the equation, PV = nRT, \(\begin{aligned}
n=\frac{P V}{R T} & =\frac{1.66 \times 4186.66 \times 10^3 \mathrm{bar} \cdot \mathrm{dm}^3}{0.083 \mathrm{bar} \cdot \mathrm{dm}^3 \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 300 \mathrm{~K}} \\
& =279.11 \times 10^3 \mathrm{~mol}
\end{aligned}\)
1 fence, the mass of f1<< gas enclosed in the balloon =4 X 279. J f X 103g = 1,110 x 106g 1116kg Therefore, the mass of the balloon filled in with He gas -(100+ 11101kg = 1216kg Volume of air displaced by the balloon = Volume of the balloon 4100,00 x 103 dm3 Density of air – 1,2 kg – m-3 = 1.2 x 10-3 kg.dm-3
Question 42. Calculate the volume occupied by 8.8 g of C02 at 31.1 °C &1 bar pressure. /t=0.083 bar -L-K-1mol-1.
Answer: \(8.8 \mathrm{~g} \mathrm{CO}_2=\frac{8.8}{44}=0.2 \mathrm{~mol} \mathrm{CO}_2\)
Applying the ideal gas equation (PV = NRT) to calculate volume, we have \(V=\frac{n R T}{P}=\frac{0.2 \times 0.083 \times(273+31.1)}{1}=5.048 \mathrm{~L}\)
∴ The volume of 8.8 g CO2 at 31.1°C and 1 bar pressure = 5.048L.
Question 43. 2.9 g of gas at 95°C occupied the same volume as 0.184g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Answer: Let the molar mass ofthe unknown gas =M g-mol-1 Number ofmoles for 2.9g ofthe gas \(=\frac{2.9}{M} \mathrm{~mol}\)
Number of moles for 0.184g of H2 gas \(=\frac{0.184}{2}=0.092 \mathrm{~mol}\)
As both the gases have the same volume at the same pressure and specified temperature, they will have the same value of PV. To calculate PV, we apply the equation PV = nRT. For unknown gas \(P V=\frac{2.9}{M} R(273+95)\)
And for He Gas, Pv [= 0.0932R(273+17)
Therefore \(\frac{2.9 \mathrm{R}}{M} \times 368=0.092 R \times 290 \quad \)
Hence, the molar mass ofthe unknown gas = 40g-mol-1
Question 44. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dlhydrogenÿ_
Answer: Let the weight of the mixture be w g. So, in the mixture, the weight of H2 gas = 0.2 w g and that of O2 gas =(w-0.2w) =0.8 w g \(0.2 w g \text { of } \mathrm{H}_2=\frac{0.2 w}{2}=0.1 \times w \mathrm{~mol} \text { of } \mathrm{H}_2\) \(0.8 \mathrm{wg} \text { of } \mathrm{O}_2=\frac{0.8 w}{32}=0.025 \times w \mathrm{~mol} \text { of } \mathrm{O}_2\)
So, the partial pressure of H2 in the mixture = JCH x total pressure of the mixture =0.8×1 bar = 0.8 bar
Question 45. What would be the SI unit for the quantity \(\frac{P V^2 T^2}{n}\)
Answer: In SI system, the units of P, V, T and n are N-m~2, m3, K and mol respectively So, the SI unit for the quantity \(\frac{V^2 T^2}{n}\) will be \(\frac{\mathrm{N} \cdot \mathrm{m}^{-2} \times\left(\mathrm{m}^3\right)^2 \cdot \mathrm{K}^2}{m o l} \text { i.e., } \mathrm{N} \cdot \mathrm{m}^4 \cdot \mathrm{K}^2 \cdot \mathrm{mol}^{-1}\)
Question 46. The critical temperature for carbon dioxide and methane arc is 31.1 °C and -81.0°C respectively. Which of these has stronger Interiuolecuhir forces and why?
Answer: The critical temperature of gas Is a measure of the intermolecular forces of attraction In the gas. A gas with stronger intermolecular forces of attraction has a higher critical temperature. Therefore, CO2., gas possesses stronger intermolecular forces of attraction than CH4 gas because CO2 has a higher critical temperature than CH4.
States Of Matter Gases And Liquids Very Short Answer Type Questions
Question 1. Mention the variables and constant quantities in Charles’ law
Answer: Variables; Volume ( V) and absolute temperature (T) and Constants: Mass (m) and pressure (P) ofthe gas.
Question 2. If P is plotted against 1/ V for 1 mol of an ideal gas at 0°C, then a straight line passing through the origin is obtained. What is the slope ofthe straight line?
Answer: For 1mol of an ideal gas, PV = RT or, P = RT/V At 0°C or 273K, P = 273R/V.
Hence, the slope of the P vsl/V plot is =273 R
Question 3. The number of molecules in an ideal gas with a volume of V at pressure P and temperature T is ‘n Write down the equation of state for this gas
Answer: No. of mole of the gas
⇒ \(=\frac{\text { Total no. of molecules of the gas }}{\text { Avogadro’s no. }}=\frac{n}{N}\)
∴ The equation of state \(P V=\frac{n}{N} R T\)
Question 4. At a definite temperature, the total pressure of a gas mixture consisting of three gases A, B and C is P. If the number of moles of A, B and C are 2, 4 and 6, respectively, then arrange these gases in increasing order of their partial pressures
Answer: Themole fractions of A, B and C in the gas mixture
⇒ \(x_{\mathrm{A}}=\frac{2}{12}=\frac{1}{6}, x_B=\frac{4}{12}=\frac{1}{3} \text { and } x_C=\frac{6}{12}=\frac{1}{2}\)
Therefore The PArticl Pressure Of Gas A \(A, p_A=\frac{P}{6}, p_B=\frac{P}{3}\) \(p_C=\frac{P}{2}\). Hence PA<PB<PC.
Question 5. Under identical conditions of temperature and pressure, it takes time t1 for the effusion of VmL of N2 gas through a porous wall and time t2 for the effusion of the same volume of O2 gas through the same porous wall. Which one is greater, t1 or t2?
Answer: At constant temperature and pressure, the rates of effusion of different gases are inversely proportional to the square roots of their molar masses. As the molar mass of N2 is smaller than that of CO2, at the given conditions the time required for the effusion of VmL of N2 gas will be less than that required for V mL CO2 of gas. Hence, t1<t2.
Question 6. Arrange the following gases in the increasing order of their densities at STP: H2, air, CO2.
Answer: The density of a gas at constant temperature and pressure \(d=\frac{P M}{R T}.\)
Thus, at a particular temperature and pressure, the density of the gas is directly proportional to the molar mass of the gas. The order of the molar mass of the given gases is: Mh1< Mair < MCO1. Therefore, the order of densities of these gases at STP would be, be < dair < dco2.
Question 7. What is the numerical value of N/n?[N and n are the number of gas molecules and number of moles of gas]
Answer: Number of gas molecules (N) = several moles of the gas (n) x Avogadro’s number or\(\frac{N}{n}\) = Avgadro’s number = 6.022 x1023.
Question 8. At a constant temperature, a container of fixed volume holds NH3 and HCl gases. Can Dalton’s law of partial pressures be applied to this gas mixture?
Answer: Dalton’s law of partial pressure applies only to a mixture composed of two or more non-reacting gases. NH3 and HC1 gases react together to produce NH4C1. So Dalton’s law of partial pressure will not be applicable in this case.
Question 9. On what factors does the value of the total kinetic energy of the molecules in a gas depend?
Answer: The total kinetic energy of the molecules of 1 mol gas \(=\frac{3}{2} R T\)
The total kinetic energy of n mole gas \(=n \times \frac{3}{2} R T\)
Therefore, the total kinetic energy of any gas depends upon the temperature and the quantity ofthe gas.
Question 10. The equation of state of a real gas is P(V-b) = RT. Can this gas be liquefied?
Answer: For the given gas, the value of van der Waals constant a = 0, indicates the absence of intermolecular forces of attraction in the gas. So this gas cannot be liquefied.
Question 11. At a low pressure, the van der Waals equation reduces What is the value of \(\left(P+\frac{a}{V^2}\right) V=R T\) compressibility factor (Z) for this case at this condition?
Answer: \(\left(P+\frac{a}{V^2}\right) V=R T\)
or, \(P V+\frac{a}{V}=R T \quad \text { or, } \frac{P V}{R T}+\frac{a}{R T V}=1 \quad \text { or, } Z=1-\frac{a}{R T V}\)
Question 12. The normal boiling points of two gases A and B are -150°C and -18°C, respectively. Which one of the two gases will behave more like an ideal gas at STP?
Answer: The very low boiling point of A implies the very weak intermolecular forces of attraction in gas A. Hence, gas A will show more ideal behaviour at STP.
Question 13. Prove that c = [E= total kinetic energy molecules of 1 mol of a gas, M=molar mass of the gas, arms = root mean square velocity of gas molecule]
Answer: we known \(c_{r m s}=\sqrt{\frac{3 R T}{M}} \quad \text { or, } c_{r m s}=\sqrt{\frac{2}{M} \times \frac{3}{2} R T}\)
The total kinetic energy of the molecules of lmol gas, \(E=\frac{3}{2} R T\)
∴ \(c_{r m s}=\sqrt{\frac{2 E}{M}} \quad \text { (Proved) }\)
Question 14. It is easier to liquefy a gas with a higher critical temperature—explain.
Answer: The critical temperature of a gas reflects the strength of intermolecular attractive forces in the gas. A gas with a higher value of critical temperature possesses a stronger intermolecular force of attraction. Now, the stronger the intermolecular forces of attraction in a gas, the easier it is to liquefy the gas. Therefore, a gas with a higher critical temperature can be liquified easily.
Question 15. Why cannot CO2 gas be liquefied above 31.1°C?
Answer: The critical temperature of CO2 is 31.1°C. Above 31.1°C, due to the very high average kinetic energy of CO2 molecules, the attraction between them becomes negligible. As a result, CO2 gas cannot be liquefied above 31.1°C.
Question 16. Among the following properties of a liquid, which one increases with the increase in temperature? Surface tension, viscosity and vapour pressure
Answer: Vapour pressure increases with an increase in temperature.
Question 17. The vapour pressures of benzene and water at 50°C are 271 and 92.5 torr, respectively. Which one would you expect to have stronger intermolecular forces of attraction?
Answer: A liquid with strong intermolecular forces of attraction has low vapour pressure. Thus, intermolecular forces of attraction are stronger in water.
Question 18. At 25°C, the vapour pressure of ethanol is 63 torr. What does it mean?
Answer: At 25°C, the pressure exerted by ethanol vapour in equilibrium with the liquid ethanol is 63 torr.
Question 19. The normal boiling point of diethyl ether is 34.6°C What will be its vapour pressure at this temperature?
Answer: At the normal boiling point of a liquid, the vapour pressure of the liquid is equal to the atmospheric pressure. Therefore, the vapour pressure of diethyl ether is at 34.6°C. temperature is 1 atm or 760 torr.
Question 20. The normal boiling points of ethanol and benzene are 78.3°C and 80°C, respectively. Is the vapour pressure of ethanol lower than, greater than or equal to the vapour pressure of benzene?
Answer: At the normal boiling point of a liquid, the vapour pressure of the liquid is equal to the atmospheric pressure. Therefore, the vapour pressure of ethanol at 78.3°C and that of benzene at 80°C are the same and equal to 1 atm.
Question 21. Why is the equilibrium established in the evaporation of a liquid in a closed vessel at a constant temperature called dynamic equilibrium?
Answer: This is so called because the process of evaporation and condensation does not stop at equilibrium and they keep on occurring equally and take
Question 22. Which one between water and ethanol has greater surface tension at a particular temperature?
Answer: As intermolecular forces of attraction are stronger in water than in ethanol, the surface tension of water is greater than that of ethanol.
Question 23. What is the value of the surface tension of a liquid at its critical temperature?
Answer: The surface tension of a liquid at critical temperature is zero because the surface of separation between the liquid and vapour disappears at this temperature.
Question 24. At a given temperature, the viscosity of liquid A is greater than that of liquid B. Which of these two liquids has stronger intermolecular forces of attraction?
Answer: The stronger the intermolecular forces of attraction of a liquid, the higher the viscosity will be. Therefore, the inter¬ molecular forces of attraction of A will be greater than B
Question 25. Give two examples where capillary action occurs.
Answer: Due to capillary action, water from the soil reaches the leaves of a tree through its stem. water comes out through the pores of a clay pot and evaporates. As a result, the water in the pot gets cooled.
States Of Matter Gases And Liquids Numerical Examples
Question 1. At 10-3 mm pressure and 300K, a 2L flask contains equal numbers of moles of N2 and water vapour, What is the total number of moles of N2 and water vapour in the mixture? What is the total mass ofthe mixture?
Answer: \(n=\frac{P V}{R T}=\frac{\left(10^{-3} / 760\right) \times 2}{0.0821 \times 300}=1.06 \times 10^{-7} \mathrm{~mol}\)
⇒ \(n_{\mathrm{N}_2}=n_{\mathrm{H}_2 \mathrm{O}}=\frac{n}{2}=\frac{1.06 \times 10^{-7}}{2}=5.3 \times 10^{-8} \mathrm{~mol}\)
∴ Total Mass \(m_{\mathrm{N}_2}+m_{\mathrm{H}_2 \mathrm{O}}\)
⇒ \(=\left(5.3 \times 10^{-8} \times 28+5.3 \times 10^{-8} \times 18\right) \mathrm{g}=2.44 \times 10^{-6} \mathrm{~g}\)
Question 2. At constant temperature and 1 atm pressure, an ideal gas occupies a volume of 3.25m3. Calculate the final pressure of the gas in the units of atm, torr and Pa if its volume is reduced to 1.25m2 while its temperature is kept constant.
Answer: \(\begin{aligned}
& P_2=\frac{P_1 V_1}{V_2}=\frac{1 \times 3.25}{1.25}=2.6 \mathrm{~atm} \\
& =2.6 \times 760 \mathrm{torr}=1.976 \times 10^3 \mathrm{torr}=2.633 \times 10^5 \mathrm{~Pa}
\end{aligned}\)
Question 3. The density of a gas at 27°C and 1 atm is 15g-mL-1.At what temperature, will the density of that gas be 10g-mL-1, at the same pressure?
Answer: \(\frac{d_2}{d_1}=\frac{T_1}{T_2} \text { or, } T_2=\frac{d_1}{d_2} \times T_1=\frac{15}{10} \times 300=450 \mathrm{~K} \text {; }\)
∴ t= 177°C
Question 4. The density of a gas at 30°C and 1.3 atm pressure is 0.027 g- mL-1. What is the molar mass ofthe gas?
Answer: \(\begin{aligned}
d=\frac{P M}{R T} \text { or, } M=\frac{d R T}{P} & =\frac{0.027 \times 0.0821 \times 10^3 \times 303}{1.3} \\
& =516.6 \mathrm{~g} \cdot \mathrm{mol}^{-1}
\end{aligned}\)
Question 5. 0.0286 g of a gas at 25°C and 76 cm pressure occupies a volume of 50 cm3. What is the molar mass ofthe gas?
Answer: \(M=\frac{g R T}{P V}=\frac{0.0286 \times 0.0821 \times 298}{1 \times 50 \times 10^{-3}}=14 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)
Question 6. The density of a gas at -135°C and 50.66 atm pressure is 2g- cm-3. What is the density ofthe gas at STP?
Answer: \(\frac{d_2}{d_1}=\left(\frac{P_2}{P_1}\right) \times\left(\frac{T_1}{T_2}\right)\)
⇒ \(\text { or, } d_2=2 \times \frac{1}{50.66} \times \frac{138}{273}=0.02 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)
Question 7. A gaseous mixture contains 336 cm3 of H2 and 224 cm3 of HE at STP. The mixture shows a pressure of 2 atm when it is kept in a container at 27°C. Calculate the volume of the gas.
Answer: \(\text { At STP, } 336 \mathrm{~cm}^3 \text { of } \mathrm{H}_2 \equiv \frac{336}{22400} \equiv 0.015 \mathrm{~mol} \text { of } \mathrm{H}_2 \text { and }\)
⇒ \(224 \mathrm{~cm}^3 \text { of } \mathrm{He} \equiv \frac{224}{22400} \equiv 0.01 \mathrm{~mol} \text { of } \mathrm{He} \text {. }\)
⇒ \(V=\frac{n R T}{P}=(0.015+0.01) \times \frac{0.0821 \times 300}{2}=307.8 \mathrm{~cm}^3\)
Question 8. At 100°C a 2-litre flask contains 0.4 g of O2 and 0.6 g of H2. What is the total pressure of this gas mixture in the flask?
Answer: \(P=\frac{n R T}{V}=\frac{\left(\frac{0.4}{32}+\frac{0.6}{2}\right) \times 0.0821 \times 373}{2}=4.78 \mathrm{~atm}\)
Question 9. 1.0 g of benzene is burnt completely in the presence of 4.0 g O2 in a completely evacuated bomb calorimeter of volume 1L. What is the pressure inside the bomb at 30°C, if the volume and pressure of water vapour produced are neglected?
Answer: lg of benzene \(=\frac{1}{78}\) 0.0128 mol of benzene
⇒ \(4 \mathrm{~g} \text { of } \mathrm{O}_2=\frac{4}{32}=0.125 \mathrm{~mol} \text { of } \mathrm{O}_2\)
⇒ \(\begin{array}{rl}
\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(g) \longrightarrow & 6 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(g) \\
0.0128 \mathrm{~mol} \frac{15}{2} \times 0.0128 & 6 \times 0.0128 \\
=0.096 \mathrm{~mol} & =0.0768 \mathrm{~mol}
\end{array}\)
Total number of moles of CO2 produced and O2 remained = (0.0768 + (0.125- 0.096)
⇒ \(P=\frac{n R T}{V}=\frac{0.1058 \times 0.0821 \times 303}{1}=2.63 \mathrm{~atm}\)
Question 10. A closed vessel of fixed volume is filled with 3.2 g O2 at a pressure of P atm and a temperature of Tk. The containers were then heated to a temperature of (T+ 30)K. To maintain the pressure of P atm inside the container at (T+30)K, a certain amount of gas is removed from the container. The gas removed is found to have a volume of 246mL at atm and 27°C. Calculate T.
Answer: Suppose the number of moles Of O2 removed = n1 mol
∴ \(n_1=\frac{P V}{R T}=\frac{1 \times 246 \times 10^{-3}}{0.0821 \times 300}=9.98 \times 10^{-3} \mathrm{~mol}\)
Number of moles of oxygen remained in the container \(\left(n_2\right)=\left(\frac{3.2}{32}-9.98 \times 10^{-3}\right) \mathrm{mol}=0.09 \mathrm{~mol}\)
Initial state: PV = NRT \(=\frac{3.2}{32} R T=0.1 R T\)
Final state: PV = n2RT = 0.09 X R(T+ 30)
∴ 0.17TT = 0.09 x R(T+ 30); hence, T = 270K
Question 11. The temperature of an ideal gas is 340K. The gas is heated to a temperature at constant pressure. As a result, its volume increases by 18%. What is the final temperature of the gas?
Answer: \(\frac{V_2}{V_1}=\frac{T_2}{T_1} \text { or, } \frac{(1.18) V_1}{V_1}=\frac{T_2}{340} ; \text { hence, } T_2=401.2 \mathrm{~K}\)
Question 12. What is the density of air at STP? Assume that air contains 78% N2 and 22% O2 by masses.
Answer: Average molar masses are –
⇒ \(\begin{aligned}
& =\frac{28 \times 78+32 \times 22}{100}=28.88 \mathrm{~g} \cdot \mathrm{mol}^{-1} \\
& d=\frac{P M}{R T}=\frac{1 \times 28.88}{0.0821 \times 273} \mathrm{~g} \cdot \mathrm{L}^{-1}=1.288 \mathrm{~g} \cdot \mathrm{L}^{-1}
\end{aligned}\)
Question 13. At 100°C and 1 atm pressure, the densities of water and water vapour are 1.0 g-mL-1 and 0.0006 g-mL-1 respectively. What is the total volume of water molecules in a litre of steam at 100°C?
Answer: Mass of1L of steam = 1000 x 0.0006 g = 0.6 g
Amount of water in 1L of steam = 0.6 g
The volume of water in 1 L of steam
Question 14. The volume of 0.44 g of a colourless oxide of N2 is 224 mL at 273°C and 1530 mm pressure. What is the compound?
Answer: \(\begin{aligned}
& P V=\frac{g}{M} R T \\
& \text { or, } M=\frac{g R T}{P V}=\frac{0.44 \times 0.0821 \times 546}{\left(\frac{1530}{760}\right) \times 224 \times 10^{-3}} \approx 44 \mathrm{~g} \cdot \mathrm{mol}^{-1}
\end{aligned}\)
The expected compound is N2O.
Question 15. A spherical balloon is filled with air at 2 atm pressure. What pressure is to be exerted on the balloon from the outside so that its diameter will be reduced to half of its initial diameter?
Answer: \(\text { Initial state, } 2 \times \frac{4}{3} \pi r^3=n R T \text {; }\)
⇒ \(\text { Final state: } P \times \frac{4}{3} \pi\left(\frac{r}{2}\right)^3=n R T[r=\text { radius of the balloon }]\)
∴ \(P V=n R T ; P \times \frac{1}{8}=2 \text { or, } P=16 \text { atm }\)
Question 16. If 3.2 g of sulphur is heated to a temperature, the sulphur vapour produced occupies a volume of 780 mL at 723 mm pressure and 450°C. What is the molecular formula of sulphur at this state?
Answer: Let the molecular formula of sulphur is Sx.
∴ \(3.2 \mathrm{~g} \text { of sulphur }=\frac{3.2}{32 \times x}=\frac{1}{10 x} \mathrm{~mol} \text { of sulphur }\)
PV=nRT
⇒ \(\text { or, } \frac{723}{760} \times 780 \times 10^{-3}=\frac{1}{10 \times x} \times 0.0821 \times(450+273)\)
∴ The molecular formula of sulphur = Sb.
Question 17. Determine partial pressures of O2 and N2 in air at 0°C and 760 mm Hg pressure. Air contains 78% N2 and 22% O2 by volume.
Answer: Mole fraction of \(\mathrm{N}_2\left(x_{\mathrm{N}_2}\right)=\frac{\text { Partial volume of } \mathrm{N}_2}{\text { Total volume }}=\frac{78}{100}=0.78\)
Similarly, mole-fraction of (xO2 ) \(=\frac{22}{100}=0.22\)
∴ \(\begin{aligned}
& p_{\mathrm{N}_2}=x_{\mathrm{N}_2} \times P=0.78 \times 760 \mathrm{~mm} \mathrm{Hg}=592.8 \mathrm{~mm} \mathrm{Hg} \\
& p_{\mathrm{O}_2}=x_{\mathrm{O}_2} \times P=0.22 \times 760 \mathrm{~mm} \mathrm{Hg}=167.2 \mathrm{~mm} \mathrm{Hg}
\end{aligned}\)
Question 18. 200 cm3 of N2 gas is collected over water at 20°C and 730 mm pressure. At this temperature, aqueous tension is 14.20 mm. What is the mass of N2 gas collected?
Answer: \(\begin{aligned}
& p_{\mathrm{N}_2}=(730-14.20) \mathrm{mm} \mathrm{Hg}=0.9418 \mathrm{~atm} \\
& p_{\mathrm{N}_2} \times V_{\mathrm{N}_2}=n R T \text { or, } 0.9418 \times 0.2=\frac{w}{28} \times 0.0821 \times 293
\end{aligned}\)
∴ w=0.219 g
Question 19. 0.5 g O2 gas is collected over water at 20°C and 730 mm pressure. If aqueous tension at that temperature is 14.20 mm, what is the volume of O2 gas collected?
Answer: \(\begin{aligned}
& p_{\mathrm{O}_2}=(730-14.20) \mathrm{mm} \mathrm{Hg}=0.9418 \mathrm{~atm} \\
& p_{\mathrm{O}_2} \times V_{\mathrm{O}_2}=n R T
\end{aligned}\)
Question 20. A vessel contains equal masses of CH4 and H2 gas at 25°C. What part of the total pressure inside the vessel is equal to the partial pressure of H2 gas?
Answer: \(W_{\mathrm{CH}_4}=W_{\mathrm{H}_2}=W ;\)
∴ \(n_{\mathrm{CH}_4}=\frac{W}{16} \mathrm{~mol} \text { and } n_{\mathrm{H}_2}=\frac{W}{2} \mathrm{~mol}\)
⇒ \(\text { Total mol, } n=\left(\frac{W}{16}+\frac{W}{2}\right) \mathrm{mol}=\frac{9 W}{16} \mathrm{~mol}\)
Question 21. 8g O2 and some quantity of CO2 are introduced at 30°C into an empty flask of volume 10L. If the total pressure of the gas, the mixture in the flask is 1520 mm. Find the amount of CO2 gas taken.
Answer: \(P V=n R T \text { or, } \frac{1520}{760} \times 10=\left(\frac{8}{32}+\frac{W}{44}\right) \times 0.0821 \times 300\)
∴ W= 24.72 g
Question 22. Partial pressures of the component gases in a gas mixture are H2 = 300 mm; CH4 – 150 mm; N2 = 250 mm. What is the percentage of N2 gas by volume in the mixture?
Answer: \(x_{\mathrm{N}_2}=\frac{\text { Partial volume of } \mathrm{N}_2}{\text { Total pressure }}=\frac{250}{300+150+250}=0.3571\)
Volume percent of N9 in the mixture = Mole-fraction of N2 in the mixture x 100 =35.71%
Question 23. At constant temperature, 2L of N2 gas at 750 mm Hg pressure is mixed with 3L of O2 gas. As a result, the pressure and volume ofthe gas mixture are found to be 732 mm Hg and 5L respectively. What is the initial pressure of O2 gas?
Answer: \(\begin{aligned}
& P_{\mathrm{N}_2} \times V=n_{\mathrm{N}_2} R T \\
& 750 \times 2=n_{\mathrm{N}_2} R T
\end{aligned} \left\lvert\, \begin{aligned}
& P_{\mathrm{O}_2} \times V=n_{\mathrm{O}_2} R T \\
& P_{\mathrm{O}_2} \times 3=n_{\mathrm{O}_2} R T
\end{aligned}\right.\)
In mixture: \(P V=n R T=\left(n_{\mathrm{O}_2}+n_{\mathrm{N}_2}\right) R T=P_{\mathrm{O}_2} \times 3+750 \times 2\)
732×5 PO2x3+1500
∴ PO2 = 720 mm Hg
Question 24. The respective mole fractions of N2 and O2 in dry air are 0.78 and 0.21. If the atmospheric pressure and temperature are 740 torr and 20°C respectively, then what will be the mass of N2 and O2 present in a room of volume 3000ft3? (Assuming the relative humidity of air as zero)
Answer: \(p_{\mathrm{N}_2}=0.78 \times 740 \text { torr }=0.78 \times \frac{740}{760}=0.76 \mathrm{~atm}\)
3000ft3 = 3000 X (30.48)3 [since ft = 30.48 cm]
= 84.95 X 106cc = 84.95 X 103L
∴ PN2 x V= nN2 x RT
or, 0.76 x 84.95 X 103 = N x 0.0821 x (273 + 20)
Or, nN2= 2.683 x 103 mol
∴ Mass of N2 = 2.683 X 103 X 28g = 75.149 kg In calculation, it can be shown that mass of 02 = 23.107 kg
Question 25. 300 cm3 of H2 gas diffuses through a fine orifice in 1 minute. At the same temperature and pressure, what volume of CO2 gas will diffuse through the same orifice in 1 minute?
Answer: \(\frac{V_{\mathrm{H}_2}}{V_{\mathrm{CO}_2}}=\sqrt{\frac{M_{\mathrm{CO}_2}}{M_{\mathrm{H}_2}}}=\sqrt{\frac{44}{2}}\)
⇒ \(\text { or, } V_{\mathrm{CO}_2}=\frac{1}{\sqrt{22}} \times 300 \mathrm{~cm}^3=63.96 \mathrm{~cm}^3\)
Question 26. At constant temperature and pressure, the rates of diffusion of two gases A and B are in the ratio of 1: 2. In a mixture of A and B gases mass ratio of A and B is 2:1, so find the mole fraction ratio of A and B in the mixture.
Answer: \(\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}} \text { or, } \frac{1}{2}=\sqrt{\frac{M_B}{M_A}} \text { or, } M_A=4 M_B\)
⇒ \(\frac{\mathrm{W}_A}{\mathrm{~W}_B}=2 \text { or, } \frac{n_A \times M_A}{n_B \times M_B}=2 \text { or, } \frac{n_A}{n_B}=\frac{1}{2} \text {; }\)
∴ \(\frac{x_A}{x_B}=\frac{n_A}{n_B}=\frac{1}{2}\)
Question 27. The average velocity of the molecules of a gas is 400m- s-1. At the same temperature, what will be the rms velocity of the molecules?
Answer: \(c_a=\sqrt{\frac{8 R T}{\pi M}} \quad c_{r m s}=\sqrt{\frac{3 R T}{M}}\)
∴ \(c_a=\sqrt{\frac{8}{3 \pi}} \times c_{r m s}\)
⇒ \(\text { or, } c_{r m s}=\sqrt{\frac{3 \pi}{8}} \times 400=434.05 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
Question 28. At a constant temperature and 1 atm pressure, the density of O2 gas is 0.0081 g- mL-1. At the same temperature, calculate the average velocity, root mean square velocity and most probable velocity of O2 molecules.
Answer: \(c_{r m s}=\sqrt{\frac{3 P}{d}}=\sqrt{\frac{3 \times 76 \times 13.6 \times .981}{0.0081}}=1.937 \times 10^4 \mathrm{~cm} / \mathrm{s}\)
⇒ \(\begin{aligned}
c_a=\sqrt{\frac{8}{3 \pi}} c_{r m s} & =\sqrt{\frac{8}{3 \pi}} \times 1.937 \times 10^4 \mathrm{~cm} / \mathrm{s} \\
& =1.785 \times 10^4 \mathrm{~cm} / \mathrm{s}
\end{aligned}\)
⇒ \(c_m=\sqrt{\frac{2 R T}{M}}=\sqrt{\frac{2}{3}} \times \sqrt{\frac{3 R T}{M}}=\sqrt{\frac{2}{3}} \times c_{r m s}\)
⇒ \(=\sqrt{\frac{2}{3}} \times 1.937 \times 10^4 \mathrm{~cm} / \mathrm{s}=1.581 \times 10^4 \mathrm{~cm} / \mathrm{s}\)
Question 29. A certain gas of mass 6.431 g occupies a volume of 5 L at a definite temperature and 750 mm pressure. At the temperature, what will be the rms velocity of the molecules of that gas?
Answer: \(P V=n R T=\frac{W}{M} R T \text { or, } \frac{R T}{M}=\frac{P V}{\mathrm{~W}}=\frac{750}{760} \times 5 \times \frac{1}{6.431} \mathrm{~g}^{-1}\)
= 0.7672 L-atm.g-1
= 0.7672 X 103 x 1.013 X 106 cm 2.s-2
[since 1 atm = 1.013 x 106g-cm-1-s-2]
∴ \(c_{r m s}=\sqrt{\frac{3 R T}{M}}=\sqrt{3 \times 0.7672 \times 1.013 \times 10^9} \mathrm{~cm} / \mathrm{s}\)
Question 30. What is the ratio of average velocity and root mean square velocity ofthe molecules of a gas?
Answer: \(c_e=\sqrt{\frac{B R T}{\pi M}}, c_{m m}=\sqrt{\frac{3 R T}{M}}\)
∴ \(\frac{c_a}{c_{m s}}=\sqrt{\frac{8}{3 \pi}}=0.9215\)
Question 31. At what temperature will the rms velocity of S02 molecules be equal to the rms velocity of O2 molecules at 25°C?
Answer: \(c_{\mathrm{rm}}\left(\mathrm{SO}_2\right)=\sqrt{\frac{3 R T}{64}}, c_{\mathrm{mav}}\left(\mathrm{O}_2, 25^{\circ} \mathrm{C}\right)=\sqrt{\frac{3 R \times 298}{32}}\)
∴ 7= 596K ie.,t = 323=C
Question 32. Show that the rms velocity of O2 molecules at 50°C is not equal to the rms velocity of N2 molecules at 25°C.
Answer: \(\begin{aligned}
& c_{\mathrm{rms}}\left(\mathrm{O}_2\right)=\sqrt{\frac{3 R \times(273+50)}{32}}, \\
& c_{\mathrm{mas}}\left(\mathrm{N}_2\right)=\sqrt{\frac{3 R \times(273+25)}{28}}
\end{aligned}\)
∴ \(c_{\mathrm{ms}}\left(\mathrm{O}_2 \text { at } 50^{\circ} \mathrm{C}\right)=c_{\mathrm{ms}}\left(\mathrm{N}_2 \text { at } 25^{\circ} \mathrm{C}\right)\)
Question 33. The average kinetic energy of the atom of Hg vapour is 1000 cal-mol-1. what will be the value of its rms velocity? [Hg = 200]
Answer: \(\begin{aligned}
E=\frac{3}{2} R T & =1000 \mathrm{cal} \cdot \mathrm{mol}^{-1} \\
& =1000 \times 4.157 \times 10^7 \mathrm{erg} \cdot \mathrm{mol}^{-1}
\end{aligned}\)
∴ 3RT = 2000 x 4.157 x 107g-cm2-s-2-mol-1
⇒ \(\begin{aligned}
n s=\sqrt{\frac{3 R T}{M}} & =\sqrt{\frac{2000 \times 4.157 \times 10^7}{200}} \mathrm{~cm} / \mathrm{s} \\
& =2.04 \times 10^4 \mathrm{~cm} / \mathrm{s}
\end{aligned}\)
Question 34. In a container of volume 1L, there are 1023 gas molecules, each of which has a mass of 10-22g. At a certain temperature, if the rms velocity of these molecules is 105cm-s-1, then what would be the pressure at the temperature inside the container?
Answer: \(\begin{aligned}
& P V=\frac{1}{3} m n c_{r m s}^2 \\
& \text { or, } P \times 1000=\frac{1}{3} \times 10^{-22} \times 10^{23} \times\left(10^5\right)^2
\end{aligned}\)
∴ P = 3.33 x 1010 g.cm-1-s.2
= 3.33 X 1010 dyne = 32.87 atm
Question 35. At a constant temperature, a vessel of 1-litre capacity contains 1023 N2 molecules. If the rms velocity of the molecules is 103m/s, then determine the total kinetic energy ofthe molecules and the temperature ofthe gas.
Answer: Average kinetic energy per molecule
⇒ \(=\frac{1}{2} m c_{r m s}^2=\frac{1}{2} \times \frac{28}{6.023 \times 10^{23}} \times\left(10^3\right)^2 \mathrm{~g} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-2}\)
= 2.324 x 10~17g-m2-s-2 = 2.324 X l0-20 J
Average kinetic energy per molecule \(=\frac{3}{2} k T\)
⇒ \(\text { or, } \frac{3}{2} \times \frac{8.314}{6.023 \times 10^{23}} \times T=2.324 \times 10^{-20} ; T=1122.6 \mathrm{~K}\)
Total kinetic energy of 1023 molecules
Question 36. At 0°C the kinetic energy of 102 molecules is 5.62 x 10-14 erg. Determine Avogadro’s number.
Answer: \(\bar{\epsilon}=\frac{3}{2} K T=\frac{3}{2} \times \frac{R}{N_A} \times T\)
⇒ \(\text { or, } 5.62 \times 10^{-14}=\frac{3}{2} \times \frac{8.314 \times 10^7}{N_A} \times 273 \text {; }\)
∴ NA= 6.058 x1023
Question 37. The volume of 2 moles of SO2 at 30°C and 55 atm pressure is 680 mL. What is the value of the compressibility factor of the gas? What is the nature of the deviation of the gas from ideal behaviour?
Answer: \(Z=\frac{P V}{n R T}=\frac{55 \times 680 \times 10^{-3}}{2 \times 0.0821 \times 303}=0.7517\)
As Z < 1, the gas shows a negative deviation from ideal behaviour.
Question 38. The compressibility factor of a real gas at 0°C and 100 atm pressure is 0.927. At this temperature and pressure, how much of this real gas is required to fill a vessel of 100L [molar mass =40g-mol-1?
Answer: \(Z=\frac{P V}{n R T} \text { or, } 0.927=\frac{100 \times 100}{n \times 0.0821 \times 273}\)
∴ n = 481.298mol
∴ Amount of gas required =481.298 x 40g = 19.2519kg
Question 39. For a van der Waals gas, b = 5.0 x 10-2 L-mol-1. What is the diameter of a molecule of this gas?
Answer: \(b=4 N_A \times \frac{4}{3} \pi r^3\)
⇒ \(\begin{aligned}
& 5 \times 10^{-2} \times 1000 \mathrm{~cm}^3 \cdot \mathrm{mol}^{-1}=4 \times 6.023 \times 10^{23} \times \frac{4}{3} \pi r^3 \\
& \text { or, } r=1.705 \times 10^{-8} \mathrm{~cm}
\end{aligned}\)
∴ Diameter of a molecule = 3.41 x 10-8 cm
Question 40. The volume of 2 moles of C02 gas at 27°C is 0.001 m3. What will be the pressure of this gas According to the van der Waals equation and ideal gas equation? [Given a(C02) =0.364 N-m4-mol-2 and b(C02) = 4.27 X 10-5 m3 -mol-1 ]
Answer: \(\left(P+\frac{n^2 a}{V^2}\right)(V-n \dot{b})=n R T\)
⇒ \(\begin{aligned}
& \text { or, }\left(P \ \frac{2^2 \times 0.364 \mathrm{~N} \cdot \mathrm{m}^4}{10^{-6} \mathrm{~m}^6}\right)\left(10^{-3}-2 \times 4.27 \times 10^{-5}\right) \mathrm{m}^3 \\
& \quad=2 \times 0.0821 \times 300 \mathrm{~L} \cdot \mathrm{atm} \\
& \text { or, }\left(P+1.456 \times 10^6 \mathrm{~N} \cdot \mathrm{m}^{-2}\right)\left(9.146 \times 10^{-4} \mathrm{~m}^3\right) \\
& \quad=2 \times 0.0821 \times 300 \times 10^{-3} \mathrm{~m}^3 \cdot \mathrm{atm}
\end{aligned}\)
Question 41. The equation of state for 1 the mole of a gas is (V-b) = RT. At STP, 1 mole of this gas occupies a volume of 28L. Calculate the compressibility factor ofthe gas at STP.
Answer: \(Z=\frac{P V}{n R T}=\frac{1 \times 28}{1 \times 0.0821 \times 273}=1.249\)
Question 42. The compressibility factor of 2 mol of NH3 gas at 27°C and 9.18 atm pressure is 0.931. If the volume of the gas molecules is not taken into consideration, then what is the value of van der Waals constant ‘a’ for NH3 gas?
Answer: \(\left(P+\frac{n^2 a}{V^2}\right) V=n R T \text { or, } P V=n R T-\frac{n^2 a}{V}\)
⇒ \(\text { or, } \frac{P V}{n R T}=1-\frac{n a}{V R T} \text { or, } Z=1-\frac{n a}{V R T}=1-\frac{2 \times a}{V R T}\)
⇒ \(\text { again, } \quad Z=\frac{P V}{n R T} \quad \text { or, } \quad 0.931=\frac{9.18 \times V}{2 \times 0.0821 \times 300}\)
∴ V= 4.991
∴ 0.931 \(=1-\frac{2 \times a}{4.99 \times 0.0821 \times 300}\)
a= 4.2412.atm.mol¯²
Question 43. The pressure exerted by 12g of an ideal gas at t°C in a vessel of volume VL is late. When the temperature is increased by 10°C at the same volume, the pressure increases by 10%. Calculate the temperature ‘f and volume V (molar mass of gas = 120).
Answer: \(12 \mathrm{~g} \text { of the gas }=\frac{12}{120}=0.1 \mathrm{~mol} \text { of the gas }\)
Given that the initial pressure of the gas, P = 1 atm and its initial temperature, T = (273 + t)K By applying ideal gas equation, PV= nRT, we get 1 x V = 1 x 0.0821 x (273 + t)…1
It is given, the final pressure of the gas, \(P=\left(1+\frac{10}{100} \times 1\right)\)
1.1 atm
and its final temperature = (273 + 1 + 10)K = (283+)K
By applying the gas equation, PV = nRT, we get
1.1 x V = IX 0.0821 X (283 + t)
From the equations [1] and [2], we have
1.1 X 0.0821 X (273)+= 1 X 0.0821 X (283 + t)
or, 1.1(273 + t) = 283 + t
∴ t = -173; So, the value off = -173°C
Putting t = -173 in equation [1] (or in equation (2),
We have, V = 1 X 0.0821(273-173) = 0.821L
Question 44. An open vessel contains air at 27°C. At what temperature should the vessel be heated so that l/4th of air escapes from the vessel? Assume that volume ofthe vessel remains the same on heating.
Answer: Initial temperature of air = (273 + 27)K = 300K. Let the amount of air in the vessel at 27°C = x mol. Suppose, the vessel is heated to a temperature of TjK so that 1 /4 th of air escapes from the vessel.
So, the amount of air in the vessel at,\(T_1 K=x-\frac{x}{4}=\frac{3}{4} x\) Since the vessel is open, the pressure of air in the vessel either at 300K or at TK is equal to atmospheric pressure, i.e., 1 atm. Again, the volume of the vessel does not change because of heating. Therefore, at temperature 300K.
⇒ \(x \times 0.0821 \times 300=\frac{3}{4} x \times 0.0821 \times T_1 \text { or, } T_1=400 \mathrm{~K}\)
Question 45. A container of fixed volume 0.4L contains 0.56g of gas at 27°C. The pressure of the gas at this temperature is 936 mmHg. If the amount ofthe gas is increased to 2.1g and its temperature is decreased to 17°C, then what will be the pressure of the gas? Assuming gas behaves ideally.
Answer: Suppose, the molar mass of the gas =M g- mol-1.
So, 0.56g of the gas \(=\frac{0.56}{M}\) mol of the gas
At the initial state \(P=\frac{936}{760}=1.23 \mathrm{~atm}, n=\frac{0.56}{M} \mathrm{~mol} \text {, }\)
V=0.41
and T = (273 + 27)K = 300K
Substituting the values of P, n, V and T in the Ideal gas equation, PV= nRT, we have
⇒ \(1.23 \times 0.4=\frac{0.56}{M} \times 0.0821 \times 300\)
∴ molar mass of the gas = 28g. mol-1
Question 46. A cylinder capable of holding 3L water contains H2 gas at 27°C and a pressure ofP atm. At STP, it is possible to fill up 10 balloons, each of which has a radius of 10cm, with the gas present in the cylinder. Find the value ofP.
Answer: The volume of each balloon
⇒ \(=\frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi(10 \mathrm{~cm})^3=4.19 \mathrm{~L} .\)
As the cylinder can hold 31. of water, the US volume Is 31. The volume of H0 gas required to 1111 up 10 balloons 10×4.19 = 41.9L.
At STP, If the volume of H2 gas present in the cylinder is \(V ., \text { then } \frac{1 \times V}{273}=\frac{P \times 3}{300} \text { or, } V=\frac{273 \times 3}{300} P=2.73 P\)
Even after the balloons have been filled up with 11, gas, the cylinder, still contains Il2 gas of its volume. So, the volume of H2 gas = (2.73P- 3)L
∴ 2.73p- 3 = 41.9 or, P = 16.44
Question 47. At room temperature, 2NO + O2 2→2NO2 → N2O4 reaction proceeds near completion. The dimer, N204, solidifies at 262K. A 250mL flask and a lOOmL flask are separated by a stop-cock. At 300K, nitric oxide in the longer flask exerts a pressure of 1.053 atm and the smaller one contains oxygen at 0.789 atm.
The gases are mixed by opening the stopcock and after the end of the reaction, the flasks are cooled to 220K. Neglecting the vapour pressure of the dimer, find out the pressure and composition ofthe gas remaining at 220K (assume that the gases behave ideally).
Answer: Number of moles of NO2 gas lit 250ml, flask
⇒ \(\frac{P V}{R T}=\frac{1.053 \times 0.25}{0.0821 \times 300}=0.01 \mathrm{~mol}\) and that of O., in 100ml. flask, \(=\frac{P V}{R T}=\frac{0.789 \times 0.1}{0.0821 \times 300}=3.2 \times 10^{-3} \mathrm{~mol} .\)
Reaction: 2NO + O2→2NO2
Thus 1 mol of O2 gns completely reacts with 2 mol of NO gns.
Therefore, 3.2 x 10-3 mol 0f O2 guns will react with
2 X 3.2 X 10-3¯6 4 x 10-3 mol if NO2
After the completion of the reaction number of moles of NO left the reaction system = (0.01 – 6.4 x 10-3 ) = 3.6 x Hi-3 mo-3
The total volume of the reaction system = (0.25 + 0.25)L = 0.351. If the pressure of NO gas left in the reaction system is then P X 0.35 m 3.6 X 10-3 x 0.0821 X 220 P= 0.185 atm.
Question 48. An LPG cylinder weighs 18.4kg when empty when full, it weighs 29.0 Kg and shows a pressure of 2.5 atm. In the course of 27°C, the mass of the full cylinder is reduced to 23.2 Kg. Find out the volume of the gas in cubic meters used up at the normal usage conditions, and the final pressure inside the cylinder. Assume LPG to be n-butane with a normal boiling point of 0°C.
Answer: Mass ofthe gas in full cylinder=(29.0 – 18.4) kg – 10,6 kg Mass of the gas after some of is used up =- (23.2 – 18.4) 4.8kg Mass ofthe gas used up = (10.6 – 4,8) = 5,8 kg. So, 5.8kg of gas = 5.8 kg of n-butane \(=\frac{5.8 \times 10^3}{58}=100\)
mol of H-butane [molar mass of-butane 58 g- mol-1]
⇒ \(V=\frac{n R T}{P}=\frac{100 \times 0.0821 \times 300}{1}=24631 .\)
[at normal usage condition, P 1 atm)
Therefore, the volume of used gas at normal usage conditions = 24631. = 2.463m3 since =10-nm³]
In the cylinder, LPG exists In a liquid state which remains in equilibrium with its vapour. So long as LPG exists in a liquid state in the cylinder, the pressure in the cylinder remains fixed. So, the pressure of the remaining gas in the cylinder will be 2.5 atm.
Question 49. An evacuated vessel weighs 50.0g when empty, 148 when filled with a liquid (d = 0.98 g-mL-1) & 50g when filled with an ideal gas at 760 mmHg at 300K. Determine the molar mass of the gas.
Answer: Mass of the evacuated vessel = 50.0g.
Mass of the vessel when filled with a liquid of density O.98gmL-1 = 148g
∴ Volume of 98g of liquid \(=\frac{98}{0.98}=100 \mathrm{~mL}=0.1 \mathrm{~L}\)
Question 50. A gas mixture composed of N2 and O2 gases has a density of 1.17 g-L¯1 at 27°C and 1 atm pressure. Calculate the mass percents of N2 and O2 in the mixture. Assume that the gas mixture behaves like an ideal gas.
Answer: Given: P = 1 atm and T = (273 + 27)K = 300K
∴ \(M=\frac{d R T}{P}=\frac{1.17 \times 0.0821 \times 300}{1}=28.8 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)
average molar mass ofdie gas mixture = 28.8g-mol 1 . T = — In the mixture, if the mole-fraction of N2 be x, then the mole-fraction of 02 is (1-x).
∴ M=28.8g-mor1 = [28xx+32(l -x)] g-mol-1 or, 4x = 32 – 28.8; hence, x = 0.8 Mole-fraction of N2 = 0.8 and that of 02 = 1- 0.8 = 0.2.
Mass percent of \(\mathrm{N}_2=\frac{0.8 \times 28}{0.8 \times 28+0.2 \times 32} \times 100=77.77 \%\) and mass present 6 Of O2 = (100-77.77)% = 22.23%
Question 51. One mole of nitrogen gas at 0.8 atm takes 38s to diffuse through a pin-hole, whereas one mole of an unknown compound of xenon with fluorine at 1.6 atm takes 57s to diffuse through the same hole. Determine the molecular formula ofthe compound.
Answer: At a given temperature and a pressure of P, the rate of diffusion of a gas \(r \propto \frac{P}{\sqrt{M}}\) [M= molar mass of the gas] If the rates of diffusion of N2 and the unknown gas are r and r.) respectively, then
⇒ \(r_2 \propto \frac{P_2}{\sqrt{M_{\text {compound }}}} \text { and } r_1 \propto \frac{P_1}{\sqrt{M_{N_2}}}\)
∴ \(\frac{r_1}{r_2}=\frac{p_1}{p_2} \times \sqrt{\frac{M_{\text {compound }}}{M_{N_2}}}\)
⇒ \(\text { Given that } r_1=\frac{1}{38} \mathrm{~mol} \cdot \mathrm{s}^{-1}, r_2=\frac{1}{57} \mathrm{~mol} \cdot \mathrm{s}^{-1} \text {, }\)
p1 = 0.8 atm and p2 = 1.6 atm
∴ \(\frac{\frac{1}{38}}{\frac{1}{57}}=\frac{0.8}{1.6} \sqrt{\frac{M_{\text {compound }}}{28}} \text { or, } \sqrt{\frac{M_{\text {compound }}}{28}}=3\)
∴ M compound = 3 X 28 = 252g-mol-1
Let the molecular formula of the compound = XeFx
∴ 131 + x x 19 = 252 or, x = 6.368 ~ 6
∴ Molecular formula ofthe compound = XeFg
Question 52. A gas bulb of capacity contains 2.0 x 1021 molecules of nitrogen exerting a pressure of 7.57 x 103 N.m-2. Calculate the rms velocity and temperature of the gas. If the ratio of the most probable speed to the root mean square speed is 0.82, calculate the most probable speed of the molecules at this temperature.
Answer: 2.0 x 1021 molecules of nitrogen are contained in
⇒ \(\frac{2 \times 10^{21}}{6.023 \times 10^{23}}=3.32 \times 10^{-3} \mathrm{~mol} \text { of } \mathrm{N}_2 \text { gas. }\)
Given that P = 7.57 x 103N.m-2 and V = 1
∴ P = 7.57 X 103N-m~2 = 7.57 X 103Pa
⇒ \(=\frac{7.57 \times 10^3}{1.013 \times 10^5} \mathrm{~atm}=0.0747 \mathrm{~atm}\)
[since lN-m-2 = IPa and latm = 1.013 x 105Pa]
∴ \(T=\frac{P V}{n R}=\frac{0.0747 \times 1}{3.32 \times 10^{-3} \times 0.0821}=274.05 \mathrm{~K}\)
∴ Temperature of the gas =274.05K
we know \(c_{r m s}=\sqrt{\frac{3 R T}{M}}\)
∴ Root mean square velocity of N2 molecules
⇒ \(=\sqrt{\frac{3 \times 8.314 \times 10^7 \times 274.05}{28}} \mathrm{~cm} \cdot \mathrm{s}^{-1}\)
= 4.94 x 104cm-s-1
⇒ \(\text { Given that } \frac{c_m}{c_{r m s}}=0.82\left[c_m=\text { most probable speed }\right]\)
∴ \(\begin{aligned}
c_m=0.82 \times c_{r m s} & =0.82 \times 4.94 \times 10^4 \\
& =4.05 \times 10^4 \mathrm{~cm} \cdot \mathrm{s}^{-1}
\end{aligned}\)
Question 53. The composition of the equilibrium mixture (Cl2 2C1), which is attained at 1200°C, is determined by measuring the rate of effusion through a pin-hole. It is observed that at 1.80 mmHg pressure, the mixture effuses 1.16 times as fast as Krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms. (Relative atomic mass of Kr = 84)
Answer: If the rates of effusion of equilibrium mixture and Kr gas are r1 and r2 then \(\frac{r_1}{r_2}=\sqrt{\frac{M_{\mathrm{Kr}}}{M}} ;\) where M= average molar mass of the equilibrium mixture. Given that rx = r2 x 1.16
∴ \(1.16=\sqrt{\frac{84}{M}}\)
∴ M = 62.42 g.mol-1
let the initial amount of Cl2 gas be 1 mol and its degree of dissociation = x. Therefore, the number of moles of Cl2 and Cl at equilibrium will be as follows: \(\mathrm{Cl}_2 \rightleftharpoons 2 \mathrm{Cl}\)
Initial number of moles: Number of moles at equilibrium
∴ Total number of moles of Cl2 and Cl in the equilibrium mixture =l-x+ 2x = 1 + x
Average molar mass of the equilibrium mixture \(=\frac{(1-x) M_{\mathrm{Cl}_2}+2 x \times M_{\mathrm{Cl}}}{1+x}=\frac{71}{1+x} \mathrm{~g} \cdot \mathrm{mol}^{-1}\)
Now, \(\frac{71}{1+x}=62.42\) or, 62.42 + x X 62.42 = 71
∴ x= 0.1374
Fraction of Cl2 molecules dissociated into atoms = 13.74%
Question 54. The density of the vapour of a substance at 1 atm and 500K is 0.36 Kg-m-3. The vapour effuses through a hole at a rate of 1.13 times faster than O2 under the same condition. Determine, the molar mass, molar volume, and Compression factor of the vapour and which forces among the gas molecules are dominating attractive or repulsive. If vapour behaves ideally at 1000K, find the average translational kinetic energy of a molecule.
Answer: If the rates of effusion ofthe vapour and O2 gas are r1 and r2 respectively then
\(\frac{r_1}{r_2}=\sqrt{\frac{M_{\mathrm{O}_2}}{M_{\text {vapour }}}}\)
∴ \(1.33=\sqrt{\frac{32}{M_{\text {vapour }}}}\)
The molar mass of the vapour = 18.09g-mol-1
The molar volume of the vapour
⇒\(\begin{aligned}
& =\frac{\text { Molar mass of the vapour }}{\text { density of the vapour }} \\
& =\frac{18.09 \mathrm{~g} \cdot \mathrm{mol}^{-1}}{0.36 \times 1000 \times 10^{-3} \mathrm{~g} \cdot \mathrm{L}^{-1}}=50.25 \mathrm{~L} \cdot \mathrm{mol}^{-1}
\end{aligned}\)
⇒ \(Z=\frac{P V_m}{R T}=\frac{1 \times 50.25}{0.0821 \times 500}=1.224\)
[Vm = molar volume] Hence, Z >1
Z> 1 indicates that the vapour shows a positive deviation from ideal behaviour. This kind of deviation occurs in gas when the effect of intermolecular repulsive forces dominates over the effect of intermolecular attractive forces.
Average translational kinetic energy of a molecule \(=\frac{3}{2} k T\) \(=\frac{3}{2} \times 1.32 \times 10^{-23} \mathrm{~J} \cdot \mathrm{K}^{-1} \times 1000 \mathrm{~K}=1.98 \times 10^{-20} \mathrm{~J}\)
Question 55. The compressibility factor for 1 mole of a van der Waals gas at 0°C and 100 atm pressure is 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals constant, a.
Answer: Compressibility factor \(Z=\frac{P V}{n R T}\)
Given that Z = 0.5, n = 1 mol, T – 273K and = 100 atm
∴ \(V=\frac{Z \times n R T}{P}=\frac{0.5 \times 1 \times 0.0821 \times 273}{100} \mathrm{~L}=0.112 \mathrm{~L}\)
For 1 mol of a real gas, the van der Waals equation is—
⇒ \(\left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=R T\left[V_m=\text { molar volume }\right]\)
As the volume of a gas molecule is assumed to be negligible, but, and hence Vm-b~ Vm
∴ \(\text { or, } \frac{P V_m}{R T}=1-\frac{a}{\mathrm{RT} V_m} \quad \text { or, } Z=1-\frac{a}{R T V_m}\)
⇒ \(\text { or, } \frac{P V_m}{R T}=1-\frac{a}{\mathrm{RT} V_m} \quad \text { or, } Z=1-\frac{a}{R T V_m}\)
⇒ \(\text { or, } 0.5=1-\frac{a}{0.0821 \times 273 \times 0.112} ; \quad \text { hence, } a=1.255\)
Question 56. The pressure in a bulb dropped from 2000 to 1500 mm of Hgin 47 min when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molar mass 79 in the molar ratio of 1: 1 at a total pressure of 4000 mm of Hg was introduced. Find the molar ratio of two gases remaining in the bulb after 75 min.
Answer: Decrease in pressure of O2 in 47 min=2000- 1500=500mmHg
Decrease in pressure of O2 in 74 min = \(=\frac{500}{47} \times 74=787.23\) mm kg
In the mixture, the mole ratio of O2 and the other gas =1: 1
So, mole-fraction of each of O2 and other gas in the mixture \(=\frac{1}{2}\)
In the mixture, the partial pressure of \(\mathrm{O}_2=\frac{1}{2} \times 4000=\) 2000 mmHg and that of the other gas \(=\frac{1}{2} \times 4000 \mathrm{mmHg}=\) = 2000 mm kg If the rates of effusion of 02 gas and the other gas are r1 and respectively, then \(\frac{r_1}{r_2}=\sqrt{\frac{M}{M_{\mathrm{O}_2}}}\) [M= molar mass of other gas]
Now, ,\(r_1=\frac{787.23}{74} \mathrm{mmHg} \cdot \mathrm{min}^{-1}\) = 10.63 mm hg. min-1.
∴ \(r_2=r_1 \times \sqrt{\frac{M_{\mathrm{O}_2}}{M}}=10.63 \sqrt{\frac{32}{79}}=6.76 \mathrm{mmHg} \cdot \mathrm{min}^{-1}\)
⇒ \(\text { Now, } r_2=\frac{\begin{array}{c}
\text { the decrease in pressure } \\
\text { of the other gas }
\end{array}}{74}\)
∴ The decrease In pressure of the other gas = r2 X 74 = 6.76 X 74 = 500.24 mmHg
Therefore, In the mixture, partial pressure of O2 gas (2000- 787.23) l212.77mmHg and that of the other gas=(2000- 500.24)- 1499.70 mm g Hence, the ratio of mole-fraction of the other gas to O2 gas =1499.70: 1212.77 = 1.236: 1 In the mixture, the molar ratio of the two gases = 1.230: 1.
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Question 1. Oxygen gas is present in a 1L flask at a pressure of 7.6 x 10¯1° mm Hg at 0°C. Calculate the number of O2 molecules in the flask
Answer: \(P=\frac{7.6 \times 10^{-10}}{760}\)
∴ \(n=\frac{P V}{R T}=\frac{7.6 \times 10^{-10} \times 1}{760 \times 273 \times 0.082}\)
[V = 1L, T = 273K, n = 4.46 x 10-14]
Question 2. Sketch PIT versus T plot for an ideal gas at constant volume. Indicate the value ofthe slope (mass fixed). Under the same conditions of temperature and pressure NH3, Cl2 and CO2 gases are allowed to diffuse through a porous wall. Arrange these gases in the increasing order ofthe rate of diffusion.
Answer: According to Gay-Lussac’s law, \(\frac{P}{T}=K\) (constant), when the mass and volume of a gas are constant. Hence, the plot of \(\frac{P}{T}=K\) versus T indicates a straight line parallel to the X -axis and the slope ofthe curve is tan 0 = 0.
The order of molar masses of NH3 Cl2 and CO2 gases are—NH3 < CO2 < Cl2.
Hence, the increasing order of their rates of diffusion is Cl2 < CO2 < NH2.
The value of ‘R ‘ in J-K-1 -mol-1 unit is 8.314
Question 3. Indicate the correct answer: The rate of diffusion of helium gas at constant temperature and pressure will be four times the rate of diffusion of the following gases
- CO2
- SO2
- NO2
- O2
Answer: 2. SO2
Question 4. A 2L flask contains 0.4 g O2 and 0.6 g H2 at 100°C. Calculate the total pressure of the gas mixture in the flask. 4
Answer: Total no. of moles of O2 & H2 in flask \(=\frac{0.4}{32}+\frac{0.6}{2}=0.3125\)
∴ \(P=\frac{n R T}{V}=\frac{0.3125 \times 0.082 \times 373}{2}=4.78 \mathrm{~atm}\)
Question 5. The equation of state of a real gas is P(V-b) = RT. Can the gas be liquified? Explain. Sketchlog P vlog V graph for a given mass of an ideal gas at constant temperature and indicate the slope.
Answer: Value of van der Waals constant V= 0 for the given gas. Hence, there exists no force of attraction among the gas molecules. So, the gas cannot be liquefied.
Question 6. Arrange CO2, SO2 and NO2 gases in increasing order of their rates of diffusion under the same condition of temperature and pressure with reason.
Answer: At constant temperature and pressure the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Hence increasing order of rates of diffusion is:
⇒ \(r_{\mathrm{SO}_2}<r_{\mathrm{NO}_2}<r_{\mathrm{CO}_2} .\)
Question 7. Any real gas behaves ideally at very low pressure and high temperature. Explain. The values of van der Waals constant ‘ a’ for N2 and NH3 are 1.37 and 4.30 L2 -atm- mol¯² respectively. Explain the difference in values.
Answer: van der Waals constant ‘a’ denotes the magnitude of attractive forces between the molecules of real gases. The value of ‘a’ is higher in the case of NH3 than N2. That means attractive forces (London force, dipole-dipole attraction force) present among the molecules of NH3 are stronger than those of N2 (London force). Consequently, NH3 can be liquefied more easily than N2.
Question 8. For a definite mass of ideal gas at constant temperature, V versus \(\frac{1}{p}\) plot is a—
- Parabola
- Straight line
- Hyperbola
- Rectangular Hyperbola
Answer: 2. Straight line
Question 9. A gas of molar mass 84.5g/mol is enclosed in a flask at 27°C has a pressure of 2atm. Calculate the density of the gas. [R = 0.082L-atm- K-1-mol-1 ]
Answer: Density of a gas,
⇒ \(\begin{aligned}
d=\frac{P M}{R T} & =\frac{2 \times 84.5}{0.082 \times 300} \mathrm{~g} \cdot \mathrm{L}^{-1} \\
& =6.87 \mathrm{~g} \cdot \mathrm{L}^{-1}
\end{aligned}\)
Question 10. Which gas among the following exhibits maximum critical temperature—
- N2
- O2
- CO2
- H2
Answer: 3. CO2
Question 11. Explain the nature of the graphs of log P versus log V and logy versus log T. What are the units of the van der Waals constants ‘a’ and ‘b’?
Answer: The plot of log P versus log V indicates a straight line with a negative slope (-1). The plot of log V versus log indicates a straight line with positive slope (+1)
Unit of is atm. L2- mol¯² a unit of ‘b’ is L-mol-1.
Question 12. The surface tension of water with the increase of temperature may—
- Increase
- Decreases
- Remain same
- Shows irregular behaviour
Answer: 2. Decreases
Question 13. For which property of the liquid the shape of a liquid drop is spherical? A 10-litre volumetric flask contains 1 gHe and 6.4 g of O2 at 27°C temperature. If the total pressure of the mixture is 1.107 atm, then what is the partial pressure ofHe and O2?
Answer: Surface tension.
⇒ \(n_{\mathrm{He}}=\frac{1}{4}=0.25 \mathrm{~mol}, n_{\mathrm{O}_2}=\frac{6.4}{32}=0.2 \mathrm{~mol}\)
∴ Total number of moles =(0.25 + 0.2) = 0.45
⇒\(x_{\mathrm{He}}=\frac{0.25}{0.45}=0.5555 \text { and } x_{\mathrm{O}_2}=\frac{0.2}{0.45}=0.4444\)
pHe = 0.5555 x 1.107 atm = 0.615 atm
Po2 = 0.4444 x 1.107 atm =0.491 atm
Question 14. Which of the following is the unit of van der Waals gas constant
- L2.mol
- L.mol¯²
- L.mol
- L.mol¯¹
Answer: 4. L.mol¯¹
Question 15. What will be the ratio of \({ }^{235} \mathrm{UF}_6\) diffusion of And \({ }^{235} \mathrm{UF}_6\)
Answer: Let, the rate of diffusion of \({ }^{235} \mathrm{UF}_6 \text { and }{ }^{238} \mathrm{UF}_6\) be r1 and r2.
According to Graham’s law, \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\)
⇒ \(\begin{aligned}
& M_1=\text { molecular mass of }{ }^{235} \mathrm{UF}_6=349 \mathrm{~g} \cdot \mathrm{mol}^{-1} \\
& M_2=\text { molecular mass of }{ }^{238} \mathrm{UF}_6=352 \mathrm{~g} \cdot \mathrm{mol}^{-1}
\end{aligned}\)
∴ \(\)
Question 16. The cause of the spherical drop of water is—
- Surface tension
- Viscosity
- Hydrogen bond
- High critical temperature of h2O vapour
Answer: 2. Viscosity
Question 17. State Gay Lussac’s law related to the pressure and temperature of a gas. 3.2 g of sulphur when vaporised, the sulphur vapour occupies a volume of 280.2 mL at STP. Determine the molecular formula of sulphur vapour under this condition. (S = 32)
Answer: 280.2 mL of sulphur weighs 3.2 g at STP. 22400 mL of sulphur weighs 255.8 g at STP. Let, the molecular formula of sulphur Sn. So, n x 32 = 255.8 or, n = 7.9 ~ 8 Molecular formula ofsulphur is Sg.
Question 18. Determine the volume of 2.2 g of carbon dioxide at 27°C and 570 mmHg pressure.
Answer:
⇒ \(2.2 \mathrm{~g} \text { of } \mathrm{CO}_2 \equiv \frac{2.2}{44}=0.05 \mathrm{~mol} \text {. }\)
According to the ideal gas equation (taking CO2 as an ideal gas), PV = nRT
⇒ \(\text { or, } \frac{570}{760} \times V=0.05 \times 0.082 \times 300\)
or, v= 1.64 The volume of CO2 at the given condition is 1.64L.
States Of Matter Gases And Liquids Multiple Choice Questions
Question 1. Equal weight of CH4 and H2 are mixed in an empty container at 25°C. The fraction of the total pressure exerted by H2 is-
- \(\frac{1}{9}\)
- \(\frac{1}{2}\)
- \(\frac{8}{9}\)
- \(\frac{16}{17}\)
Answer: 3. \(\frac{8}{9}\)
The gas mixture contains equal masses of CH and H2. Suppose, mass of each of these gases = u>g so, in the mixture \(x_{\mathrm{H}_2}=\frac{w / 2}{\frac{w}{2}+\frac{w}{16}}=\frac{8}{9} \text { and } x_{\mathrm{CH}_4}=1-x_{\mathrm{H}_2}=1-\frac{8}{9}=\frac{1}{9}\)
∴ The partial pressure of \(\mathrm{H}_2, p_{\mathrm{H}_2}=x_{\mathrm{H}_2} \times P=\frac{8}{9} P\) [p= total pressure of the mixture]
∴ \(\frac{p_{\mathrm{H}_2}}{P}=\frac{8}{9}\)
Question 2. Avan der Waals gas may behave ideally when—
- Volume is very low
- Temperature is very high
- The pressure is very
- The temperature, pressure, and volume all are very high
Answer: 3. The pressure is very
A van der Waals gas behaves ideally when its temperature is very high or pressure is very low. At either of these two conditions, the volume of the gas becomes very large, which results in a large separation of gas molecules.
Thus, intermolecular forces of attraction become negligible, and gas behaves approximately like an ideal gas.
Question 3. Two gases X (mol. wt. Mx) and Y (mol. wt. My; My> Mx) are at the same temperature, in two different containers. Their root mean square velocities are Cx and CY respectively. If average kinetic energies per molecule of two gases X and Y are Ex and Ey respectively, then which of the following relation(s) is (are) true
- Ex>EY
- CX>CY
- \(E_X=E_Y=\frac{3}{2} R T\)
- \(E_X=E_Y=\frac{3}{2} k_B T\)
Answer: 3. \(E_X=E_Y=\frac{3}{2} R T\)
For 1 mol of a gas, van der Waals equation: \(\left(P+\frac{a}{V^2}\right)(v-b)=R T\) If the value of is negligible, then \(P+\frac{a}{V^2} \approx P\)
∴ P(V-b) = RT or, PV = RT + Pb \(\text { or, } \frac{P V}{R T}=1+\frac{P b}{R T} \quad \text { or, } Z=1+\frac{P b}{R T}\)
Question 4. The compressibility factor (Z) of one mole of a van der Waals gas of negligible ‘a’ value is-
- 1
- \(\frac{b P}{R T}\)
- \(1+\frac{b P}{R T}\)
- \(1-\frac{b P}{R T}\)
Answer: 3. \(1+\frac{b P}{R T}\)
For 1 mol of a gas, van der Waals equation: \(\left(p+\frac{a}{V^2}\right)(v-b)=R T\)
If the value of ‘a’ is negligible, the \(P+\frac{a}{V^2} \approx P\)
∴ P(V-b) = RT
⇒ \(\text { or, } P V=R T+P b \quad \text { or, } \frac{P V}{R T}=1+\frac{P b}{R T} \quad \text { or, } Z=1+\frac{P b}{R T}\)
Question 5. For one mole of an ideal gas, the slope of the V vs. T curve at a constant pressure of 2 atm is XL-moH-K¯¹. The value of the ideal universal gas constant ‘R’ in terms of X is
- \(X \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
- \(\frac{X}{2} \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
- \(2 X \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
- \(2 X \mathrm{~atm} \cdot \mathrm{L}^{-1} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
Answer: 3. \(2 X \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
⇒ \(P V=n R T \text { or, } P\left(\frac{V}{n}\right)=R T\)
or,PVm = RT [Vm = molar volume \(\text { or, } V_m=\frac{R}{P} T\)
At constant pressure, \(\frac{R}{P}\) = constant = K. So, at constant pressure, for 1 mol of an ideal gas, Vm = KT. This relation represents a straight-line equation passing through the origin. So, for 1 mol of an ideal gas at constant pressure, the graph of Vm vs. Twill be a straight line with slope = K.
Given,K= XL.mol¯¹.K-1o
⇒ \(\text { or, } \frac{R}{P}=X \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \quad \text { or, } \frac{R}{2 \mathrm{~atm}}=X \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
∴ R = 2XL-atm-mol-1.K-1
Question 6. At a certain temperature, the time required for the complete diffusion of 200 mL of H2 gas is 30 minutes. The time required for the complete diffusion of 50 mL of O2 gas at the same temperature will be—
- 60 mins.
- 30 mins.
- 45 mins.
- 15 mins.
Answer: 2. 30 mins.
According to Graham’s law \(\frac{V_{\mathrm{H}_2} / t_1}{V_{\mathrm{O}_2} / t_2}=\sqrt{M_{\mathrm{O}_2} / M_{\mathrm{H}_2}}\)
⇒ \(\text { or, } \frac{V_{\mathrm{H}_2}}{V_{\mathrm{O}_2}} \times \frac{t_2}{t_1}=\sqrt{\frac{M_{\mathrm{O}_2}}{M_{\mathrm{H}_2}}} \text { or, } \frac{200}{50} \times \frac{t_2}{30 \mathrm{~min}}=\sqrt{\frac{32}{2}}=4\)
∴ T2 = 30 min
Question 7. Four gases P, Q, R, and S have almost the same values but their ‘a! values (a, b are van der Waals constants) are in the order Q<R<S<P. At particular temperatures, among the four gases, the most easily liquefiable one is
- P
- Q
- R
- S
Answer: 1. P
The van der Waals constant ‘a’ of a gas is a measure of the intermolecular forces of attraction in the gas. The larger the value of ‘a’, the stronger the intermolecular forces of attraction. Now, a gas with strong intermolecular forces of attraction can easily be liquefied. So, the most easily liquefiable gas is P.
Question 8. Units of surface tension and viscosity are—
- \(\mathrm{kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}, \mathrm{~N} \cdot \mathrm{m}^{-1}\)
- \(\mathrm{kg} \cdot \mathrm{s}^{-2}, \mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}\)
- \(\mathrm{N} \cdot \mathrm{m}^{-1}, \mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-2}\)
- \(\mathrm{kg} \cdot \mathrm{s}^{-1}, \mathrm{~kg} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}^{-1}\)
Answer: 2. \(\mathrm{kg} \cdot \mathrm{s}^{-2}, \mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}\)
⇒ \(\text { Surface tension }=\frac{\text { Force }}{\text { Length }}=\frac{\mathrm{N}}{\mathrm{m}}=\frac{(\mathrm{kg} \cdot \mathrm{m} \cdot \mathrm{s})^{-2}}{\mathrm{~m}}=\mathrm{kg} \cdot \mathrm{s}^{-2} \text {. }\)
⇒ \(\begin{aligned}
\text { Coefficient of viscosity }=\mathrm{N} \cdot \mathrm{m}^{-2} \cdot \mathrm{s} & =\mathrm{kg} \cdot \mathrm{m} \mathrm{s}^{-2} \cdot \mathrm{m}^{-2} \cdot \mathrm{s} \\
& =\mathrm{kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}
\end{aligned}\)
Question 9. A gas can be liquefied at temperature T and pressure P if-
- T= TC, P<PC
- T<TC,P>PC
- T>TC,P>PC
- T>TC,P<PC
Answer: 2. Two important conditions for liquefying a gas are— temperature should be lower than critical temperature (T<TC) and pressure should be greater than critical pressure (P > Pc).
Question 10. The rms velocity of CO2 gas molecules at 27°C is approximately 1000 m/s. For N2 molecules at 600K the rms velocity approximately
- 2000 m/s
- 1414 m/s
- 1000 m/s
- 1500 m/s
Answer: 3. 1000 m/s \(c_{r m s}=\sqrt{\frac{3 R T}{M}} \text {, so, } \frac{c_{r m s}(\mathrm{CO})}{c_{r m s}\left(\mathrm{~N}_2\right)}=\sqrt{\frac{3 R T_{300}}{M_{\mathrm{CO}}} \times \frac{M_{N_2}}{3 R T_{600}}}\)
Question 11. Among tire following which should have the highest rms speed at the same temperature—
- SO2
- CO2
- O2
- H2
Answer: 4. \(c_{r m s}=\sqrt{\frac{3 R T}{M}} \quad \text { or, } c_{r m s} \propto \frac{1}{\sqrt{M}}\)
Question 12. Which of the following has the dimension of ML°T-2 —
- Coefficient of viscosity
- Surface tension
- Vapour pressure
- Kinetic energy
Answer: 2. Surface tension \(\gamma=\frac{\text { force }}{\text { length }}\)
⇒ \(\text { or, } \gamma=\frac{\text { mass } \times \text { acceleration }}{\text { length }}\)
⇒ \(=\frac{\mathrm{M} \times \mathrm{LT}^{-2}}{\mathrm{~L}}=\mathrm{ML}^0 \mathrm{~T}^{-2}\)
Question 13. For the same mass of two different ideal gases molecular weights M1 and M2, plots of logV vs logP at a given constant temperature are shown. Identify the correct option-
- M1>M2
- M1=M2
- M1<M2
- Can be predicted only if the temperature is known
Answer: 1. M1>M2 From the ideal gas equation
- \(P V=n R T=\frac{W}{M} R T\)
- \(\text { or, } \left.P V=\frac{k}{M} \text { (where, } k=W R T\right)\)
- \(\text { or, } \log P+\log V=\log \frac{k}{M}\)
- \(\text { or, } \log V=-\log P+\log \frac{\kappa}{M}(y=m x+c)\)
According to the intercepts in the graph \(\log \frac{k}{M_2}>\log \frac{k}{M_1}\) \(\text { or, } \frac{k}{M_2}>\frac{k}{M_1} \quad \text { or, } M_1>M_2\)
Question 14. Equal weights of ethane and hydrogen are mixed in an empty container at 25°C. The fraction of total pressure exerted by hydrogen is—
- 1: 2
- 1:1
- 1:16
- 15:16
Answer: 4. 15:16
Let, wc2H6 = wH2 = w
⇒ \(n_{\mathrm{C}_2 \mathrm{H}_6}=\frac{w}{30} \text { and } n_{\mathrm{H}_2}=\frac{w}{2}\)
∴ \(n_{\mathrm{C}_2 \mathrm{H}_6}=\frac{w}{30} \text {. and } n_{\mathrm{H}_2}=\frac{w}{2}\)
∴ \(x_{\mathrm{C}_2 \mathrm{H}_6}=\frac{n_{\mathrm{C}_2 \mathrm{H}_6}}{n_{\mathrm{C}_2 \mathrm{H}_6}+n_{\mathrm{H}_2}}=\frac{\frac{1}{30}}{\frac{1}{30}+\frac{1}{2}}=\frac{1}{16}\)
[n= number of moles, x = mole fraction]
Similarly \(x_{\mathrm{H}_2}=\frac{15}{16}\)
According to Dalton’s law of partial pressure \(\begin{aligned}
p_{\mathrm{H}_2} & =x_{\mathrm{H}_2} \times P \quad[P=\text { total pressure }] \\
\text { or, } \frac{P_{\mathrm{H}_2}}{P} & =x_{\mathrm{H}_2}=\frac{15}{16}
\end{aligned}\)
Question 15. Compressibility factor for a real gas at high pressure—
- 1
- \(1+\frac{P b}{R T}\)
- \(1-\frac{P b}{R T}\)
- \(1+\frac{R T}{P b}\)
Answer: 2. \(1+\frac{P b}{R T}\)
van der Waals equation \(\left(P+\frac{a}{V^2}\right)(V-b)=R T\)
At high pressure \(P \gg \frac{a}{V^2}\) hence, P(V-b) = RT
⇒ \(\text { or, } P V=R T+P b \text { or, } \frac{P V}{R T}=1+\frac{P b}{R T} \text { or, } Z=1+\frac{P b}{R T}\)
Question 16. ‘a’ and ‘b’ are van der Waals constant for gases. Chlorine is more easily liquefied them ethane because—
- A and B for CL2 < A and B for C2H6
- A for CL2 < A for C2H6 but B for CL2 > B for C2H6
- A for CL2 > A for C2H6 but B for CL2< B for C2H6
- A and B for CL2 > A and B for C2H6
Answer: 3. A for CL2 > A for C2H6 but B for CL2 < B for C2H6
Van der Waals constants ‘a’ is a measure of intermolecular forces of attraction of a gas whereas ‘b’ is a measure of the size of gas molecules. Hence, more is the value of a more easily the gas will be liquefied.
Question 17. For the gaseous state, if the most probable speed is denoted by c, average speed by c, and mean square speed by c, then for a large number of molecules the ratios ofthese speeds are
- c*:c:c = 1.225: 1.128: 1
- c*:c:c = 1.128: 1.225: 1
- c*:c:c = 1:1.128:1.225
- c*:c:c = 1:1.225:1.128
Answer: 4. c*:c:c = 1:1.225:1.128
Question 18. If Z is a compressibility factor, the van der Waals equation at low pressure can be written as
- \(Z=1+\frac{P b}{R T}\)
- \(Z=1+\frac{R T}{P b}\)
- \(Z=1-\frac{a}{R T V}\)
- \(Z=1-\frac{P b}{R T}\)
Answer: 3. \(Z=1-\frac{a}{R T V}\)
van der Waals constants ‘a’ is a measure of intermolecular forces of attraction of a gas whereas ‘b’ is a measure of the size of gas molecules. Hence, more is the value of a more easily the gas will be liquefied.
Question 19. The ratio of masses of oxygen and nitrogen in a gaseous mixture is 1:4 ratio of the number of their molecule is
- 3:16
- 1:4
- 7:32
- 1:8
Answer: 4. 1:8
Question 20. Intermolecular interaction that is dependent on the inverse cube of the distance between the molecules—
- London force
- Hydrogen bond
- ion ion interaction
- ion-dipole interaction
Answer: 1. London force
1, 3, and 4, are not applicable as the interaction is intermolecular. 2 is not the correct choice as the hydrogen bond does not follow the relation mentioned above.
Question 21. Two closed bulbs of equal volume ( V) containing an Ideal gas Initially at pressure pt and temperature T1 are connected through a narrow tube of negligible volume. The temperature of one of the bulbs is then raised to 7 2. The final pressure p1 Is-
- \(p_1\left(\frac{T_1 T_2}{T_1+T_2}\right)\)
- \(2 p_i\left(\frac{T_1}{T_1+T_2}\right)\)
- \(2 p_1\left(\frac{T_2}{T_1+T_2}\right)\)
- \(2 p_i\left(\frac{T_1 T_2}{T_1+T_2}\right)\)
Answer: 3. \(2 p_1\left(\frac{T_2}{T_1+T_2}\right)\)
n1+n2=n1‘+n’2
∴ \(\frac{p_i V}{R T_1}+\frac{p_i V}{R T_1}=\frac{p_f V}{R T_1}+\frac{p_f V}{R T_2}\)
Question 22. A gas mixture was prepared by taking equal moles of CO and N2. If the total pressure of the mixture was 1 atm, the partial pressure of N2 in the mixture is—
- 0.5 atm
- 0.8 atm
- 0.9 atm
- 1 atm
Answer: 1. 0.5 atm
In the mixture \(x_{\mathrm{N}_2}=\frac{1}{2} \text { and } x_{\mathrm{CO}}=\frac{1}{2}\)
∴ \(p_{\mathrm{N}_2}=x_{\mathrm{N}_2} \times P=\frac{1}{2} \times 1 \mathrm{~atm}=0.5 \mathrm{~atm}\)
Question 23. Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. Molar mass ofA is 49u. The molecular mass will be—
- 50.00u
- 12.25u
- 6.50u
- 25.00u
Answer: 2. 12.25u
⇒ \(\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}} \text { or, } \frac{V / 20}{V / 10}=\sqrt{\frac{M_B}{49}} \text { or, } \frac{1}{2}=\sqrt{\frac{M_B}{49}}\)
∴ Mb = 12.25u
Question 24. By what factors does the average velocity of a gas molecule increase when the temperature (inK) is doubled
- 2.0
- 2.8
- 4.0
- 1.4
Answer: 4. 1.4
⇒ \(\bar{c}=\sqrt{\frac{8 R T}{\pi M}}\)
When Pis doubled, \(\bar{c}_1=\sqrt{\frac{8 R \times 2 T}{\pi M}}=\sqrt{2} \sqrt{\frac{8 R}{\pi M}}\)
∴ MB= 12.25u
Question 25. 50mL of each gas A and B takes 150s and 200s respectively for effusing through a pinhole under similar conditions. If the molar mass of gas B is 36, the molar mass of gas A—
- 20.25
- 64
- 96
- 128
Answer: 1. 20.25
⇒ \(\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}} \text { or, } \frac{V_A}{t_A} \times \frac{t_B}{V_B}=\sqrt{\frac{M_B}{M_A}}\)
⇒ \(\text { or, } \frac{50}{150} \times \frac{200}{50}=\sqrt{\frac{36}{M_A}} \text { or, } \frac{4}{3}=\sqrt{\frac{36}{M_A}} \text { or, } \frac{16}{9}=\frac{36}{M_A}\)
∴ \(M_A=\frac{36 \times 9}{16}=20.25\)
Question 26. Set-1: O2, CO2, H2 and He, Set-2: CH4, O2 and H2. The gases in set-I in increasing order of ’b’ and gases given in set-II in decreasing order of ‘a’ are arranged below here ‘a’ and ‘b’ are van der Waals constants. Select the Correct order from the following—
- O2 < He < H2 < CO2 ; H2 >O2 > CH2
- H2 < He < O2 < CO2; CH4 >O2 > H2
- H2 < O2< He < CO2 ; O2 > CH4 > H2
- He < H2 < CO2 < O2 ; CH4 > H2 > O2
Answer: 2. H2 < He < O2 < CO2 ; CH4 > O2 > H2
A gas with strong intermolecular forces of attraction has a large value of a and a gas has a large value of ‘b’ if its molecules are big. The increasing order of sizes of H2, He, O2, and CO2 molecules is H2 < He <O2 < CO2. So, the increasing order of values for these gases is H2 < He < O2 < CO2. CH4, O2, and H2 are all non-polar molecules.
The only intermolecular forces of attraction that act in CH4, O2, and H2 gases are London forces. The strength of London forces increases with molecular size.
So, the increasing order of intermolecular forces in CH4, O2, and H2 gases will be CH4 > O2 > H2. Again, the stronger the intermolecular forces of attraction in a gas, the larger the value of the gas. Therefore, the decreasing order of ‘b’ values for these gases will be CH4 > O2 > H2.
Question 27. A certain gas takes three times as long to effuse out ns helium. Its molecular mass will be
- 36u
- 64u
- 9u
- 27u
Answer: 1. 36u
⇒ \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}} \text { or, } \frac{v / t_1}{v / t_2}=\sqrt{\frac{M_2}{M_1}} \text { or, } \frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}\)
or, \(\frac{3}{1}=\sqrt{\frac{M_2}{4 \mathrm{u}}} \text { or, } M_2=36 \mathrm{u}\)
[t1 and t2 are the times for the diffusion of VmLHe and VmL unknown gas respectively. Mj = molar mass ofHe, M2 = molar mass of unknown gas]
Question 28. Maximum deviation from Ideal gas is expected in case of—
- CH4(g)
- NH3(g)
- H2(g)
- N2(g)
Answer: 2. NH3(g) CH4, H2 and N2 are non-polar molecules. Only intermolecular forces that operate in CH4, H2, and N2 gases are weak London forces. As NH3 is a polar molecule, besides weak London forces, relatively stronger dipole-dipole attractive forces also act among the molecules in NH3 gas. So, among the given gases, intermolecular forces of attraction will be strongest in NH3, and hence it will show maximum deviation from ideal behaviour.
Question 29. Equal masses of H2, O2, and methane have been taken in a container of volume V at a temperature of 27°C in identical conditions. The ratio of the volumes of gases H2: O2: methane would be –
- 8:16:1
- 16:8:1
- 16:1:2
- 8:1:2
Answer: 3. 16:1:2
Let the mass of each of the gases be wg. In the mixture \(n_{\mathrm{H}_2}=\frac{w}{2} \mathrm{~mol}, n_{\mathrm{O}_2}=\frac{w}{32} \mathrm{~mol} \text { and } n_{\mathrm{CH}_4}=\frac{w}{16} \mathrm{~mol} \text {. }\)
Total number of moles in the mixture \(=\frac{w}{2}+\frac{w}{32}+\frac{w}{16}=\frac{19}{32} w\)
So,in the mixture, \(x_{\mathrm{H}_2}=\frac{w / 2}{19 w / 32}=\frac{16}{19}, x_{\mathrm{O}_2}=\frac{1}{19}\) and \(x_{\mathrm{CH}_4}=\frac{2}{19} .\)
The volume fraction of a component in the mixture = mole fraction ofthe component x total volume ofthe mixture
∴ \(V_{\mathrm{H}_2}=\frac{16}{19} \times V, V_{\mathrm{O}_2}=\frac{1}{19} \times V \text { and } V_{\mathrm{CH}_4}=\frac{2}{19} \times V\)
∴ \(V_{\mathrm{H}_2}: V_{\mathrm{O}_2}: V_{\mathrm{CH}_4}=16: 1: 2\)
Question 30. A gas such as carbon monoxide would be most likely to obey the ideal gas law at—
- High temperatures and low pressures
- Low temperatures and high pressures
- High temperatures and high pressures
- Low temperatures and low pressures
Answer: 1. High temperatures and low pressures
At high temperatures and low pressures, real gases show ideal behaviour
Question 31. Equal moles ofhydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape—
- \(\frac{1}{2}\)
- \(\frac{1}{8}\)
- \(\frac{1}{4}\)
- \(\frac{3}{8}\)
Answer: 2. \(\frac{1}{8}\)
⇒ \(\text { } \frac{r_{\mathrm{O}_2}}{r_{\mathrm{H}_2}}=\sqrt{\frac{M_{\mathrm{H}_2}}{M_{\mathrm{O}_2}}} \text { or, } \frac{n_{\mathrm{O}_2} / t}{0.5 / t} \quad \text { or, } n_{\mathrm{O}_2}=\frac{1}{8}\)
Question 32. The correction factor ‘a’ to the ideal gas equation corresponds to—
- Forces of attraction between the gas molecules
- Density of the gas molecules
- The electric field present between the gas molecules
- The volume of the gas molecules
Answer: 1. Forces of attraction between the gas molecules
In the real gas equation \(\left(P+\frac{a n^2}{V^2}\right)(V-n b)=n R T \text {; }\) van der Waals constant ‘ a ‘ represents the intermolecular forces attraction between the molecules.
Question 33. Given van der Waals constant of NH3, H2, O2, and CO2 are respectively 4.17, 0.244, 1.36, and 3.59. Which one of the following gases is most easily liquefied—
- NH3
- H2
- O2
- CO2
Answer: 1. NH3 The gases having strong intermolecular attraction have a value of van der Waals constant Such gases can be liquefied easily. Among the given gases NH3 has the highest value of a.
Question 34. In the van der Waals equation, ‘ a ‘ signifies
- Intermolecular attraction
- Intramolecular attraction
- The attraction between molecules and walls of the container
- Volume of molecules
Answer: 1. Intermolecular attraction
In van der Waals equation, a signifies the intermolecular forces of attraction
Question 35. Arrange the following gases in order of their critical temperature: NH3, H2, CO2, O2—
- NH3 > H2O > CO2 > O2
- O2>CO2>H2O>NH3
- H2O > NH3 > CO2 > O2
- CO2 >O2 > H2O > NH3
Answer: 3. H2O > NH3 > CO2> O2
The greater the intermolecular forces of attraction, the higher the critical temperature
Question 36. The density of gas A is thrice that of a gas B at the same temperature. The molecular weight of gas B is twice that of A. What will be the ratio of the pressures acting on B and A —
- \(\frac{1}{4}\)
- \(\frac{7}{8}\)
- \(\frac{2}{5}\)
- \(\frac{1}{6}\)
Answer: 4. \(\frac{1}{6}\)
⇒ \(\frac{d}{p}=\frac{M}{R T}\)
⇒ \(\frac{1}{6}\)
Let density ofgas B be d
∴ The density of gas A = 3d and molecular weight of A be M.
∴ Molecular weight of B = 2M Since, R is gas constant and T is the same for gases, so
⇒ \(\begin{aligned}
& p_A=\frac{d_A R T}{M_A} \text { and } p_B=\frac{d_B R T}{M_B} \\
& \frac{p_B}{p_A}=\frac{d_B}{d_A} \times \frac{M_A}{M_B}=\frac{d}{3 d} \times \frac{M}{2 M}=\frac{1}{6}
\end{aligned}\)
Question 37. In van der Waals equation at constant temperature 300k, if a = 1.4 atm-L2-mol-2, V = 100 mL, n = 1 mole, then what is the pressure of the gas
- 42 atm
- 210 atm
- 500 atm
- 106 atm
Answer: 4. 106 atm
At moderate pressure, the van der Waals equation is given
as \(\text { as: } \begin{aligned}
& \left(P+\frac{a n^2}{V^2}\right)(V)=n R T \\
& \left(P+\frac{1.4}{(0.1)^2}\right)(0.1)=1 \times 0.082 \times 300
\end{aligned}\)
or, (P+ 140) X 0.1 = 24.6 or, 0.1P+ 14 = 24.6
or, 0.1 P = 10.6 or, P = 106 atm
Question 38. When 1 g of gas A at 4 bar pressure is added to 2 g of gas B, the total pressure inside the container becomes 6 bar. Which of the following is true
- MA= 2MB
- MB=2MA
- MA=4MB
- MB=4MA
Answer: 4. MB=4MA
⇒ \(\frac{n_1}{p_1}=\frac{n_2}{p_2}\)
∴ \(\frac{\frac{1}{M_A}}{4}=\frac{\frac{1}{M_A}+\frac{2}{M_B}}{6}\)
⇒ \(\text { or, } \frac{6}{4 M_A}-\frac{1}{M_A}=\frac{2}{M_B}\)
⇒ \(\text { or, } \frac{6-4}{4 M_A}=\frac{2}{M_B} \quad \text { or, } \frac{1}{4 M_A}=\frac{1}{M_B} \quad \text { or, } M_B=4 M_A\)
Question 39. Gas in a cylinder is maintained at 10 atm pressure and 300 K temperature. The cylinder will explode if the pressure of the gas goes beyond 15 atm. What is the maximum temperature to which gas can be heated
- 400k
- 500k
- 450k
- 250k
Answer: 3. 450k
⇒ \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)
∴ \(\frac{10}{300}=\frac{15}{T_2} \quad \text { or, } T_2=450 \mathrm{~K}\)
Question 40. Two separate bulbs contain gas A and gas B. The density of gas A is twice that of B. The molecular mass of A is half that of B. If temperature is constant, the ratio of the pressure of A and B is-
- 1:1
- 1:2
- 4:1
- 2:1
Answer: 3. 4:1
⇒ \(d=\frac{P M}{R T}\)
⇒ \(\begin{aligned}
& \text { Given, } \frac{d_A}{d_B}=2, \frac{M_A}{M_B}=\frac{1}{2} \\
& \frac{d_A}{d_B}=\frac{P_A M_A}{R T} \times \frac{R T}{P_B M_B}=2 \\
&
\end{aligned}\)
⇒ \(\begin{aligned}
& \text { or, } \frac{P_A}{P_B} \times \frac{M_A}{M_B}=2 \\
& \text { or, } \frac{P_A}{P_B} \times \frac{1}{2}=2 \\
& \text { or, } \frac{P_A}{P_B}=4: 1
\end{aligned}\)
Question 41. Which of the following does not change during compression of a gas at a constant temperature—
- Density of a gas
- Distance between molecules
- The average speed of molecules
- The number of collisions
Answer: 3. Average speed of molecules
Question 42. For which of the following gaseous mixtures, Dalton’s law of partial pressure is not applicable—
- SO2, HE, NE
- NH3, HBr, HC1
- O2,N2,CO2
- N2,H2,O2
Answer: 2. NH3, HBr, HC1
Question 43. The volume of a given mass of an ideal gas is VL at 27°C and 1 atm pressure. If the volume of the gas is reduced by 80% at constant pressure, the temperature of the gas will have to be—
- -50°C
- -127°C
- -200°C
- -213°C
Answer: 4. -213°C
Question 44. AT STP, the density of air is 1.3 x 10¯³g.cm¯³. The vapour density of air is—
- 1.3
- 14.6
- 2.56
- 10.8
Answer: 2. 14.6
Question 45. At a given temperature; the molar concentration of N2 is greater than that of H2 in a mixture of N2 and H2 gases present in a closed container. If the average kinetic energies of N2 and H2 molecules are xj and yj respectively, then—
- x>y
- x<y
- x = y
- Impossible To Predict
Answer: 3. Impossible To Predict
Question 46. The density of gas A is dA at a temperature of TAK, and the density of gas B is dB at a temperature of TBK. The molar mass of A is 4 times that of B. If TA : TB = 2:1 and dA : dB = 1:2, the ratio of pressures of A to B is—
- 2:1
- 1:8
- 3:2
- 1:4
Answer: 4. 1:4
Question 47. Two gases A and B have respective van der Waals constants a2, bx and a2, b2. If ‘ A ’ is more compressible than ‘B,’ then which of the following conditions has to be satisfied—
- a1 = a2 and b1> b2
- a1 < a2 and b1> b2
- a1 < a2 and b1 = b2
- a1 > a2and b1 < b2
Answer: 4. a1 > a2and b1 < b2
Question 48. The dimension of the coefficient of viscosity—
- MLT
- ML-1T-1
- MLT-1
- MLT-2
Answer: 2. ML-1T-1
Question 49. The densities of water and water vapour are 1.0 g.cm¯³ and
0.0006 g.cm¯³ respectively at 100°C and 1 atm pressure. At this temperature, the total volume occupied by water molecules in 1L of water vapour is—
- 2.24 cc
- 0.6 cc
- 0.12 cc
- 1.72 cc
Answer: 2. 0.6 cc
Question 50. The most probable velocities of the molecules of gas A (molar mass 16 g.mol¯¹) and that of the molecules of gas B (molar mass 28 g.mol¯¹) are the same. If the absolute temperatures of the gases A and B are T(A) and T{B) respectively, then—
- T{A) = 2T{B)
- T(B) = 3T(A)
- T(B) = 1.75 T (A)
- T(B) = 2.5 T (A)
Answer: 3. T(B) = 1.75 T (A)
Question 51. At a given temperature and pressure, the volume of 1 mol of an ideal gas is 10L. At the same temperature and pressure, the volume of 1 mol of a real gas is VL. At this temperature and pressure, if the compressibility factor of the real gas is greater than 1, then—
- V- 10L
- V< 10L
- V> 10L
- V< 10L
Answer: 3. V> 10L
Question 52. The pressure of a gas increases when its temperature is increased at constant volume. This is because with an increase in temperature—
- The collision frequency of the gas molecules increases.
- Motions of the gas molecules become more random
- Gas molecules make more collisions with the walls of the container
- The compressibility factor of the gas increases
Answer: 3. Gas molecules make more collisions with the walls of the container
Question 53. Under given conditions, the rate of diffusion of CH4 gas is times that of f2 gas. Gas 2 reacts with element A to form gaseous compounds AB2 and AB3. Under a given condition, the rate of diffusion of AB2 is 1.12 times that of AB3 The atomic mass of A (in g-mol-1) is—
- 32
- 16
- 8
- 24
Answer: 1. 32
Question 54. Two flasks are connected by a valve: One of them with volume 5L contains 0.1 mol of H2 at 27°C and the other with volume 2L contains 0.1 mol of N2 at the same temperature. If the valve is opened keeping temperature constant, then at equilibrium the contribution of H2 gas to the total pressure of the gas mixture—
- Is the same as that of n2 gas
- Is greater than that of n2 gas
- Is less than that of n2 gas
- Cannot be predicted
Answer: 1. Is the same as that of n2 gas
Question 55. A balloon filled with acetylene is pricked by a pin and dropped readily in a tank of H2 gas under identical conditions. After a while the balloon will—
- Enlarge
- Shrink completely
- Collapse remain
- Unchanged in size
Answer: 1. Enlarge
Question 56. At STP, the density of a gas is 1.25g-L-1. The molar concentration (mol-L-1) of 0.7g of this gas at 27°C and a pressure of 2 atm is—
- 0.27
- 0.08
- 0.19
- 0.64
Answer: 2. 0.08
Question 57. 100 persons are sitting at equal distances in a row XY. Laughing gas (N2O) is released from side X and tear gas (mol. mass = 176) from side Y at the same moment and the same pressure. The person who will tend to laugh and weep simultaneously is—
- 34th from side X
- 67th from side X
- 76th from side X
- 67th from side Y
Answer: 2. 67th from side X
Question 58. van der Waals constant, b of a gas is 4.42 centilitre – mol 1. How near can the centres of 2 molecules approach each other—
- 127.2pm
- 427.2pm
- 327.2pm
- 627.2pm
Answer: 3. 327.2pm
Question 59. Which of the following liquids has the least surface tension—
- Acetic acid
- Diethyl ether
- Chlorobenzene
- Benzene
Answer: 2. Diethyl ether
Question 60. At P atm pressure and TK, a spherical air bubble is rising from the depth of a lake. When it comes to the surface of the lake the percentage increase in the radius will be (assume pressure and temperature at the surface to be PI4 atm and 27TC respectively)—
- 100%
- 50%
- 40%
- 200%
Answer: 1. 100%
Question 61. A given mass of a perfect gas is first heated in a small and then in a large vessel, such that their volumes remain unaltered. The P- T curves are—
- Parabolic with the same curvature
- Linear with the same slope
- Linear with different slopes
- Parabolic with different curvatures
Answer: 3. Linear with different slopes
Question 62. At a given temperature, most of the molecules in a sample of oxygen gas move with a velocity of 4.08 x 104 cm-s-1. The average velocity of the molecules of the gas at the same temperature is—
- 1.7 x 104 cm.s¯¹
- 4.6 x 104 cm.s¯¹
- 5.0 x 104 cm.s¯¹
- 8.9 x 103 cm.s¯¹
Answer: 2. 4.6 x 104 cm.s¯¹
Question 63. There is a depression in the surface of the liquid inside a capillary tube when—
- The cohesive force is greater than
- The adhesive force the adhesive force is greater than
- The cohesive force both adhesive and cohesive forces are equal
- None of the above is true
Answer: 1. The cohesive force is greater than
Question 64. One mol of a real gas following the equation, P(V-b) = RT, has a compressibility factor of 1.2 at 0°C and 200 atm pressure. The value of ‘b’ for this gas is—
- 0.03521 L-mol-1
- 0.0224 L-mol-1
- 0.04610 L-mol-1
- 0.01270 L-mol-1
Answer: 2. 0.0224 L-mol-1
Question 65. At a given temperature, the root mean square velocity of O2 molecules is times that of the molecules of a gas. The molar mass of the gas (in g-mol-1) is—
- 8
- 64
- 96
- 16
Answer: 2. 64
Question 66. At a given condition, 20L of SO2 gas takes 60 for its effusion. At the same condition, the volume of 09 gas that will effuse out in 30 seconds is—
- 12.4L
- 10.9L
- 14.1L
- 6.8L
Answer: 3. 14.1L
Question 67. The average velocity of the molecules of a gas at T1K will be the same as the most probable velocity of the molecules of the gas at T2K when—
- T1 > r2
- t2 >T1
- t1=t2
- t1> r2
Answer: 2. t2 >T1
Question 68. Two ideal gases A and B have molar masses MA and MB g-mol-1 respectively. Volumes of the same mass of A and B are the same, and the rms velocity of A molecules is twice that of the molecules of B. If MB: MA = 2:1, then the ratio of the pressures of A to B is—
- 4:1
- 8:1
- 2:1
- 1:6
Answer: 1. 4:1
Question 69. Containing gas molecules, the percentage of molecules moving with velocities 2x104cm*s-1 and l x 10-1 cut-s.1 are 30% and 45% respectively, and the rest one moving with velocity 5 x 104 cm-s.1. The root mean square velocity of the molecules is—
- 3.7 X 104cm.s-1
- 1.8 x 104 cm.s-1
- 6.2 X 103 cms.-1
- 2.8 x 104 cms-1
Answer: 4. 2.8 x 104 cms-1
Question 70. An open vessel has a temperature of TK. When the vessel is heated at 477°C, three-fifths of the air in the vessel escapes out. What fraction of air in the vessel would have been expelled out if the vessel were heated at 900K (assume that the volume of the vessel remains unchanged on heating)—
- 4
- 3
- 2
- 5
Answer: 2. 2
Question 71. Critical temperatures of the gases A, B, C and D are 126K, 155K, 304K and 356K respectively. Among these gases, the one with the strongest intermolecular forces of attraction is—
- A
- B
- C
- D
Answer: 4. D
Question 72. The volumes of two gases A and B at 0°C and 200 atm pressure are 0.112L and 0.09L respectively. Which of the following comments is true for these gases at this temperature and pressure—
- The compressibility of gases a and b are the same
- The compressibility of a is less than that of b
- The compressibility of a is more than that of b
- Both gases show positive deviation from ideality
Answer: 3. Compressibility of a is more than that of b
Question 73. Which of the following correctly represents the relation between capillary rise (h) and radius of the capillary (r) —
Answer: 2.
Question 74. For CO2 gas the P vs V isotherms at temperatures above 31.1°C are—
- Straight line
- Rectangular hyperbolic
- Elliptical
- Hyperbolic
Answer: 2. Rectangular hyperbolic
Question 75. At a certain temperature, lmol of chlorine gas at 1.2 atm takes 40 sec to diffuse while 1 mol of its oxide at 2 atm takes 26.5 sec. The oxide is—
- C12O
- C1O2
- Cl2O6
- Cl2O7
Answer: 1. C12O
Question 76. At 10 bar pressure, a 4:1 mixture of He and CH4 is contained in a vessel. The gas mixture leaks out through a hole present in the vessel. The mixture effusing out has an initial composition of—
- 1:1
- 2:1
- 4:1
- 8:1
Answer: 4. 8:1
Question 77. A gas mixture consisting of N2 and 3 mol of O2 had a pressure of 2 atm at 0 °C. Keeping the volume and the temperature of the mixture constant, some amount of O2 was removed from the mixture. As a result, the total pressure of the mixture and the partial pressure of N2 in the mixture became 1.5 atm and 0.5 atm respectively. The amount of oxygen gas removed was—
- 8g
- 16g
- 32g
- 64g
Answer: 3. 32g
Question 78. The quantity — represents—
- Mass of a gas
- Translation energy of a gas
- Number of moles of a gas
- Number of molecules in a gas
Answer: 4. Number of molecules in a gas
Question 79. At STP, O2 gas present in a flask was replaced by SO2 under similar conditions. The mass of SO2 present in the flask will be—
- Twice that of O2
- Half that of O2
- Equal to that of O2
- One-third of O2
Answer: 1. Twice that of O2
Question 80. The relative densities of oxygen and carbon dioxide are 16 and 22 respectively. If 37.5cm³ of oxygen effuses out in 96s, what volume of carbon dioxide will effuse out in 75s under similar conditions—
- 25cm³
- 37.5cm³
- 14cm³
- 30.8cm³
Answer: 1. 25cm³
Question 81. At 27°C, the average translational kinetic energies of the molecules in 8g of CH4,8g of 02 and 8g of He are, e2 and e2 respectively and the total kinetic energies of the molecules in these gases are E1, E2 and E3 respectively. Which of the following is true—
- \(\bar{\epsilon}_1=\bar{\epsilon}_2=\bar{\epsilon}_3\)
- \(\bar{\epsilon}_3=\bar{\epsilon}_2=\bar{\epsilon}_1\)
- E1 = E2 = E3
- E2<E1<E3
Answer: 1. \(\bar{\epsilon}_1=\bar{\epsilon}_2=\bar{\epsilon}_3\)
Question 82. Several molecules of an ideal gas present in a flask of volume 2L are 1023. The mass of each gas molecule is 6.64 x 10-23 g and the root mean square velocity of the molecules is 4.33 x 104 cm-s-1. Hence—
- The pressure of the gas is 3.27 atm
- The average kinetic energy of each molecule is 6.23 x ltr14J
- The total kinetic energy of the molecules is 6.23 x 109J
- The total kinetic energy of the molecules is 1.492 x 109 J
Answer: 2. Average kinetic energy of each molecule is 6.23 x ltr14J
Question 83. In which conditions does the most probable velocity of O2 molecules have maximum value and in which conditions does it have minimum value—
- O2 : P = 1 atm, d (density) = 0.0081 g mL-1
- O2 : P = 4 atm, V = 2L and w (mass) = 4g
- O2 : r=300K
- O2: STP
Answer: 1. O2 : P = 1 atm, d (density) = 0.0081 g mL-1
Question 84. The time required to effuse V mL of H2 gas through a porous wall at a constant temperature and pressure is 20 min. Under the same conditions time required to effuse V mL of the following gases is—
- He:28.28min
- CO2:90.82min
- CH4:60.52 min
- N2:74.83 min
Answer: 1. He:28.28min
Question 85. At a particular temperature and pressure, if the number of moles of an ideal gas is increased by 50%, then—
- The final volume of the gas will be 1.5 times its initial volume
- The most probable velocity of gas molecules becomes 1.5 times its initial value
- The total kinetic energy of the gas molecules becomes 1.5 times its initial value
- The density of gas becomes 1.5 times its initial value.
Answer: 1. Final volume of the gas will be 1.5 times its initial volume
Question 86. The pressure and temperature of a gas are P and T respectively. If the critical pressure and critical temperature of the gas are Pc and Tc respectively, then liquefaction will be possible when—
- P<PC,T<TC
- P = PC,T=TC
- P = PC,T>TC
- P>PC,T=TC
Answer: 2. P = PC,T=TC
Question 87. If the orders of the values of van der Waals constants a and b for three gases X, Y and Z are X < Y < Z and Z < Y < X respectively, then—
- Liquefaction will be easier for gas than gases and z.
- The size of the molecule, y will be in between the sizes of the molecules x and z.
- The order of the critical temperatures of these three gases is: x< y<z.
- The gas, z, at 0°c and 1 atm will behave most ideally.
Answer: 2. The size of the molecule, y will be in between the sizes of the molecules x and z.
Question 88. Identify the correct statements—
- At a particular temperature, the vapour pressure of dimethyl ether is greater than water because the molar mass of dimethyl ether is higher than that of water.
- The vapour pressure of a liquid remains the same when the surface area of the liquid is increased at a given temperature.
- The correct order of viscosity coefficient is ethylene glycol < glycerol.
- The surface tension of water at 30°c is greater than that at 20°c.
Answer: 2. The vapour pressure of a liquid remains the same when the surface area of the liquid is increased at a given temperature.
Question 89. P(V-b) = RT equation of state is obeyed by a particular gas. Which of the given statements is correct—
- For this gas, the isochoric curves have slope = \(=\frac{R}{V-b}\)
- The compressibility factor of the gas is less than unity.
- For this gas, the isobaric curves have slope = r/p
- In this gas, the attraction forces are overcome by repulsive forces.
Answer: 1. For this gas, the isochoric curves have slope = \(=\frac{R}{V-b}\)
Question 90. Four gas balloons P, Q, R and S of equal volumes containing H2, N2O, CO, and CO2 respectively were pricked with a needle and immersed in a tank containing CO2. Which of them will shrink after some time—
- P
- Q
- R
- S
Answer: 1. P
Question 91. A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have—
- Equal total energy and potential energy.
- Equal kinetic energy different total
- Energy and potential
- Energy is different from kinetic energy.
Answer: 2. Equal kinetic energy different total
Question 92. The root mean square velocity of an ideal gas in a closed vessel of fixed volume is increased from 5 x 104cm-s-1 to 10 x 104cm-s-1. Which of die following statements clearly explains how the change is accomplished —
- By heating the gas, the die temperature is quadrupled.
- By heating die gas, the temperature is doubled by heating the gas,
- The pressure is quadrupled by heating the gas,
- The pressure is doubled
Answer: 1. By heating the gas, the die temperature is quadrupled.
Question 93. Which of the following pairs of gases have the same type of intermolecular force of attraction—
- Ch4, CI2
- SO2,CO2
- HC1, CHCI3
- N2,NH2
Answer: 1. Ch4, CI2
Question 94. Select the correct orders—
- Critical temperature < boyleg’s temperature < inversion temperature
- Van der waals constant ‘a’: H2O> nh3 > N2 > ne
- Van der waals constant ‘b’: CH4> O2 >H2
- Mean free path: he > H2 >O2 > N2 > CO2
- All the above
Answer: 4. All the above
Question 95. Which are responsible for the liquefaction of H2 —
- Coulombic forces
- London forces
- Hydrogen bonding
- Van der Waals forces
Answer: 2. London forces
Question 96. Which of the following gases will have the same rate of effusion under identical conditions—
- CO
- N2O
- C2H4
- CO2
Answer: 2. N2O
Question 97. Select the correct statements—
- The presence of impurities invariably increases the viscosity of a liquid.
- In the presence of impurities, the viscosity of a liquid remains unaltered
- The viscosity coefficient of associated liquids is larger than that of non-associated liquids.
- Viscosity coefficients of non-associated liquids are larger than those of associated liquids.
Answer: 1. Presence of impurities invariably increases the viscosity of a liquid.
Question 98. Select the correct statements—
- Surface energy of a liquid = \(=\frac{\text { force } \times \text { distance }}{\text { area }}\)
- Surface energy can be represented as force/area
- The addition of NaCl increases and the addition of acetone decreases the surface tension of water.
- The addition of NaCl decreases and the addition of acetone increases the surface tension of water.
Answer: 1. Surface energy of a liquid = \(=\frac{\text { force } \times \text { distance }}{\text { area }}\)
Question 99. Precisely lmol of He and lmol of Ne are placed in a container. Select correct statements about the system—
- Molecules of the strike the wall more frequently
- Molecules of he have greater average
- The molecular speed molecule of the two gases strikes the wall of the container with the same frequency
- He has a larger pressure
Answer: 1. Molecules of he strikes the wall more frequently
Question 100. Which of the following is correct for different gases under the same condition of pressure and temperature—
- Hydrogen diffuses 6 times faster than oxygen
- Hydrogen diffuses 2.83 times faster than methane
- Helium escapes at a rate 2 times as fast as sulphur dioxide does
- Helium escapes at a rate 2 times as fast as methane does
Answer: 2. Hydrogen diffuses 2.83 times faster than methane.