Kinetic Theory Of Gases – Principle Of Equipartition Of Energy
The principle of equipartition of energy in kinetic theory came essentially from the concept of degrees of freedom. we have seen that the pressure of an ideal gas is p = \(\frac{1}{3}\left(p_x+p_y+p_z\right)\)
The number of degrees of freedom of an ideal gas molecule = 3. Clearly, it led to a factor of 1/3.
So, we get \(p=\frac{1}{3} \rho c^2 \text { or, } c^2=\frac{3 p}{\rho}\)(c = rms speed of the gas molecules).
Again, the total kinetic energy of the molecules in 1 mol of an ideal gas is \(\frac{3}{2} p V=\frac{3}{2} R T\). So, it may be said that the molecular kinetic energy of 1 mol of an ideal gas for each degree of freedom is \(\frac{1}{2}\) RT Scientist Ludwig Boltzmann analysed the motion of a single molecule in an ideal gas and theoretically established the principle of equipartition of energy
Statement of the principle of equipartition of energy: The average molecular kinetic energy of any substance is equally shared among the degrees of freedom; the average kinetic energy of a single molecule associated with each degree of freedom is \(\frac{1}{2} k T\)(T= absolute temperature, k = Boltzmann constant = 1.38 x 10-23 J · K-1)
The number of degrees of freedom of an ideal gas molecule = 3. So, from the equipartition principle, the average kinetic energy of a molecule = 3 x \(\frac{1}{2}kT\) = \(\frac{3}{2}\)kT. The molecule has no potential energy. So, the average total energy of a molecule is e = \(\frac{3}{2}\)kT
Again, the total energy of 1 mol of an ideal gas is E = \(\frac{3}{2}\)RT.
So, the number of molecules in 1 mol of a gas = \(\frac{E}{e}\) = \(\frac{R}{k}\).
Clearly, this is the Avogadro number NA, i.e., \(N_A=\frac{R}{k} \text { or, } R=N_A k\)
This is the relation between the universal gas constant R, the Avogadro number NA, and the Boltzmann constant k. Then the total energy of 1 mol of an ideal gas is \(E=\frac{3}{2} N_A k T\)
For any amount of ideal gas containing N molecules, \(E=\frac{3}{2} N_A k T\). This relation is widely used particularly in chemistry.
It is to be noted that the equipartition principle is applicable to all substances, not only to gases. For any substance, solid, liquid, or gas, the average molecular kinetic energy associated with each degree of freedom is \(\frac{1}{2}\)kT.
For ideal gases, the molecular potential energy is zero; so it is easy to calculate the total energy. But for real gases, liquids, or solids, the potential energy calculations are not easy. However, for these substances, the kinetic energy strictly follows the principle of equipartition.
Specific heat of a gas: The first law of thermodynamics is written as
dQ = dE+ dW [E is taken for internal energy]
At constant volume, dV = 0; so dW = pdV = 0.
Then, dQ = dE.
So, the heat absorbed or released at constant volume for a temperature change dT of 1 mol of gas is dQ = CvdT.
Here, Cv = molar specific heat at constant volume.
Then, \(C_v=\frac{d Q}{d T}=\frac{d E}{d T}\)
In the case of monatomic gas: The kinetic theory assumes that the ideal gas molecules are monatomic. Actually, gases like helium, neon, and argon are monatomic. For 1 mol of such a gas, E = \(\frac{3}{2}\)RT
So, \(C_v=\frac{d E}{d T}=\frac{3}{2} R\)
The molar-specific heat at constant pressure is Cp.
∴ \(C_p-C_\nu=R \quad \text { or, } C_p=C_\nu+R=\frac{3}{2} R+R=\frac{5}{2} R\)
The ratio between the two specific heats is \(\gamma=\frac{C_p}{C_v}=\frac{5}{3}=1.67\)
This value tallies with the experimentally determined values of γ in the case of helium, neon, etc.
In Case Of diatomic gas: The molecules of gases like oxygen, nitrogen, hydrogen, etc., are diatomic. The number of degrees of freedom of a diatomic molecule = 5; so for 1 mol of such a gas, E = \(\frac{5}{2}\) RT.
Then, \(C_\nu=\frac{d E}{d T}=\frac{5}{2} R ; C_p=C_\nu+R=\frac{5}{2} R+R=\frac{7}{2} R\)
∴ \(\gamma=\frac{C_p}{C_v}=\frac{7}{5}=1.4\)
This value of γ is also supported by experiments.
In general cases: if f is the number of degrees of free-dom of an ideal gas molecule then, the energy of 1 mol of such a gas E = \(\frac{f}{2}\)RT
⇒ \(C_\nu=\frac{f_R}{2} ; \quad C_p=\frac{f_2}{2} R+R=\left(\frac{f+2}{2}\right) R\)
So, \(\gamma=\frac{C_p}{C_v}=\frac{f+2}{f}=1+\frac{2}{f}\).
Specific heats of helium and hydrogen gases: We know that, R ≈ 2 cal • mol-1 • °C-1 For helium gas,
⇒ \(C_v=\frac{3}{2} R=\frac{3}{2} \times 2=3 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)
⇒ \(C_p=\frac{5}{2} R=\frac{5}{2} \times 2=5 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)
The molecular weight of helium =4.
So the specific heats are, \(c_\nu=\frac{C_v}{4}=\frac{3}{4}=0.75 \mathrm{cal} \cdot \mathrm{g}^{-1 \cdot{ }^{\circ} \mathrm{C}^{-1}}\)
⇒ \(c_p=\frac{C_p}{4}=\frac{5}{4}=1.25 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)
For hydrogen gas,
⇒ \(C_v=\frac{5}{2} R=\frac{5}{2} \times 2=5 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)
⇒ \(C_p=\frac{7}{2} R=\frac{7}{2} \times 2=7 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)
The molecular weight of hydrogen = 2.
So, the specific heats of hydrogen gas are,
⇒ \(c_\nu=\frac{C_\nu}{2}=\frac{5}{2}=2.5 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)
and \(c_p=\frac{C_p}{2}=\frac{7}{2}=3.5 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)
It may be noted that the value of cp for helium gas and the values of both cv and cp for hydrogen gas are greater than the specific heat of water (1 cal · g-1 · C-1).
cv, cp, and γ of a gas mixture: Let n1 mol of a gas (molar specific heat at constant volume and constant pressure be \(C_{v_1} \text { and } C_{p_1}\) respectively) is mixed with n2 mol of a gas (molar specific heat at constant volume and constant pressure be \(C_{\nu_2} \text { and } C_{p_2}\) respectively) such that they do not react chemically.
Therefore, the thermal capacity of (n1 + n2) mol gas of the mixture at constant volume is \(n_1 C_{v_1}+n_2 C_{v_2}\)
Hence, at constant volume effective molar specific heat of the mixture is, \(C_v=\frac{n_1 C_{v_1}+n_2 C_{v_2}}{n_1+n_2}\)
Similarly, at constant pressure, the effective molar specific heat of the mixture is \(C_p=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}\)
∴ Ratio of the two molar specific heats of the mixure, \(\gamma=\frac{C_p}{C_v}=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1 C_{v_1}+n_2 C_{v_2}}\)
Now, if f1 and f2 be the degress of freedom of the molecules of two gases respectively then,
⇒ \(C_{v_1}=\frac{f_1}{2} R ; C_{p_1}=C_{v_1}+R=\left(\frac{f_1}{2}+1\right) R\)
and \(C_{v_2}=\frac{f_2}{2} R ; C_{p_2}=C_{v_2}+R=\left(\frac{f_2}{2}+1\right) R\)
Putting these values in equation(1), we can find γ of the gas mixure.
Kinetic Theory Of Gases – Principle Of Equipartition Of Energy Numerical Examples
Example 1. The rms speed of the molecules of an ideal gas at STP is 0.5 km · s-1. Find the density of the gas. What will be the density at 21°C if the pressure remains the same? Given, atmospheric pressure = 105 N · m-2
Solution:
rms speed, \(c=\sqrt{\frac{3 p}{\rho}}; so \rho=\frac{3 p}{c^2}\).
Here, p = \(10^5 \mathrm{~N} \cdot \mathrm{m}^{-2}\),
c = \(0.5 \mathrm{~km} \cdot \mathrm{s}^{-1}=0.5 \times 1000 \mathrm{~m} \cdot \mathrm{s}^{-1}=500 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
∴ \(\rho=\frac{3 \times 10^5}{(500)^2}=1.2 \mathrm{~kg} \cdot \mathrm{m}^{-3}\)
At constant pressure, \(c \propto \frac{1}{\sqrt{\rho}}; also c \propto \sqrt{T}. So, \rho \propto \frac{1}{T}\)
Then, \(\frac{\rho_0}{\rho_{21}}=\frac{T_{21}}{T_0}\)
or, \(\rho_{21}= \rho_0 \frac{T_0}{T_{21}}\)
(\(T_0=0^{\circ} \mathrm{C}=273 \mathrm{~K}, T_{21}=21^{\circ} \mathrm{C}=(21+273) \mathrm{K}=294 \mathrm{~K}\))
= \(1.2 \times \frac{273}{294}=1.11 \mathrm{~kg} \cdot \mathrm{m}^{-3}\).
Example 2. Find out the energy of 1 mol of a gas and its average molecular kinetic energy at 27°C. Given, R = 8.3×107 erg mol.Kl and NA = 6.02 x 1023 mol-1.
Solution:
The energy of 1 mol of the gas is
E = \(\frac{3}{2} R T=\frac{3}{2} \times\left(8.3 \times 10^7\right) \times 300\left[T=27^{\circ} \mathrm{C}=300 \mathrm{~K}\right]\)
= 3.735 x 1010 erg
Average molecular kinetic energy is
e = \(\frac{3}{2} k T=\frac{3}{2} \frac{R}{N_A} T=\frac{3}{2} \times \frac{\left(8.3 \times 10^7\right) \times 300}{6.02 \times 10^{23}}\)
= \(6.2 \times 10^{-14} \mathrm{erg} .\)
Example 3. The average kinetic energy of a molecule in a gas at STP is 5.6 x 10-14 erg. Find out the number of molecules per volume of the gas. Given, density of mercury = 13.6 g · cm-3
Solution:
Pressure of the gas, p
= \(\frac{2}{3}\) x energy of gas molecules per unit volume
= \(\frac{2}{3}\) x n x average kinetic energy of 1 molecule
where n = number of molecules per unit volume of the gas
∴ n = \(\frac{3 p}{2 \times \text { average kinetic energy of } 1 \text { molecule }}\)
= \(\frac{3 \times(76 \times 13.6 \times 980)}{2 \times\left(5.6 \times 10^{-14}\right)}=2.71 \times 10^{19} .\)
Example 4. Find out the temperature at which the rms speed of nitrogen molecules will be equal to the escape velocity from the earth’s gravity. Given, the mass of a nitrogen atom = 23.24 x 10-24 g; average radius of the earth = 6390 km; g = 980 cm · s-2; Boltzmann constant = 1.37 x 10-16 erg · °C-1.
Solution:
rms speed of a molecule = \(\sqrt{\frac{3 R T}{M}} ;\)
escape velocity = \(\sqrt{2 g R_1}\), where Rl = radius of the earth.
According to the question, \(\sqrt{\frac{3 R T}{M}}=\sqrt{2 g R_1}\)
T = \(\frac{2}{3} \frac{g M R_1}{R}\)
Now, R = Nk and M = mN
where m = mass of a nitrogen molecule
= 2 x (23.24 X 10-24) g ;
N = number of nitrogen molecules
So, T = \(\frac{2}{3} \cdot \frac{g m N R_1}{N k}=\frac{2}{3} \cdot \frac{g m R_1}{k}\)
= \(\frac{2}{3} \times \frac{980 \times\left(2 \times 23.24 \times 10^{-24}\right) \times\left(6390 \times 10^5\right)}{1.37 \times 10^{-16}}\)
= \(1.42 \times 10^5 \mathrm{~K} .\)
Example 5. Find out the temperature at which the average kinetic energy of a gas molecule will be equal to the energy gained by an electron on acceleration across a potential difference of 1 V. Given, Boltzmann constant = 1.38 x 10-23 J · K-1; charge of an electron = 1.6 x 10-19 C.
Solution:
Energy gained by the electron
= 1 eV = (1.6 x 10~19 C) x 1 V = 1.6 x 1019 J.
Average kinetic energy of a gas molecule
= \(\frac{3}{2}\)kT = \(\frac{3}{2}\) x (1.38 x 1-23) x T J
∴ 1.6 x 10-19 = \(\frac{3}{2}\) x (1.38 x 10-23) x T
or, \(T=\frac{2}{3} \times \frac{1.6 \times 10^{-19}}{1.38 \times 10^{-23}}=7729 \mathrm{~K}=7456^{\circ} \mathrm{C}\)
Example 6. Find out the molecular kinetic energy of 1 mol of oxygen gas at STP. Given, the molecular weight of oxygen = 32 density of oxygen at STP = 1.43 g · L-1 density of mercury = 13.6 g · cm-3.
Solution:
The volume of 1 mol or 32 g oxygen
= \(\frac{32}{1.43} \mathrm{~L}=\frac{32 \times 10^3}{1.43} \mathrm{~cm}^3 \text {; }\)
At STP, pressure p = 76 x 13.6 x 981 dyn · cm-2; temperature, T = 0°C = 273 K.
∴ The molecular kinetic energy of 1 mol oxygen gas at STP (diatomic gas) is
E = \(\frac{5}{2} R T=\frac{5}{2} \frac{p V}{T} T=\frac{5}{2} p V\)
= \(\frac{5}{2} \times(76 \times 13.6 \times 981) \times \frac{32 \times 10^3}{1.43}\)
= \(5.67 \times 10^{10} \mathrm{erg} .\)
Example 7. At what temperature will the rms speed of a hydrogen molecule be equal to that of an oxygen molecule at 47°C?
Solution:
The molecular weights of oxygen and hydrogen, respectively, are M1 = 32 and M2 = 2.
rms speed, c = \(\sqrt{\frac{3 R T}{M}}\)
∴ \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2} \cdot \frac{M_2}{M_1}}\)
Here, \(c_1=c_2 and T_1=47^{\circ} \mathrm{C}=320 \mathrm{~K}\)
∴ \(\frac{T_1}{T_2} \cdot \frac{M_2}{M_1}=1\)
or, \(T_2 =T_1 \cdot \frac{M_2}{M_1}=320 \times \frac{2}{32}\)
= \(20 \mathrm{~K}=(20-273)^{\circ} \mathrm{C}=-253^{\circ} \mathrm{C} .\)
Example 8. 0.76 g of a mixture of hydrogen and oxygen gases has a volume of 2 L, temperature of 300 K, and pressure of 105 N · m-2. Find out the individual masses of hydrogen and oxygen in the mixture.
Solution:
Let the number of moles of hydrogen and oxygen be n1 and n2 respectively.
Then, the pressure of the mixture
p = \(\frac{n_1 R T}{V}+\frac{n_2 R T}{V}=\left(n_1+n_2\right) \frac{R T}{V}\)
or, \(n_1+n_2=\frac{p V}{R T}=\frac{10^5 \times\left(2 \times 10^{-3}\right)}{8.3 \times 300}=0.08\)…(1)
Now, the mass of hydrogen gas = 2n1 and the mass of oxygen gas = 32n2.
So, 2n1 + 32n2 = 0.76
or, n + 16n2 = 0.38….(2)
(2)-(1)
15n2 = 0.3 or, n2 = 0.02
Then from (1), n1 = 0.08 – 0.02 = 0.06
∴ Mass of hydrogen = 2 x 0.06 = 0.12 g and mass of oxygen = 32 x 0.02 = 0.64g
Example 9. Find out the number of molecules in a gas of volume 20cm³ at a pressure of 76 cm of mercury, and at 27°C. Given, the average molecule kinetic energy at 27°C = 2 x 10-14 erg.
Solution:
The pressure of the gas,
p = \(\frac{1}{3} \rho c^2=\frac{1}{3} \frac{m N}{V} {c^2}\)
[m = mass of a molecule, N = number of molecules]
or, N = \(\frac{3 p V}{m c^2}=\frac{3 p V}{2 \times \frac{1}{2} m c^2}\)
[average molecular kinetic energy = \(\frac{1}{2}\) = 2 x 10-14 erg]
= \(\frac{3 \times(76 \times 13.6 \times 980) \times 20}{2 \times\left(2 \times 10^{-14}\right)}\)
= 1.52 x 1021
Example 10. Find the temperature at which the average kinetic energy of a gas molecule will be equal to the energy of a photon in 600Å radiation. Given, the Boltzmann constant, k = 1.38 X 10-23 J · K-1; Planck’s constant, h = 6.625 x 10-34 J · s.
Solution:
Let the required temperature = T.
Average molecular kinetic energy = \(\frac{3}{2}\)kT;
Energy of a photon = hv = \(\frac{h c}{\lambda}\)
[Here, λ = 6000Å = 6000 x 10-10 m]
According to the question, \(\frac{3}{2} k T=\frac{h c}{\lambda}\)
∴ T = \(\frac{2}{3} \frac{h c}{\lambda k}=\frac{2}{3} \times \frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\left(6000 \times 10^{-10}\right) \times\left(1.38 \times 10^{-23}\right)}\)
= 1.6 x 104 K.
Example 11. Some amount of oxygen gas contained in a vessel has a density of 1.429 kg · m-3 at STP. The temperature is increased until the pressure is doubled. Neglecting the change in volume of the vessel, find the rms speed of the oxygen molecules.
Solution:
Mass and volume of the gas remain the same; so the density also remains the same.
So, ρ =1.429 kg · m-3 = 1.429 x 10-3 g · cm-3
In the first case, rms speed of oxygen molecules,
⇒ \(c_1=\sqrt{\frac{3 p_1}{\rho}}=\sqrt{\frac{3 \times(76 \times 13.6 \times 980)}{1.429 \times 10^{-3}}}=46114 \mathrm{~cm} \cdot \mathrm{s}^{-1} .\)
In the second case, p2 = 2p1
∴ \(\frac{c_2}{c_1}=\sqrt{\frac{p_2}{p_1}}=\sqrt{2}\)
or, \(c_2=\sqrt{2} c_1=46114 \times \sqrt{2}=65215 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)
Example 12. Two ideal gases at absolute temperatures T1 and T2 are mixed with each other. If the molecular mass and the number of molecules are m1, n1, and m2, n2, respectively, find out the temperature of the mixture.
Solution:
Molecular kinetic energy of the first gas = \(n_1 \cdot \frac{3}{2} k T_1\) and that of the second gas = \(n_2 \cdot \frac{3}{2} k T_2\)
So the net energy = \(\frac{3}{2} k\left(n_1 T_1+n_2 T_2\right)\)
The number of molecules in the mixture = n1 + n2.
Let the temperature of the mixture be T.
The total molecular kinetic energy = \(\left(n_1+n_2\right) \cdot \frac{3}{2} k T\)
From the principle of energy conservation \(\left(n_1+n_2\right) \cdot \frac{3}{2} k T=\frac{3}{2} k\left(n_1 T_1+n_2 T_2\right)\)
or, T = \(\frac{n_1 T_1+n_2 T_2}{n_1+n_2}\) .
Example 13. 2 mol of a monatomic gas is mixed with 1 mol of a diatomic gas, Find out the value of γ of the mixture.
Solution:
For the monatomic gas, \(\gamma_1=1+\frac{2}{f}=1+\frac{2}{3}=\frac{5}{3}\)
where f = number of degrees of freedom
For the diatomic gas, \(\gamma_2=1+\frac{2}{f}=1+\frac{2}{5}=\frac{7}{5}\)
[f = 3 for a monatomic gas and f = 5 for a diatomic gas]
So, for the mixture, \(\gamma=\frac{n_1 \gamma_1+n_2 \gamma_2}{n_1+n_2}=\frac{2 \times \frac{5}{3}+1 \times \frac{7}{5}}{2+1}=1.58\)
Example 14. The mean free path for the collision of nitrogen molecules at STP is 6.44 x 10-6 cm. What is the mean time interval between collisions? Given, R = 8.31 x 107 erg · mol-1 · K-1; molecular mass of nitrogen = 28.
Solution:
rms speed, c = \(\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 \times\left(8.31 \times 10^7\right) \times 273}{28}} \mathrm{~cm} \cdot \mathrm{s}^{-1}\)
∴ Mean time interval between collisions = \(\frac{\text { mean free path }}{\text { rms velocity }}\)
= \(6.44 \times 10^{-6} \times \sqrt{\frac{28}{3 \times\left(8.31 \times 10^7\right) \times 273}}\)
= \(1.306 \times 10^{-10} \mathrm{~s}\)
Example 15. The mass of a hydrogen molecule is 3.32 x 10-27 kg. 1023 such molecules hit every second on a rigid wall of area 2 cm² at an angle of 45° with horizontal with a velocity of 103 m · s-1. If the molecules are reflected with the same velocity, then what is the pressure exerted on the wall?
Solution:
Suppose the molecules are incident along PO and reflected along OQ.
Here, ∠PON = ∠NOQ = 45°.
Change of momentum normal to the wall for each hydrogen molecule = 2mvcos45°.
∴ Net change of momentum in a second
= exerted force = n· 2mvcos45°
(where, n = number of molecules)
∴ Pressure exerted on the wall = \(\frac{\text { force }}{\text { area }}=\frac{n \cdot 2 m \nu \cos 45^{\circ}}{\text { area }}\)
= \(\frac{10^{23} \times 2 \times\left(3.32 \times 10^{-27}\right) \times 10^3 \times \frac{1}{\sqrt{2}}}{2 \times 10^{-4}}\)
= \(2.35 \times 10^3 \mathrm{~N} \cdot \mathrm{m}^{-2}\)
Example 16. 22 g of CO2 gas at 27°C is mixed with 16g of O2 gas at 37° C. What will be the temperature of the mixture?
Solution:
22 g CO2= \(\frac{22}{44}\) or \(\frac{1}{2}\) mol of CO2
16 g O2 = \(\frac{16}{32}\) or \(\frac{1}{2}\) mol O2
Temperature of the mixture, T = \(\frac{n_1 T_1+n_2 T_2}{n_1+n_2}=\frac{\frac{1}{2}(27+273)+\frac{1}{2}(37+273)}{\frac{1}{2}+\frac{1}{2}}\)
= \(305 \mathrm{~K}=32^{\circ} \mathrm{C}\)
Example 17. A mixture of 8g oxygen, 14g nitrogen, and 22g carbon dioxide is contained in a vessel of volume 4L. What will be the pressure of the gas mixture if the temperature of the mixture is 27°C? [Given R = 8.315 J · mol-1 K-1]
Solution:
We know, pV = nRT [n =number of moles]
or, \(p=\frac{n R T}{V}=\frac{g}{M} \cdot \frac{R T}{V}\)
[g = mass of the gas, M = atomic weight of the gas]
Pressure exerted by \(\mathrm{O}_2 \text { is } p_{\mathrm{O}_2}=\frac{8}{32} \cdot \frac{R T}{V}=\frac{1}{4} \frac{R T}{V}\)
Pressure exerted by N is \(p_{\mathrm{N}_2}=\frac{14}{28} \cdot \frac{R T}{V}=\frac{1}{2} \frac{R T}{V}\)
and pressure exerted by \(\mathrm{CO}_2 \text { is } p_{\mathrm{CO}_2}=\frac{22}{44} \cdot \frac{R T}{V}=\frac{1}{2} \frac{R T}{V}\)
According to Dalton’s law, the pressure of the gas mixture,
p = \(p_{\mathrm{O}_2}+p_{\mathrm{N}_2}+p_{\mathrm{CO}_2}=\frac{R T}{V}\left(\frac{1}{4}+\frac{1}{2}+\frac{1}{2}\right)=\frac{5}{4} \frac{R T}{V} \)
= \(\frac{5}{4} \times \frac{8.315 \times 300}{4 \times 10^{-3}}=7.795 \times 10^5 \mathrm{~N} / \mathrm{m}^2\)
Example 18. 1 mol of He at 57°C is mixed with 1 mol of Ar at 27°C. Find the temperature of the gas mixture.
Solution:
Average kinetic energy of 1 mol gas
= \(\frac{3}{2} R T=\frac{3}{2} k N_A T\)
Average kinetic energy of 1 mol of He =
= \(\frac{3}{2} k N_A(273+57)=\frac{3}{2} k N_A \times 330\)
Average kinetic energy of 1 mol of Ar
= \(\frac{3}{2} k N_A(273+27)=\frac{3}{2} k N_A \times 330\)
After mixing the two gases, the number of atoms in the mixture = 2 NA.
Let the temperature of the mixture is T.
Average kinetic energy of the mixture = \(\frac{3}{2} \times 2 N_A k T\)
∴ From the principle of conservation of energy,
⇒ \(\frac{3}{2} \times 2 N_A k T=\frac{3}{2} k N_A \times 330+\frac{3}{2} k N_A \times 300\)
or, T = \(\frac{1}{2}\)(330 + 300) = 315K = (315 – 273) = 42°C.
Example 19. Find the minimum radius of the planet of density 5.5 x 103 kg/m³ and temperature 427°C which can hold O2 in its atmosphere. [Given G = 6.67 x 10-11 N · m-2 · kg-2 and R = 8.3 J · mol-1 · K-1
Solution:
The escape velocity of any object at the surface of the planet of radius r and mass M1 is
v = \(\sqrt{\frac{2 G M_1}{r}}=\sqrt{\frac{2 G}{r} \cdot \frac{4}{3} \pi r^3 \rho}]\)
[where ρ is the material density of the planet]
= \(\sqrt{\frac{8}{3} G \pi r^2 \rho}\)
rms speed of a gas of molecular weight M at an absorb temperature T is \(c=\sqrt{\frac{3 R T}{M}}\)
Since the planet holds O2, hence vmin = c
∴ \(\sqrt{\frac{8}{3} G \pi r_{\min } \rho}=\sqrt{\frac{3 R T}{M}}\)
(\(r_{\min }\)= minimum radius of the planet)
or, \(\frac{8}{3} G \pi r_{\min }^2 \rho=\frac{3 R T}{M}\)
or, \(r_{\min }^2=\frac{9 R T}{8 G \pi \rho M}\)
or, \(r_{\min }=\sqrt{\frac{9 R T}{8 G \pi \rho M}}\)
= \(\sqrt{\frac{9 \times 8.3 \times(427+273) \times 7}{8 \times 6.67 \times 10^{-11} \times 22 \times 5.5 \times 10^3 \times 32 \times 10^{-3}}}\)
= \(421 \times 10^3 \mathrm{~m}=421 \mathrm{~km} .\)
Example 20. 1 mol O2 at temperature 27°C at STP (1.01 x 105 N/m²) IJS kept in a vessel. Find the number of collisions the molecules experience (in SI) per second per unit area with the wall of the vessel. [Given Boltzmann constant k = 138 x 1013 J/K]
Solution:
Number of molecules in 1 mol of 02 = 6.023 x 1023
∴ Mass of a molecule, m = \(\frac{32}{6.023 \times 10^{23} \times 1000}\)
= \(5.316 \times 10^{-26} \mathrm{~kg}\)
Momentum of the molecule, P = mv = \(m \sqrt{\frac{3 k T}{m}}=\sqrt{3 k T m}\)
Change in momentum due to each collision,
ΔP = \(2 P=2 \sqrt{3 k T m}\)
= \(2 \sqrt{3 \times 1.38 \times 10^{-23} \times 300 \times 5.316 \times 10^{-26}}\)
= \(5.139 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}\)
If the molecules experience ‘n’ number of collisions per second per square meter of the wall, then the pressure of the gas will be
p = nΔP or, 1.01 x 105 = n x 5.139 x 10-23
or, n = 1.965 x 1027.
Example 21. 0.014kg N2 gas at 27°C is kept in a dosed vessel, How much heat is required to double the rms of the N2 molecules?
Solution:
Heat received, ΔQ = nCvΔT
[Cv = molar specific heat at constant volume
Here, \(n=\frac{0.014 \times 1000}{28}=\frac{1}{2}\)
For diatomic molecule Cv = \(\frac{5}{2}\)R
and rms speed of N2 molecule, c ∝ √T
∴ To double the velocity c, the temperature should be 4 T.
∴ \(\Delta Q=\frac{1}{2} \times \frac{5}{2} R \times(4 T-T)\)
= \(\frac{5}{4} \times 2 \times(273+27) \times 3\left[because R=2 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\right]\)
= 2250 cal
Example 22. If 2 mol of a gas at constant pressure, requires 70 cal heat to increase its temperature from 30°C to 35°C, then find its degrees of freedom.
Solution:
Work done at constant pressure,
ΔW = pΔV = nRΔT =2x2x5
[ΔT = (308-303) = 5K] = 20 cal
Given that, heat received, ΔQ = nCpΔT = 70 cal.
∴ \(C_p=\frac{70}{2 \times 5}=7 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
∴ \(C_v=C_p-R=7-2=5 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
∴ \(\gamma=\frac{C_p}{C_v}=\frac{7}{5}\)
Again, \(\gamma=1+\frac{2}{f}\)
∴ 1 + \(\frac{2}{f}=\frac{7}{5}\)
or, \(\frac{2}{f}=\frac{7}{5}-1=\frac{7-5}{5}=\frac{2}{5}\)
∴ f = 5
Degree of freedom of the gas molecule is 5.
