Important Questions For Class 11 Physics Expansion Of Gases

Expansion Of Gases Long Answer Type Questions

Question 1. In case of volume expansion of a gas, mention of both the pressure and the temperature is necessary, whereas for expansion of solids and liquids, only the temperature is mentioned. Why?
Answer:

The effect of change in pressure on the volume of a solid or of a liquid is practically insignificant because of compactness of molecules. But, inter-molecular attraction in gases is weak so change in pressure significantly changes the volume of a gas.

All substances, solid, liquid or gas, expand on increase in temperature in general. Hence, the mention of pressure along with temperature is necessary in case of volume expansion of a gas.

Question 2. When a balloon is inflated both its volume and its pressure increase. Is there any violation of Boyle’s law?
Answer:

According to Boyles law, the volume of a gas varies inversely with pressure at constant temperature, only if the mass of the gas is fixed. During the inflation of a balloon, the mass of a gas is increasing. So there is no violation of Boyle’s law.

Question 3. Unlike a liquid, there is no coefficient of apparent expansion in case of a gas—why? Or, During the expansion of a liquid, volume expansion of the container is taken into account, but not for a gas—why?
Answer:

When a liquid in a container is heated, both the liquid and the container expand. The coefficient of expansion of any solid is less than that of any liquid but not negligible.

• So the expansion of the solid is not neglected. Thus in case of liquids we get two coefficients of expansion. One is the coefficient of apparent expansion and the other is the coefficient of real expansion.
• In apparent expansion the expansion of the container is ignored, whereas in real expansion, the coefficient of expansion of the container is added with the coefficient of apparent expansion.
• A mass of a gas is also heated by heating the container. This is similar to the heating of a liquid. But for the same rise in temperature, the expansion of a gas is nearly 100 times more than that of the container.
• Unless a very accurate measurement is required, expansion of the container is neglected. Hence, no apparent expansion needs to be considered in case of a gas.

Question 4. Two identical spherical bulbs contain air and are connected by a short horizontal glass tube. The tube contains a short mercury thread in it. Temperatures of the two bulbs are 0°C and 20°C respectively. If the temperature of each bulb is increased by 10°C, what will be the change in the position of the mercury thread?
Answer:

Given

Two identical spherical bulbs contain air and are connected by a short horizontal glass tube. The tube contains a short mercury thread in it. Temperatures of the two bulbs are 0°C and 20°C respectively. If the temperature of each bulb is increased by 10°C

The mercury thread shift⇒ s towards the bulb at higher temperature. As the mercury thread is in equilibrium before increasing the temperature, the initial pressure on both sides of the thread should be the same, say p.

On heating, the equilibrium is disturbed and the pressures change to p₁ and p₂ respectively for the bulbs at (0+ 10)°C or 10°C, and (20 + 10)°C or 30°C respectively.

Applying Charles’ law for the first bulb, \(\frac{p_1}{p}=\frac{T_1}{T}=\frac{273+10}{273}=\frac{283}{273}=1.037\)

and for the second bulb \(\frac{p_2}{p}=\frac{T_2}{T}=\frac{273+30}{273+20}=\frac{303}{293}=1.034 \)

⇒ \(\frac{p_1}{p_2}=\frac{283 \times 293}{273 \times 303}=1.0024\)

As p₁ is greater than p₂, the thread shifts towards the bulb at higher temperature.

Question 5. To define the coefficient of expansion of gases, the initial volume or pressure is always taken at 0°C. But for the coefficients of expansion of solids and liquids, the initial temperature need not be taken as 0°C. Why?
Answer:

The values of the coefficients of expansion of solids and liquids are very small. Hence, in this case, the volume at any temperature can be taken as the initial volume. The coefficient of volume expansion of a gas is relatively higher.

So if we consider volumes or pressures at different temperatures as the initial volume or pressure, the coefficient of expansion differs considerably. Hence, to define the coefficient of expansion of gases, the initial volume or pressure should always be taken at 0°C.

Question 6. Air pressure in a car tyre increases during driving. Explain why.
Answer:

Air pressure in a car tyre increases during driving.

Due to friction between the road and the car tyre, the temperature of air inside the tyre increases reasonably. This increases the air pressure during driving.

Thermal Expansion vs. Gas Expansion Questions

Question 7. The expansion of a gas follows the condition pV² = constant. Show that such an expansion causes cool- ingof the gas.
Answer:

Given

The expansion of a gas follows the condition pV² = constant.

Assume that the initial volume of the gas is V₁ at pressure p₁ and at temperature T₁. After the expansion, the corresponding quantities are V₂, p₂ and T₂ (say). From ideal gas equation,

⇒ \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \text { or, } \frac{p_1 V_1}{p_2 V_2}=\frac{T_1}{T_2}\)

or, \(\frac{p_1 V_1^2}{p_2 V_2^2}=\frac{T_1 V_1}{T_2 V_2}\) multiplying both sides by \(\frac{V_1}{V_2}\)

As the condition is, \(p_1 V_1^2=p_2 V_2^2\), we have

⇒ \(\frac{V_1 T_1}{V_2 T_2}=1 \text { or, } \frac{T_1}{T_2}=\frac{\dot{V}_2}{V_1}\)

As the gas expands, \(V_2>V_1\) therefore \(\frac{T_1}{T_2}>1\) or, \(T_2<T_1\).

∴ The gas cools down upon expansion.

Question 8. For a fixed mass of a gas at constant volume, draw p-t°C and p-TK graphs. How can the value of, absolute zero be obtained from the 1st graph?
Answer:

The graphs are shown in Figures respectively for 3 different volumes.

The graphs are straight lines in nature.

In the p-t°C graph, the three lines corresponding to the three volumes, converge at a point on the negative t axis, where pressure is zero. This point gives the value of absolute zero as per definition.

In the p-TK graph, the three straight lines pass through the origin, showing p = 0 when T = 0.

Question 9. For a fixed mass of a gas at constant pressure, draw V-t° C and V- T K graphs. How can the value of absolute zero be obtained from the 1st graph?
Answer:

The graphs are shown in Figures respectively for three different pressures.

The graphs are straight lines in nature.

In the V-t°C graph shown, the three lines corresponding to the three pressures, converge at a point on the negative t-axis, where volume is zero. This point gives the value of absolute zero as per definition.

In the V-T K graph the three straight lines pass through the origin, showing V = 0 when T = 0.

Question 10. Determine the value of universal gas constant JR and gas constant k for lg of air. (At STP the density of air =1.293 g · L^-1 and that of mercury = 13.6 g · cm^-3)
Answer:

Pressure p – 76 x 13.6 x 980 dyn · cm^-2 and temperature = 0°C = 273 K

Volume of 1 mol of any gas at STP = 22.4 L = 22400 cm³

∴ R = \(\frac{p V}{T}=\frac{76 \times 13.6 \times 980 \times 22400}{273}\)

= \(8.31 \times 10^7 \mathrm{erg} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

Now, volume of 1 g air,

v = \(\frac{1}{1.293} l=\frac{1000}{1.293} \mathrm{~cm}^3\)

∴ k = \(\frac{p v}{T}=\frac{76 \times 13.6 \times 980 \times 1000}{1.293 \times 273}\)

= \(0.287 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

Question 11. A given mass of an Ideal gas is heated in a vessel. The same amount of gas is then heated by keeping it in a larger vessel. Assume that the volumes of both vessel remain the same during heating. What will be the nature of the pressure-temperature (p- T) graphs in the two cases?
Answer:

Given

A given mass of an Ideal gas is heated in a vessel. The same amount of gas is then heated by keeping it in a larger vessel. Assume that the volumes of both vessel remain the same during heating.

In each case the graph is a straight line, For a Fixed mass of gas, when the volume is less, the pressure increases with temperature at a higher rate.

So the slope of the p- T graph for the smaller container is greater than that for the bigger container.

Alternative method: For a fixed mass of an ideal gas, \(p V=k T \text { or, } p=\frac{k}{V} T \text {. }\)

pV = kT or, p = k/V T

If V is fixed, this relation resembles, y = mx.

So the p – T graph is a straight line passing through the origin.

Also, the slope k/V is higher for a smaller V. So the graph for the smaller vessel has a higher slope.

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Sample Problems on Ideal Gas Expansion

Question 12. Draw p- T graphs for masses m and 2m of the same gas, when heated in a container of constant volume. Interpret the slopes.
Answer:

The desired graphs are shown in Figure.

The equation of state for mass m of a gas of molecular weight M is, \(p V=\frac{m}{M} R T \quad \text { or } p=\frac{m R}{M V} T\)

Given, V = constant; so for any fixed mass, p = constant x T.

Thus, the p- T graph is a straight line passing through the origin.

Also, the slope \(\frac{m R}{M V}\) is higher for a higher m.

∴ The slope of the graph for mass 2 m is more than that for mass m.

For any temperature the value of p in the first case will be half of second case. \(\left(B C=\frac{1}{2} A C\right)\)

Question 13. Shows the V- T graph for a fixed mass of an ideal gas at pressures p₁ and p₂. Can you infer from the graph whether p₁ is greater than p₂?
Answer:

We have, pV= nRT,

or, V = \(\frac{n R}{p} T .\)

This shows that the V- T graph is a straight line passing through the origin, for any fixed pressure p (n = constant for a fixed mass).

The slope \(\frac{n R}{p}\) is greater for a smaller value of p.

In Figure, the graph for p₂ has a higher slope.

So, p₂ < p₁.

Thus, p₁ is greater than p₂.

Question 14. In a faulty barometer, some air is occupying the space over mercury column. How can the air pressure be correctly determined with this faulty barometer?
Answer:

Let the length of mercury column = h₁, and that of air column over mercury =l₁.

Now the barometer tube is raised a little still keeping the open end dipped into mercury. Let in this case, the length of mercury column = h₂, and that of air column over mercury = l₁.

If the true reading of the air pressure is H, and area of the cross-section of the tube is α, then from Boyle’s law

⇒ \(\left(H-h_1\right) l_1 \alpha=\left(H-h_2\right) l_2 \alpha \quad \text { or, } H=\frac{h_1 l_1-h_2 l_2}{l_1-l_2}\)

Hence, measuring h₁, h₂, l₁, l₂ the correct atmospheric pressure can be found out.

Question 15. A container filled with oxygen is taken to the moon’s surface from the earth. How will the volume and pressure of the gas change when the container is

1. A rubber balloon,
2. A steel cylinder?

Answer:

1. Atmospheric pressure on moon’s surface is practically zero. As pressure of oxygen inside the rubber balloon is high, the gas will expand in volume and finally the balloon will burst.

2. Volume of the gas in the steel cylinder would remain unchanged. There will be no effect of zero atmospheric pressure.

But the pressure of the gas would change due to a different temperature of the surroundings.

Question 16. What is meant by specific gas constant? Is the value of this constant same for all gases?
Answer:

Value of \(\frac{p V}{T}\) for 1 g of a gas is the specific heat constant of that gas. Value of the constant is different for different gases because of the difference in molecular weight M, since \(\frac{p V}{T}=n R=\frac{1}{M} R \text {. }\)

Question 17. Equal number of hydrogen and helium molecules are kept in two identical gas jars at the same temperature. What will be the ratio of the pressures of the gases in the two jars?
Answer:

Given

Equal number of hydrogen and helium molecules are kept in two identical gas jars at the same temperature.

Pressure is the same in both the jars. As per Avaga- dro’s law, all gases of equal volume contain the same number of molecules under identical values of temperature and pressure. In this example, volume, temperature and number of molecules are the same. So, the pressure will be equal in the two jars and the ratio is 1:1.

Question 18. A gas container contains 1 mol of O₂ gas (specific molar mass 32) at pressure p and temperature T. In a similar container one mol of He gas (specific molar mass 4) is kept at temperature 2 T. What is the pressure of this He gas?
Answer:

Given

A gas container contains 1 mol of O₂ gas (specific molar mass 32) at pressure p and temperature T. In a similar container one mol of He gas (specific molar mass 4) is kept at temperature 2 T.

Since number of moles and volume are the same for both the gases, using pV = nRT, it can be said that the pressure is directly proportional to the absolute temperature. Since the temperature of the second container is double, pressure will also be double, i.e., 2p.

Question 19. Same ideal gas is kept in two containers A and B fitted with frictionless pistons. Volume and temperature of the gas in both containers are the same. m_A and m_B are the masses of the gas in A and B respectively. The volume of the gases in the two containers are changed to 2 V keeping their temperature constant.

Corresponding changes in pressure in A and B are Δp and 1.5 Δp. Find the ratio of the masses of the gas kept in A and in B.

Answer:

Given

Same ideal gas is kept in two containers A and B fitted with frictionless pistons. Volume and temperature of the gas in both containers are the same. m_A and m_B are the masses of the gas in A and B respectively. The volume of the gases in the two containers are changed to 2 V keeping their temperature constant.

Corresponding changes in pressure in A and B are Δp and 1.5 Δp.

Volumes of the gases in both the containers become double. So, according to Boyle’s law, the corresponding pressures of the gases become half of their initial pressures.

∴ Initial pressure in A and B are,

p_A = 2Δp and p_B = 2 x 1.5Δp = 3Δp

∴ \(\frac{p_A}{p_B}=\frac{2}{3} \text { or, } \frac{n_A \frac{R T}{V}}{n_B \frac{R T}{V}}=\frac{2}{3}\)

or, \(\frac{m_A \frac{R T}{M V}}{m_B \frac{R T}{M V}}=\frac{2}{3} \text { or, } \frac{m_A}{m_B}=\frac{2}{3}\)

Question 20. An ideal gas is found to obey a gas law, Vp² = constant Initial temperature and volume of the gas are T and V respectively. If the gas expands to a volume 2 V, what will be the effect on temperature?
Answer:

Given

An ideal gas is found to obey a gas law, Vp² = constant Initial temperature and volume of the gas are T and V respectively. If the gas expands to a volume 2 V,

Here Vp² = constant…..(1)

For an ideal gas pV = RT or \(p=\frac{R T}{V}\)

Substituting for p, equation (1) gives

⇒ \(\frac{V \times R^2 T^2}{V^2}=\text { constant or, } \frac{T^2}{V}=\text { constant }\)

When the volume of the gas expands to 2V, suppose the temperature is T’.

∴ \(\frac{T^2}{V}=\frac{T^{\prime 2}}{2 V} \text { or, } \frac{T^{\prime 2}}{T^2}=2 \text { or, } T^{\prime}=T \sqrt{2} \text {. }\)

Question 21. Shows the p- T graphs for a fixed mass of an ideal gas at volumes V₁ and V₂. Can it be concluded from the graphs that V₁ is greater than V₂?
Answer:

∴ \(p V=n R T=\frac{m}{M} R T\)

or, p = \(\frac{m R}{M V} T\) [M= molecular weights]

So p- T graphs are straight lines passing through the origin, for any fixed V. The slope \(\frac{m R}{M V}\) is smaller for a greater value of V.

The slope for V₁ is smaller than that for V₂. So, V₁>V₂.

Question 22. An ideal gas is Initially at temperature T and volume V. Its volume increases by dV due to an increase in temperature dT, while pressure remains constant. Here \(\gamma=\frac{1}{V} \frac{d V}{d T}\).What will be the nature of the graph between γ and T1
Answer:

In case of an ideal gas,

pV= RT or, \(V=\frac{R T}{p}\)

When pressure remains constant, we have \(\frac{d V}{d T}=\frac{R}{p}\)

∴ \(\gamma=\frac{1}{V} \frac{d V}{d T}=\frac{1}{V} \frac{R}{p}=\frac{R}{R T}=\frac{1}{T}\)

So, the graph γ-T will be a rectangular hyperbola.

Question 23. An ideal gas is initially at pressure p and volume V. Its pressure is increased by dp, so that its volume decreases by dV, while the temperature remains constant. Here \(\beta=-\frac{1}{V} \frac{d V}{d p}\). What will be the nature of the graph between β and p?
Answer:

Given

An ideal gas is initially at pressure p and volume V. Its pressure is increased by dp, so that its volume decreases by dV, while the temperature remains constant.

In case of an ideal gas, pV = RT

If the temperature remains constant, p V = constant

∴ pdV+ Vdp = 0

or, \(\frac{d V}{d p}=-\frac{V}{p}\)

∴ \(\beta=-\frac{1}{V} \frac{d V}{d p}=\frac{1}{p}\)

i. e., βp = 1 = constant So, the graph β-p will be a rectangular hyperbola.

Question 24. Pressure coefficient of a gas is \(\frac{1}{273}{ }^{\circ} \mathrm{C}^{-1}\) Explain.
Answer:

Pressure coefficient of a gas is \(\frac{1}{273}{ }^{\circ} \mathrm{C}^{-1}\)

It means that when the temperature increases 1°C, at constant volume the increase in pressure of the gas of fixed mass per unit pressure at 0°C is equal to 1/273 of the original pressure of that gas at °C.