JEE Advanced PYQ Kinetic Theory Miscellaneous Mixed Concept Problems

Q 1 :

Two moles of ideal helium gas are in a rubber balloon at $30 ° C$ . The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to $35 ° C$ . The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol.K) [2012]

62 J
104 J
124 J
208 J

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(4)

The heat is supplied at constant pressure, i.e., the process is isobaric.

$∴$ $Q = n C_{p} Δ t$

$= 2 [\frac{5}{2} R] \times Δ t$ $= 2 \times \frac{5}{2} \times 8.31 \times 5 = 208 J$ $(∵ C_{p} = \frac{5}{2} R for mono-atomic gas)$

Q 2 :

An ideal gas is expanding such that $P T^{2} = constant .$ The coefficient of volume expansion of the gas is [2008]

$\frac{1}{T}$
$\frac{2}{T}$
$\frac{3}{T}$
$\frac{4}{T}$

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(3)

$p T^{2} = constant$ $(given)$

$∴$ $(\frac{n R T}{V}) T^{2} = constant$ or $T^{3} V^{- 1} = constant$

$(∴ P V = n R T)$

Differentiating the equation, we get

$\frac{3 T^{2}}{V} d T - \frac{T^{3}}{V^{2}} d V = 0$ or $3 d T = \frac{T}{V} d V \dots (i)$

From the equation

$d V = V γ d T$

$γ = coefficient of volume expansion of gas = \frac{d V}{V \cdot d T}$

From eq. (i) $γ = \frac{d V}{V \cdot d T} = \frac{3}{T}$

Q 3 :

The left and right compartments of a thermally isolated container of length L are separated by a thermally conducting, movable piston of area A. The left and right compartments are filled with $\frac{3}{2}$ and 1 moles of an ideal gas, respectively. In the left compartment the piston is attached by a spring with spring constant $k$ and natural length $\frac{2 L}{5}$ . In thermodynamic equilibrium, the piston is at a distance $\frac{L}{2}$ from the left and right edges of the container as shown in the figure. Under the above conditions, if the pressure in the right compartment is $P = \frac{k L}{A} α,$ then the value of $α$ is ________. [2025]

[IMAGE 513]

(0.20)

$Extension in spring$

$x = 0.5 L - 0.4 L = 0.1 L$

FBD of piston

[IMAGE 514]

$k x + P_{2} A = P_{1} A$

$P_{2} A = P_{1} A - k x$

$P_{2} = P_{1} - \frac{k L}{A (10)} \dots (i)$

$P_{1} V = n_{1} R T and P_{2} V = n_{2} R T$

$\frac{P_{1}}{P_{2}} = \frac{n_{1}}{n_{2}} = \frac{3}{2}$

$P_{1} = \frac{3}{2} P_{2} \dots (i i)$

$P_{2} = \frac{3}{2} P_{2} - \frac{k L}{10 A}$

$\frac{P_{2}}{2} = \frac{k L}{10 A}$

$P_{2} = \frac{k L}{5 A} = \frac{k L}{A} α$

$∴$ $α = \frac{1}{5} = 0.2$

Q 4 :

A soft plastic bottle, filled with water of density $1 gm/cc$ , carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density $2.5 gm/cc$ . Initially the bottle is sealed at atmospheric pressure $p_{0} = 10^{5} Pa$ so that the volume of the trapped air is $v_{0} = 3.3 cc .$ When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure $p_{0} + Δ p$ without changing its orientation. At this pressure, the volume of the trapped air is $v_{0} - Δ v .$

Let $Δ v = X cc$ and $Δ p = Y \times 10^{3} Pa .$

[IMAGE 515]

Q. The value of X is _____. [2021]

(0.30)

[IMAGE 516]

$When tube + air system starts sinking$

$F_{B} = m g$

$\Rightarrow ρ_{0} (V_{glass} + V_{gas}) = m$

$1 (2 + V_{gas}) = 5$

$\Rightarrow V_{gas} = 3 cc$

Hence, $Δ V = V_{0} - V_{gas}$

$= 3.3 cc - 3 cc = 0.3 cc$

$∴$ $x = Δ V = 0.3$

Q 5 :

A soft plastic bottle, filled with water of density $1 gm/cc$ , carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density $2.5 gm/cc$ . Initially the bottle is sealed at atmospheric pressure $p_{0} = 10^{5} Pa$ so that the volume of the trapped air is $v_{0} = 3.3 cc .$ When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure $p_{0} + Δ p$ without changing its orientation. At this pressure, the volume of the trapped air is $v_{0} - Δ v .$

Let $Δ v = X cc$ and $Δ p = Y \times 10^{3} Pa .$

[IMAGE 517]

Q. The value of Y is _____. [2021]

(10)

$Isothermal process for air, temperature is constant.$

$∴$ $From P_{1} V_{1} = P_{2} V_{2}$

$10^{5} \times (3.3) = P_{2} (3)$ $\Rightarrow P_{2} = 1.1 \times 10^{5}$

$Δ P = P_{2} - P_{1}$ $= 1.1 \times 10^{5} - 10^{5} = 0.1 \times 10^{5}$

or, $Δ P = 10 \times 10^{3} Pascal = Y \times 10^{3} Pascal$

$∴$ $Y = 10$

Q 6 :

One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is $100 K$ and the universal gas constant $R = 8.0 {J mol}^{- 1} K^{- 1},$ the decrease in its internal energy, in Joule, is ________. [2018]

(900)

$Given: T_{i} = 100 K, V_{f} = 8 V_{i}$

For an adiabatic process, $T V^{γ - 1} = constant$

or, $T_{i} V_{i}^{γ - 1} = T_{f} V_{f}^{γ - 1}$

$\Rightarrow \frac{T_{i}}{T_{f}} = {(\frac{V_{f}}{V_{i}})}^{γ - 1}$ $\Rightarrow \frac{T_{i}}{T_{f}} = {(\frac{8 V_{i}}{V_{i}})}^{γ - 1}$

For monoatomic gas, $γ = \frac{5}{3}$

$∴$ $T_{f} = \frac{T_{i}}{8^{(\frac{5}{3} - 1)}} = \frac{T_{i}}{4}$

Change in internal energy, $Δ u = n C_{v} Δ T$

$= 1 \times \frac{3}{2} R (\frac{T_{i}}{4} - T_{i})$ $= \frac{3}{2} \times 8 \times (\frac{- 3}{4}) \times 100 = - 900 J$

Q 7 :

As shown schematically in the figure, two vessels contain water solutions (at temperature $T$ ) of potassium permanganate $({KMnO}_{4})$ of different concentrations $n_{1}$ and $n_{2}$ $(n_{1} > n_{2})$ molecules per unit volume with $Δ n = (n_{1} - n_{2}) ≪ n_{1} .$ When they are connected by a tube of small length $l$ and cross-sectional area $S$ , ${KMnO}_{4}$ starts to diffuse from the left to the right vessel through the tube. Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion. The speed $v$ of the molecules is limited by the viscous force $- β v$ on each molecule, where $β$ is a constant. Neglecting all terms of the order ${(Δ n)}^{2}$ , which of the following is/are correct? ( $k_{B}$ is the Boltzmann constant) [2013]

[IMAGE 518]

the force causing the molecules to move across the tube is $Δ n k_{B} T S$
force balance implies $n_{1} β v l = Δ n k_{B} T$
total number of molecules going across the tube per sec is $(\frac{Δ n}{l}) (\frac{k_{B} T}{β}) S$
rate of molecules getting transferred through the tube does not change with time

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4

Select one or more options

(1, 2, 3)

$Force = Pressure \times Area$

$p_{1} = \frac{n_{1} R T}{N_{A}} and p_{2} = \frac{n_{2} R T}{N_{A}}$

$F = Δ p \cdot A = (\frac{n_{1} R T}{N_{A}} - \frac{n_{2} R T}{N_{A}}) S$

$F = (n_{1} - n_{2}) k_{B} T S = Δ n k_{B} T S$ $(∵ \frac{R}{N_{A}} = K_{B} (Boltzmann constant))$

Hence, option (1) is correct.

$V = \frac{Δ n k_{B} T S}{β}$

[IMAGE 519]

Force balance = pressure $\times$ area = total number of molecules $\times β v$

$Δ n k_{B} T S = ℓ n_{1} S β v$ $[β v = viscous force (given)]$

$\Rightarrow n_{1} β v ℓ = Δ n k_{B} T$

So option (2) is correct.

Total number of molecules/sec, $\frac{Δ N}{Δ t} = \frac{(n_{1} v d t) S}{Δ t}$

$= n_{1} v S = \frac{Δ n k_{B} T v S}{β v ℓ} = (\frac{Δ n}{ℓ}) (\frac{k_{B} T}{β}) S$

Option (3) is correct.

As $Δ n$ will decrease with time, so rate of molecules getting transferred through the tube decreases with time.

Hence option (4) is incorrect.

Q 8 :

The figure below shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the following statement(s) is (are) correct to a reasonable approximation. [2013]

[IMAGE 520]

The rate at which heat is absorbed in the range 0–100 K varies linearly with temperature T.
Heat absorbed in increasing the temperature from 0–100 K is less than the heat required for increasing the temperature from 400–500 K.
There is no change in the rate of heat absorption in the range 400–500 K.
The rate of heat absorption increases in the range 200–300 K.

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Select one or more options

(2, 3, 4)

$(1) As we know, Q = m c Δ T$

$\Rightarrow \frac{d Q}{d t} = m c \frac{d T}{d t}$ or, $\frac{d Q}{d t} \propto C$ i.e., rate of heat absorption $\propto C$ .

In the range 0 to 100K from the graph, C increases with temperature but not linearly therefore the rate at which heat is absorbed varies with temperature. But not linearly.

(2) As the value of C is greater in the temperature range 400-500K, the heat absorbed in increasing the temperature from 0-100K is less than the heat required for increasing the temperature from 400-500K.

(3) From the graph the value of C does not change in the temperature range 400-500K, therefore there is no change in the rate of heat absorption in this range.

(4) As the value of C increases from 200-300K, the rate of heat absorption increases in the range 200-300K.

Q 9 :

In the figure, a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulated material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are $C_{V} = \frac{3}{2} R, C_{P} = \frac{5}{2} R$ and those for an ideal diatomic gas are $C_{V} = \frac{5}{2} R, C_{P} = \frac{7}{2} R .$

[IMAGE 521]

Q. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be [2014]

550 K
525 K
513 K
490 K

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(4)

$Let T be the final temperature of the gases when equilibrium is achieved.$

$Heat lost by monoatomic gas at constant volume$ $= Heat gained by diatomic gas at constant pressure$

$∴$ $n C_{v 1} (700 - T) = n C_{p 2} (T - 400)$

$\frac{3}{2} R (700 - T) = \frac{7}{2} R (T - 400)$

$\Rightarrow 2100 - 3 T = 7 T - 2800$

$\Rightarrow 10 T = 4900$

$∴$ $T = 490 K$

Q 10 :

In the figure, a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulated material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are $C_{V} = \frac{3}{2} R, C_{P} = \frac{5}{2} R$ and those for an ideal diatomic gas are $C_{V} = \frac{5}{2} R, C_{P} = \frac{7}{2} R .$

[IMAGE 522]

Q. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. The total work done by the gases till the time they achieve equilibrium will be [2014]

250 R
200 R
100 R
−100 R

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(4)

$As the pressure of gases in both compartments is the same$

$∴$ $n C_{p 1} (700 - T) = n C_{p 2} (T - 400)$

$\frac{5}{2} R (700 - T) = \frac{7}{2} R (T - 400)$

$3500 - 5 T = 7 T - 2800$ $\Rightarrow 12 T = 6300$

$∴$ $T = 525 K$

Applying first law of thermodynamics,

$Δ W_{1} + Δ U_{1} = Δ Q_{1}$

and $Δ W_{2} + Δ U_{2} = Δ Q_{2}$

$Δ Q_{1} + Δ Q_{2} = 0$

or, $- (Δ W_{1} + Δ W_{2}) = Δ U_{1} + Δ U_{2}$

$= n_{1} C_{v 1} (525 - 700) + n_{2} C_{v 2} (525 - 400)$

$= - 2 \times \frac{3 R}{2} \times 175 + 2 \times \frac{5 R}{2} \times 125$

$= - 525 R + 625 R = - 100 R$

Therefore, total work done $= - 100 R$

Q 11 :

A small spherical mono-atomic ideal gas bubble $(γ = 5 / 3)$ is trapped inside a liquid of density $ρ$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_{0}$ , the height of the liquid is $H$ and the atmospheric pressure is $P_{0}$ (Neglect surface tension). [2008]

[IMAGE 523]

Q. As the bubble moves upwards, besides the buoyancy force the following forces are acting on it

Only the force of gravity
The force due to gravity and the force due to the pressure of the liquid
The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid
The force due to gravity and the force due to viscosity of the liquid

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(4)

The forces acting besides buoyancy force which is the force due to pressure of the liquid are

(i) Force of gravity (vertically downwards)

(ii) Viscous force (vertically downwards)

Q 12 :

A small spherical mono-atomic ideal gas bubble $(γ = 5 / 3)$ is trapped inside a liquid of density $ρ$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_{0}$ , the height of the liquid is $H$ and the atmospheric pressure is $P_{0}$ (Neglect surface tension). [2008]

[IMAGE 524]

Q. When the gas bubble is at a height $y$ from the bottom, its temperature is

$T_{0} {(\frac{P_{0} + ρ_{ℓ} g H}{P_{0} + ρ_{ℓ} g y})}^{2 / 5}$
$T_{0} {(\frac{P_{0} + ρ_{ℓ} g (H - y)}{P_{0} + ρ_{ℓ} g H})}^{2 / 5}$
$T_{0} {(\frac{P_{0} + ρ_{ℓ} g H}{P_{0} + ρ_{ℓ} g y})}^{3 / 5}$
$T_{0} {(\frac{P_{0} + ρ_{ℓ} g (H - y)}{P_{0} + ρ_{ℓ} g H})}^{3 / 5}$

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(2)

$As the bubble does not exchange any heat with the liquid.$

$So, the process is adiabatic.$

$Applying relation,$ $T^{γ} P^{1 - γ} = constant$

$T_{2} = T_{1} {[\frac{P_{1}}{P_{2}}]}^{\frac{1 - γ}{γ}}$

Here $T_{1} = T_{0}, P_{1} = P_{0} + H ρ_{ℓ} g, T_{2} = T, P_{2} = P$

$P = P_{0} + (H - y) ρ_{ℓ} g, γ = \frac{5}{3}$

$∴$ $T = T_{0} {[\frac{P_{0} + H ρ_{ℓ} g}{P_{0} + (H - y) ρ_{ℓ} g}]}^{\frac{1 - \frac{5}{3}}{\frac{5}{3}}}$

$= T_{0} {[\frac{P_{0} + H ρ_{ℓ} g}{P_{0} + (H - y) ρ_{ℓ} g}]}^{- \frac{2}{3} \times \frac{3}{5}}$

$T = T_{0} {[\frac{P_{0} + (H - y) ρ_{ℓ} g}{P_{0} + H ρ_{ℓ} g}]}^{\frac{2}{5}}$

Q 13 :

A small spherical mono-atomic ideal gas bubble $(γ = 5 / 3)$ is trapped inside a liquid of density $ρ$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_{0}$ , the height of the liquid is $H$ and the atmospheric pressure is $P_{0}$ (Neglect surface tension). [2008]

[IMAGE 525]

Q. The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant)

$ρ_{ℓ} n R g T_{0} \frac{{(P_{0} + ρ_{ℓ} g H)}^{2 / 5}}{{(P_{0} + ρ_{ℓ} g y)}^{7 / 5}}$
$\frac{ρ_{ℓ} n R g T_{0}}{{(P_{0} + ρ_{ℓ} g H)}^{2 / 5} {[P_{0} + ρ_{ℓ} g (H - y)]}^{3 / 5}}$
$ρ_{ℓ} n R g T_{0} \frac{{(P_{0} + ρ_{ℓ} g H)}^{3 / 5}}{{(P_{0} + ρ_{ℓ} g y)}^{8 / 5}}$
$\frac{ρ_{ℓ} n R g T_{0}}{{(P_{0} + ρ_{ℓ} g H)}^{3 / 5} {[P_{0} + ρ_{ℓ} g (H - y)]}^{2 / 5}}$

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(2)

$∵$ $P V = n R T$ $\Rightarrow V = \frac{n R T}{P} = \frac{n R T}{P_{0} + (H - y) ρ_{ℓ} g}$

Where P is pressure of the bubble at an arbitrary location distant y from the bottom.

Substituting the value of P and T from above we get

$V = \frac{n R}{[P_{0} + (H - y) ρ_{ℓ} g]} \times \frac{T_{0} {[P_{0} + (H - y) ρ_{ℓ} g]}^{\frac{2}{5}}}{{[P_{0} + {H ρ}_{ℓ} g]}^{\frac{2}{5}}}$

$= \frac{n R T_{0}}{{[P_{0} + (H - y) ρ_{ℓ} g]}^{\frac{3}{5}} {[P_{0} + H ρ_{ℓ} g]}^{\frac{2}{5}}}$

$∴$ $Buoyancy force$

$= V ρ_{ℓ} g = \frac{n R T_{0} ρ_{ℓ} g}{{[P_{0} + (H - y) ρ_{ℓ} g]}^{\frac{3}{5}} {[P_{0} + H ρ_{ℓ} g]}^{\frac{2}{5}}}$

Topic Question Set

Q 2 :

An ideal gas is expanding such that $P T^{2} = constant .$ The coefficient of volume expansion of the gas is [2008]

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Topic Question Set

Q 2 : An ideal gas is expanding such that PT2=constant. The coefficient of volume expansion of the gas is [2008]

Q 6 : One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R=8.0 J mol-1K-1, the decrease in its internal energy, in Joule, is ________. [2018]

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Q 2 :

An ideal gas is expanding such that $P T^{2} = constant .$ The coefficient of volume expansion of the gas is [2008]