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PHXI13:KINETIC THEORY

360073 A cylindrical container is shown in figure in which a gas is enclosed. Its initial volume is $V$ and temperature is $T$ . As no external pressure is applied on the light piston shown, gas pressure must be equal to the atmospheric pressure. If gas temperature is doubled, find its final volume. In its final state if piston is clamped and temperature is again doubled, the ratio of final pressure and initial pressure of the gas is
supporting img

1 6

2 4

3 7

4 2

Explanation:

In the initial state the pressure, volume and temperature of gas $P_{0} V$ and T respectively if $P_{0}$ is the atmospheric pressure. It is given that temperature of gas is increased to double its value $i . e .$ up to $2 T$ . As in initial and final state pressure of gas remains constant as initially as well as finally the piston is exposed to atmospheric pressure. Thus we have from Charles and Gay Loussac law $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
Here, $V_{2} = \frac{V_{1} T_{2}}{T_{1}} = 2 V$
$[As$ $V_{1} = V, T_{1} = T$ and $T_{2} = 2 T]$
Thus after increasing the gas temperature to 27, its volume becomes $2 V$ . Now the piston is clamped that means, now the volume of gas remains constant. Again its temperature is doubled from $2 T$ to $4 T$ , let the pressure changes from $P_{0}$ to $P^{'}$ . Thus we have
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
or $P_{2} = P^{'} = \frac{P_{1} T_{2}}{T_{1}}$
$= 2 P_{0} [As P_{1} = P_{0}, T_{1} = 2 T and T_{2} = 4 T]$
Thus in final state gas pressure becomes $2 P_{0}$ .

PHXI13:KINETIC THEORY

360074 Find the minimum attainable pressure of ideal gas in the process $T = T_{0} + α V^{2}$ , where $T_{0}$ , and $α$ are $+ v e$ constants and $V$ is the volume of one mole of gas.

1

R \sqrt{T_{0}}

2

4 R \sqrt{T_{0}}

3

2 R \sqrt{α T_{0}}

4

\frac{\sqrt{α T_{0}}}{R}

Explanation:

Given that
$T = T_{0} + α V^{2}$
For one mole of the gas, $V = R T / p$
$T = T_{0} + α (\frac{R^{2} T^{2}}{p^{2}})$
Solving it for $p,$ we get
$p = \sqrt{α} R T {(T - T_{0})}^{- 1 / 2} (1)$
Differentiating it with respect to $T$ , we get
$\frac{d p}{d T} = \sqrt{α} \cdot R [{(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2}]$
Conditions for minimum pressure is that
$(d p / d T) = 0 . Hence,$
$\begin{matrix} \sqrt{α} R [{(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2}] = 0 \\ {(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2} = 0 (∵ \sqrt{α} R \neq 0) \end{matrix}$
Now, Solving we get,
$T = 2 T_{0}$
Substituting the value of $T = 2 T_{0}$ in Eq. (1) we get
$p_{min} = \sqrt{α} R (2 T_{0}) {[2 T_{0} - T_{0}]}^{- 1 / 2}$
$o r p_{min} = 2 R \sqrt{(α T_{0})}$

PHXI13:KINETIC THEORY

360076 From the following $P - T$ graph what inference can be drawn
supporting img

1

V_{2} < V_{1}

2

V_{2} > V_{1}

3

V_{2} = V_{1}

4 None of these

Explanation:

As
$θ_{2} > θ_{1} \Rightarrow \tan θ_{2} > \tan θ_{1} \Rightarrow {(\frac{T}{P})}_{2} > {(\frac{T}{P})}_{1}$
From $P V = n R T; \frac{T}{P} \propto V \Rightarrow V_{2} > V_{1}$ .

PHXI13:KINETIC THEORY

360077 The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at $1.38 a t m$ is
(Given, $k = 1.38 \times 10^{- 23} J K^{- 1}$ ) is

1

100 K

2

300 K

3

500 K

4

200 K

Explanation:

Given, $N = 2 \times 10^{25} / m^{3}$
$P = 1.38 a t m, k = 1.38 \times 10^{- 23} J / K$
By ideal gas equation
$P V = n R T$
$\Rightarrow P V = \frac{N}{N_{A}} R T$
$\Rightarrow P = \frac{N}{V} \cdot k T$
$\Rightarrow 1.38 \times 1.01 \times 10^{5}$
$= 2 \times 10^{25} \times 1.38 \times 10^{- 23} \times T$
$\Rightarrow T ≃ 500 K$
So, correct option is (3).

JEE - 2024

PHXI13:KINETIC THEORY

360073 A cylindrical container is shown in figure in which a gas is enclosed. Its initial volume is $V$ and temperature is $T$ . As no external pressure is applied on the light piston shown, gas pressure must be equal to the atmospheric pressure. If gas temperature is doubled, find its final volume. In its final state if piston is clamped and temperature is again doubled, the ratio of final pressure and initial pressure of the gas is
supporting img

1 6

2 4

3 7

4 2

Explanation:

In the initial state the pressure, volume and temperature of gas $P_{0} V$ and T respectively if $P_{0}$ is the atmospheric pressure. It is given that temperature of gas is increased to double its value $i . e .$ up to $2 T$ . As in initial and final state pressure of gas remains constant as initially as well as finally the piston is exposed to atmospheric pressure. Thus we have from Charles and Gay Loussac law $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
Here, $V_{2} = \frac{V_{1} T_{2}}{T_{1}} = 2 V$
$[As$ $V_{1} = V, T_{1} = T$ and $T_{2} = 2 T]$
Thus after increasing the gas temperature to 27, its volume becomes $2 V$ . Now the piston is clamped that means, now the volume of gas remains constant. Again its temperature is doubled from $2 T$ to $4 T$ , let the pressure changes from $P_{0}$ to $P^{'}$ . Thus we have
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
or $P_{2} = P^{'} = \frac{P_{1} T_{2}}{T_{1}}$
$= 2 P_{0} [As P_{1} = P_{0}, T_{1} = 2 T and T_{2} = 4 T]$
Thus in final state gas pressure becomes $2 P_{0}$ .

PHXI13:KINETIC THEORY

360074 Find the minimum attainable pressure of ideal gas in the process $T = T_{0} + α V^{2}$ , where $T_{0}$ , and $α$ are $+ v e$ constants and $V$ is the volume of one mole of gas.

1

R \sqrt{T_{0}}

2

4 R \sqrt{T_{0}}

3

2 R \sqrt{α T_{0}}

4

\frac{\sqrt{α T_{0}}}{R}

Explanation:

Given that
$T = T_{0} + α V^{2}$
For one mole of the gas, $V = R T / p$
$T = T_{0} + α (\frac{R^{2} T^{2}}{p^{2}})$
Solving it for $p,$ we get
$p = \sqrt{α} R T {(T - T_{0})}^{- 1 / 2} (1)$
Differentiating it with respect to $T$ , we get
$\frac{d p}{d T} = \sqrt{α} \cdot R [{(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2}]$
Conditions for minimum pressure is that
$(d p / d T) = 0 . Hence,$
$\begin{matrix} \sqrt{α} R [{(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2}] = 0 \\ {(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2} = 0 (∵ \sqrt{α} R \neq 0) \end{matrix}$
Now, Solving we get,
$T = 2 T_{0}$
Substituting the value of $T = 2 T_{0}$ in Eq. (1) we get
$p_{min} = \sqrt{α} R (2 T_{0}) {[2 T_{0} - T_{0}]}^{- 1 / 2}$
$o r p_{min} = 2 R \sqrt{(α T_{0})}$

PHXI13:KINETIC THEORY

360076 From the following $P - T$ graph what inference can be drawn
supporting img

1

V_{2} < V_{1}

2

V_{2} > V_{1}

3

V_{2} = V_{1}

4 None of these

Explanation:

As
$θ_{2} > θ_{1} \Rightarrow \tan θ_{2} > \tan θ_{1} \Rightarrow {(\frac{T}{P})}_{2} > {(\frac{T}{P})}_{1}$
From $P V = n R T; \frac{T}{P} \propto V \Rightarrow V_{2} > V_{1}$ .

PHXI13:KINETIC THEORY

360077 The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at $1.38 a t m$ is
(Given, $k = 1.38 \times 10^{- 23} J K^{- 1}$ ) is

1

100 K

2

300 K

3

500 K

4

200 K

Explanation:

Given, $N = 2 \times 10^{25} / m^{3}$
$P = 1.38 a t m, k = 1.38 \times 10^{- 23} J / K$
By ideal gas equation
$P V = n R T$
$\Rightarrow P V = \frac{N}{N_{A}} R T$
$\Rightarrow P = \frac{N}{V} \cdot k T$
$\Rightarrow 1.38 \times 1.01 \times 10^{5}$
$= 2 \times 10^{25} \times 1.38 \times 10^{- 23} \times T$
$\Rightarrow T ≃ 500 K$
So, correct option is (3).

JEE - 2024

PHXI13:KINETIC THEORY

360073 A cylindrical container is shown in figure in which a gas is enclosed. Its initial volume is $V$ and temperature is $T$ . As no external pressure is applied on the light piston shown, gas pressure must be equal to the atmospheric pressure. If gas temperature is doubled, find its final volume. In its final state if piston is clamped and temperature is again doubled, the ratio of final pressure and initial pressure of the gas is
supporting img

1 6

2 4

3 7

4 2

Explanation:

In the initial state the pressure, volume and temperature of gas $P_{0} V$ and T respectively if $P_{0}$ is the atmospheric pressure. It is given that temperature of gas is increased to double its value $i . e .$ up to $2 T$ . As in initial and final state pressure of gas remains constant as initially as well as finally the piston is exposed to atmospheric pressure. Thus we have from Charles and Gay Loussac law $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
Here, $V_{2} = \frac{V_{1} T_{2}}{T_{1}} = 2 V$
$[As$ $V_{1} = V, T_{1} = T$ and $T_{2} = 2 T]$
Thus after increasing the gas temperature to 27, its volume becomes $2 V$ . Now the piston is clamped that means, now the volume of gas remains constant. Again its temperature is doubled from $2 T$ to $4 T$ , let the pressure changes from $P_{0}$ to $P^{'}$ . Thus we have
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
or $P_{2} = P^{'} = \frac{P_{1} T_{2}}{T_{1}}$
$= 2 P_{0} [As P_{1} = P_{0}, T_{1} = 2 T and T_{2} = 4 T]$
Thus in final state gas pressure becomes $2 P_{0}$ .

PHXI13:KINETIC THEORY

360074 Find the minimum attainable pressure of ideal gas in the process $T = T_{0} + α V^{2}$ , where $T_{0}$ , and $α$ are $+ v e$ constants and $V$ is the volume of one mole of gas.

1

R \sqrt{T_{0}}

2

4 R \sqrt{T_{0}}

3

2 R \sqrt{α T_{0}}

4

\frac{\sqrt{α T_{0}}}{R}

Explanation:

Given that
$T = T_{0} + α V^{2}$
For one mole of the gas, $V = R T / p$
$T = T_{0} + α (\frac{R^{2} T^{2}}{p^{2}})$
Solving it for $p,$ we get
$p = \sqrt{α} R T {(T - T_{0})}^{- 1 / 2} (1)$
Differentiating it with respect to $T$ , we get
$\frac{d p}{d T} = \sqrt{α} \cdot R [{(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2}]$
Conditions for minimum pressure is that
$(d p / d T) = 0 . Hence,$
$\begin{matrix} \sqrt{α} R [{(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2}] = 0 \\ {(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2} = 0 (∵ \sqrt{α} R \neq 0) \end{matrix}$
Now, Solving we get,
$T = 2 T_{0}$
Substituting the value of $T = 2 T_{0}$ in Eq. (1) we get
$p_{min} = \sqrt{α} R (2 T_{0}) {[2 T_{0} - T_{0}]}^{- 1 / 2}$
$o r p_{min} = 2 R \sqrt{(α T_{0})}$

PHXI13:KINETIC THEORY

360076 From the following $P - T$ graph what inference can be drawn
supporting img

1

V_{2} < V_{1}

2

V_{2} > V_{1}

3

V_{2} = V_{1}

4 None of these

Explanation:

As
$θ_{2} > θ_{1} \Rightarrow \tan θ_{2} > \tan θ_{1} \Rightarrow {(\frac{T}{P})}_{2} > {(\frac{T}{P})}_{1}$
From $P V = n R T; \frac{T}{P} \propto V \Rightarrow V_{2} > V_{1}$ .

PHXI13:KINETIC THEORY

360077 The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at $1.38 a t m$ is
(Given, $k = 1.38 \times 10^{- 23} J K^{- 1}$ ) is

1

100 K

2

300 K

3

500 K

4

200 K

Explanation:

Given, $N = 2 \times 10^{25} / m^{3}$
$P = 1.38 a t m, k = 1.38 \times 10^{- 23} J / K$
By ideal gas equation
$P V = n R T$
$\Rightarrow P V = \frac{N}{N_{A}} R T$
$\Rightarrow P = \frac{N}{V} \cdot k T$
$\Rightarrow 1.38 \times 1.01 \times 10^{5}$
$= 2 \times 10^{25} \times 1.38 \times 10^{- 23} \times T$
$\Rightarrow T ≃ 500 K$
So, correct option is (3).

JEE - 2024

PHXI13:KINETIC THEORY

360073 A cylindrical container is shown in figure in which a gas is enclosed. Its initial volume is $V$ and temperature is $T$ . As no external pressure is applied on the light piston shown, gas pressure must be equal to the atmospheric pressure. If gas temperature is doubled, find its final volume. In its final state if piston is clamped and temperature is again doubled, the ratio of final pressure and initial pressure of the gas is
supporting img

1 6

2 4

3 7

4 2

Explanation:

In the initial state the pressure, volume and temperature of gas $P_{0} V$ and T respectively if $P_{0}$ is the atmospheric pressure. It is given that temperature of gas is increased to double its value $i . e .$ up to $2 T$ . As in initial and final state pressure of gas remains constant as initially as well as finally the piston is exposed to atmospheric pressure. Thus we have from Charles and Gay Loussac law $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
Here, $V_{2} = \frac{V_{1} T_{2}}{T_{1}} = 2 V$
$[As$ $V_{1} = V, T_{1} = T$ and $T_{2} = 2 T]$
Thus after increasing the gas temperature to 27, its volume becomes $2 V$ . Now the piston is clamped that means, now the volume of gas remains constant. Again its temperature is doubled from $2 T$ to $4 T$ , let the pressure changes from $P_{0}$ to $P^{'}$ . Thus we have
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
or $P_{2} = P^{'} = \frac{P_{1} T_{2}}{T_{1}}$
$= 2 P_{0} [As P_{1} = P_{0}, T_{1} = 2 T and T_{2} = 4 T]$
Thus in final state gas pressure becomes $2 P_{0}$ .

PHXI13:KINETIC THEORY

360074 Find the minimum attainable pressure of ideal gas in the process $T = T_{0} + α V^{2}$ , where $T_{0}$ , and $α$ are $+ v e$ constants and $V$ is the volume of one mole of gas.

1

R \sqrt{T_{0}}

2

4 R \sqrt{T_{0}}

3

2 R \sqrt{α T_{0}}

4

\frac{\sqrt{α T_{0}}}{R}

Explanation:

Given that
$T = T_{0} + α V^{2}$
For one mole of the gas, $V = R T / p$
$T = T_{0} + α (\frac{R^{2} T^{2}}{p^{2}})$
Solving it for $p,$ we get
$p = \sqrt{α} R T {(T - T_{0})}^{- 1 / 2} (1)$
Differentiating it with respect to $T$ , we get
$\frac{d p}{d T} = \sqrt{α} \cdot R [{(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2}]$
Conditions for minimum pressure is that
$(d p / d T) = 0 . Hence,$
$\begin{matrix} \sqrt{α} R [{(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2}] = 0 \\ {(T - T_{0})}^{- 1 / 2} - \frac{1}{2} T {(T - T_{0})}^{- 3 / 2} = 0 (∵ \sqrt{α} R \neq 0) \end{matrix}$
Now, Solving we get,
$T = 2 T_{0}$
Substituting the value of $T = 2 T_{0}$ in Eq. (1) we get
$p_{min} = \sqrt{α} R (2 T_{0}) {[2 T_{0} - T_{0}]}^{- 1 / 2}$
$o r p_{min} = 2 R \sqrt{(α T_{0})}$

PHXI13:KINETIC THEORY

360076 From the following $P - T$ graph what inference can be drawn
supporting img

1

V_{2} < V_{1}

2

V_{2} > V_{1}

3

V_{2} = V_{1}

4 None of these

Explanation:

As
$θ_{2} > θ_{1} \Rightarrow \tan θ_{2} > \tan θ_{1} \Rightarrow {(\frac{T}{P})}_{2} > {(\frac{T}{P})}_{1}$
From $P V = n R T; \frac{T}{P} \propto V \Rightarrow V_{2} > V_{1}$ .

PHXI13:KINETIC THEORY

360077 The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at $1.38 a t m$ is
(Given, $k = 1.38 \times 10^{- 23} J K^{- 1}$ ) is

1

100 K

2

300 K

3

500 K

4

200 K

Explanation:

Given, $N = 2 \times 10^{25} / m^{3}$
$P = 1.38 a t m, k = 1.38 \times 10^{- 23} J / K$
By ideal gas equation
$P V = n R T$
$\Rightarrow P V = \frac{N}{N_{A}} R T$
$\Rightarrow P = \frac{N}{V} \cdot k T$
$\Rightarrow 1.38 \times 1.01 \times 10^{5}$
$= 2 \times 10^{25} \times 1.38 \times 10^{- 23} \times T$
$\Rightarrow T ≃ 500 K$
So, correct option is (3).

JEE - 2024