QBManager

Kinetic Theory of Gases

138978 A certain amount of gas of volume $V$ at $27^{\circ}$ temperature and pressure $2 \times 10^{7} {Nm}^{- 2}$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (User $γ = 1.5$ )

1

3.536 \times 10^{5} Pa

2

3.536 \times 10^{6} Pa

3

1.25 \times 10^{6} Pa

4

1.25 \times 10^{5} Pa

Explanation:

B Given that,
$P_{1} = 2 \times 10^{7} N / m^{2}, V_{1} = V, T = 27^{\circ} C = 300 K$
$P_{2} = ? V_{2} = 2 V$
For,
isothermal process-
$PV = constant,$
$P_{1} V_{1} = P_{2} V_{2}$
$P_{2} = \frac{P_{1} V_{1}}{V_{2}} = \frac{2 \times 10^{7} \times V}{2 V}$
$P_{2} = 10^{7}$
Similearlly in Adiabatic process-
$P_{3} = ?, V_{3} = 4 V, P_{2} = 10^{7}, V_{2} = 2 V$
${PV}^{γ} = constant$
$P_{2} V_{2}^{γ} = P_{3} V_{3}^{γ}$
$P_{3} = P_{2} {(\frac{V_{2}}{V_{3}})}^{γ} {∵ γ = 1.5 = 3 / 2}$
$P_{3} = 10^{7} {(\frac{2 V}{4 V})}^{3 / 2} \Rightarrow 10^{7} \times {(\frac{1}{2})}^{3 / 2}$
$P_{3} = 3.536 \times 10^{6} Pascal$

Shift-I]

Kinetic Theory of Gases

138979 The net work done by an ideal gas going through the cycle as shown in the $P - V$ diagram below is

1 0

2

P_{1} V_{1}

3

\frac{3}{2} P_{1} V_{1}

4

\frac{1}{2} P_{1} V_{1}

Explanation:

B In the P-V diagram -

Net work done is equal to the area of triangle.
So, $W = \frac{1}{2} \times$ Base $\times$ height
$W = \frac{1}{2} \times (2 V_{1} - V_{1}) (3 P_{1} - P_{1})$
$W = \frac{1}{2} \times V_{1} \times 2 P_{1}$
$W = P_{1} V_{1}$

Shift-II

Kinetic Theory of Gases

138980 An ideal gas at pressure $P$ is enclosed in a container that is placed in a reservoir at constant temperature $T$ . If the volume of the gas is increased to two times its original value, then the new pressure $P^{'} =$ P

1

\frac{1}{2}

2 2

3 1

4 Cannot be determined

Explanation:

A Given that, $V_{1} = V, V^{'} = 2 V, P_{1} = P, P^{'} =$ ?
According to Boyle's law -
$P \propto \frac{1}{V}$
$P_{1} V_{1} = P^{'} V^{'}$
$P^{'} = \frac{P_{1} V_{1}}{V^{'}} = \frac{PV}{2 V}$
$P^{'} = P / 2$
$P^{'} = 1 / 2 P$

Shift-II

Kinetic Theory of Gases

138981 For a monoatomic ideal gas following the cyclic process ABCA shown in the $U$ vs $ρ$ plot, identify the incorrect option.

1 Molar heat capacity of the process

AB

is

\frac{R}{2}

2 Heat is rejected by the system in path

BC

3 Molar heat capacity for the process

BC

is

\frac{2 R}{3}

4 Work done by the system in the process CA is

\frac{2 U_{\circ}}{3} \ln 4

Explanation:

C For process $AB$
$TV =$ constant
${PV}^{2} =$ constant
Comparing with polytropic equation ${PV}^{m} =$ constant $m = 2$
for option (a)
molar heat capacity $C = \frac{R}{γ - 1} - \frac{R}{m - 1}, γ = \frac{5}{3}$
$C = \frac{3 R}{2} - R = \frac{R}{2}$
Hence option (a) is correct.
For option (b)
The process is isochoric and temperature is decreasing. Hence (b) is correct.
For option (c)
$C_{V} = \frac{3}{2} R$
Hence (c) is incorrect.
For option (d).
The process is isothermal-
$W = \frac{2}{3} (\frac{3}{2} nRT) \ln (4) = \frac{2}{3} U_{0} \ln (4)$

Shift-I]

Kinetic Theory of Gases

138978 A certain amount of gas of volume $V$ at $27^{\circ}$ temperature and pressure $2 \times 10^{7} {Nm}^{- 2}$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (User $γ = 1.5$ )

1

3.536 \times 10^{5} Pa

2

3.536 \times 10^{6} Pa

3

1.25 \times 10^{6} Pa

4

1.25 \times 10^{5} Pa

Explanation:

B Given that,
$P_{1} = 2 \times 10^{7} N / m^{2}, V_{1} = V, T = 27^{\circ} C = 300 K$
$P_{2} = ? V_{2} = 2 V$
For,
isothermal process-
$PV = constant,$
$P_{1} V_{1} = P_{2} V_{2}$
$P_{2} = \frac{P_{1} V_{1}}{V_{2}} = \frac{2 \times 10^{7} \times V}{2 V}$
$P_{2} = 10^{7}$
Similearlly in Adiabatic process-
$P_{3} = ?, V_{3} = 4 V, P_{2} = 10^{7}, V_{2} = 2 V$
${PV}^{γ} = constant$
$P_{2} V_{2}^{γ} = P_{3} V_{3}^{γ}$
$P_{3} = P_{2} {(\frac{V_{2}}{V_{3}})}^{γ} {∵ γ = 1.5 = 3 / 2}$
$P_{3} = 10^{7} {(\frac{2 V}{4 V})}^{3 / 2} \Rightarrow 10^{7} \times {(\frac{1}{2})}^{3 / 2}$
$P_{3} = 3.536 \times 10^{6} Pascal$

Shift-I]

Kinetic Theory of Gases

138979 The net work done by an ideal gas going through the cycle as shown in the $P - V$ diagram below is

1 0

2

P_{1} V_{1}

3

\frac{3}{2} P_{1} V_{1}

4

\frac{1}{2} P_{1} V_{1}

Explanation:

B In the P-V diagram -

Net work done is equal to the area of triangle.
So, $W = \frac{1}{2} \times$ Base $\times$ height
$W = \frac{1}{2} \times (2 V_{1} - V_{1}) (3 P_{1} - P_{1})$
$W = \frac{1}{2} \times V_{1} \times 2 P_{1}$
$W = P_{1} V_{1}$

Shift-II

Kinetic Theory of Gases

138980 An ideal gas at pressure $P$ is enclosed in a container that is placed in a reservoir at constant temperature $T$ . If the volume of the gas is increased to two times its original value, then the new pressure $P^{'} =$ P

1

\frac{1}{2}

2 2

3 1

4 Cannot be determined

Explanation:

A Given that, $V_{1} = V, V^{'} = 2 V, P_{1} = P, P^{'} =$ ?
According to Boyle's law -
$P \propto \frac{1}{V}$
$P_{1} V_{1} = P^{'} V^{'}$
$P^{'} = \frac{P_{1} V_{1}}{V^{'}} = \frac{PV}{2 V}$
$P^{'} = P / 2$
$P^{'} = 1 / 2 P$

Shift-II

Kinetic Theory of Gases

138981 For a monoatomic ideal gas following the cyclic process ABCA shown in the $U$ vs $ρ$ plot, identify the incorrect option.

1 Molar heat capacity of the process

AB

is

\frac{R}{2}

2 Heat is rejected by the system in path

BC

3 Molar heat capacity for the process

BC

is

\frac{2 R}{3}

4 Work done by the system in the process CA is

\frac{2 U_{\circ}}{3} \ln 4

Explanation:

C For process $AB$
$TV =$ constant
${PV}^{2} =$ constant
Comparing with polytropic equation ${PV}^{m} =$ constant $m = 2$
for option (a)
molar heat capacity $C = \frac{R}{γ - 1} - \frac{R}{m - 1}, γ = \frac{5}{3}$
$C = \frac{3 R}{2} - R = \frac{R}{2}$
Hence option (a) is correct.
For option (b)
The process is isochoric and temperature is decreasing. Hence (b) is correct.
For option (c)
$C_{V} = \frac{3}{2} R$
Hence (c) is incorrect.
For option (d).
The process is isothermal-
$W = \frac{2}{3} (\frac{3}{2} nRT) \ln (4) = \frac{2}{3} U_{0} \ln (4)$

Shift-I]

Kinetic Theory of Gases

138978 A certain amount of gas of volume $V$ at $27^{\circ}$ temperature and pressure $2 \times 10^{7} {Nm}^{- 2}$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (User $γ = 1.5$ )

1

3.536 \times 10^{5} Pa

2

3.536 \times 10^{6} Pa

3

1.25 \times 10^{6} Pa

4

1.25 \times 10^{5} Pa

Explanation:

B Given that,
$P_{1} = 2 \times 10^{7} N / m^{2}, V_{1} = V, T = 27^{\circ} C = 300 K$
$P_{2} = ? V_{2} = 2 V$
For,
isothermal process-
$PV = constant,$
$P_{1} V_{1} = P_{2} V_{2}$
$P_{2} = \frac{P_{1} V_{1}}{V_{2}} = \frac{2 \times 10^{7} \times V}{2 V}$
$P_{2} = 10^{7}$
Similearlly in Adiabatic process-
$P_{3} = ?, V_{3} = 4 V, P_{2} = 10^{7}, V_{2} = 2 V$
${PV}^{γ} = constant$
$P_{2} V_{2}^{γ} = P_{3} V_{3}^{γ}$
$P_{3} = P_{2} {(\frac{V_{2}}{V_{3}})}^{γ} {∵ γ = 1.5 = 3 / 2}$
$P_{3} = 10^{7} {(\frac{2 V}{4 V})}^{3 / 2} \Rightarrow 10^{7} \times {(\frac{1}{2})}^{3 / 2}$
$P_{3} = 3.536 \times 10^{6} Pascal$

Shift-I]

Kinetic Theory of Gases

138979 The net work done by an ideal gas going through the cycle as shown in the $P - V$ diagram below is

1 0

2

P_{1} V_{1}

3

\frac{3}{2} P_{1} V_{1}

4

\frac{1}{2} P_{1} V_{1}

Explanation:

B In the P-V diagram -

Net work done is equal to the area of triangle.
So, $W = \frac{1}{2} \times$ Base $\times$ height
$W = \frac{1}{2} \times (2 V_{1} - V_{1}) (3 P_{1} - P_{1})$
$W = \frac{1}{2} \times V_{1} \times 2 P_{1}$
$W = P_{1} V_{1}$

Shift-II

Kinetic Theory of Gases

138980 An ideal gas at pressure $P$ is enclosed in a container that is placed in a reservoir at constant temperature $T$ . If the volume of the gas is increased to two times its original value, then the new pressure $P^{'} =$ P

1

\frac{1}{2}

2 2

3 1

4 Cannot be determined

Explanation:

A Given that, $V_{1} = V, V^{'} = 2 V, P_{1} = P, P^{'} =$ ?
According to Boyle's law -
$P \propto \frac{1}{V}$
$P_{1} V_{1} = P^{'} V^{'}$
$P^{'} = \frac{P_{1} V_{1}}{V^{'}} = \frac{PV}{2 V}$
$P^{'} = P / 2$
$P^{'} = 1 / 2 P$

Shift-II

Kinetic Theory of Gases

138981 For a monoatomic ideal gas following the cyclic process ABCA shown in the $U$ vs $ρ$ plot, identify the incorrect option.

1 Molar heat capacity of the process

AB

is

\frac{R}{2}

2 Heat is rejected by the system in path

BC

3 Molar heat capacity for the process

BC

is

\frac{2 R}{3}

4 Work done by the system in the process CA is

\frac{2 U_{\circ}}{3} \ln 4

Explanation:

C For process $AB$
$TV =$ constant
${PV}^{2} =$ constant
Comparing with polytropic equation ${PV}^{m} =$ constant $m = 2$
for option (a)
molar heat capacity $C = \frac{R}{γ - 1} - \frac{R}{m - 1}, γ = \frac{5}{3}$
$C = \frac{3 R}{2} - R = \frac{R}{2}$
Hence option (a) is correct.
For option (b)
The process is isochoric and temperature is decreasing. Hence (b) is correct.
For option (c)
$C_{V} = \frac{3}{2} R$
Hence (c) is incorrect.
For option (d).
The process is isothermal-
$W = \frac{2}{3} (\frac{3}{2} nRT) \ln (4) = \frac{2}{3} U_{0} \ln (4)$

Shift-I]

Kinetic Theory of Gases

138978 A certain amount of gas of volume $V$ at $27^{\circ}$ temperature and pressure $2 \times 10^{7} {Nm}^{- 2}$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (User $γ = 1.5$ )

1

3.536 \times 10^{5} Pa

2

3.536 \times 10^{6} Pa

3

1.25 \times 10^{6} Pa

4

1.25 \times 10^{5} Pa

Explanation:

B Given that,
$P_{1} = 2 \times 10^{7} N / m^{2}, V_{1} = V, T = 27^{\circ} C = 300 K$
$P_{2} = ? V_{2} = 2 V$
For,
isothermal process-
$PV = constant,$
$P_{1} V_{1} = P_{2} V_{2}$
$P_{2} = \frac{P_{1} V_{1}}{V_{2}} = \frac{2 \times 10^{7} \times V}{2 V}$
$P_{2} = 10^{7}$
Similearlly in Adiabatic process-
$P_{3} = ?, V_{3} = 4 V, P_{2} = 10^{7}, V_{2} = 2 V$
${PV}^{γ} = constant$
$P_{2} V_{2}^{γ} = P_{3} V_{3}^{γ}$
$P_{3} = P_{2} {(\frac{V_{2}}{V_{3}})}^{γ} {∵ γ = 1.5 = 3 / 2}$
$P_{3} = 10^{7} {(\frac{2 V}{4 V})}^{3 / 2} \Rightarrow 10^{7} \times {(\frac{1}{2})}^{3 / 2}$
$P_{3} = 3.536 \times 10^{6} Pascal$

Shift-I]

Kinetic Theory of Gases

138979 The net work done by an ideal gas going through the cycle as shown in the $P - V$ diagram below is

1 0

2

P_{1} V_{1}

3

\frac{3}{2} P_{1} V_{1}

4

\frac{1}{2} P_{1} V_{1}

Explanation:

B In the P-V diagram -

Net work done is equal to the area of triangle.
So, $W = \frac{1}{2} \times$ Base $\times$ height
$W = \frac{1}{2} \times (2 V_{1} - V_{1}) (3 P_{1} - P_{1})$
$W = \frac{1}{2} \times V_{1} \times 2 P_{1}$
$W = P_{1} V_{1}$

Shift-II

Kinetic Theory of Gases

138980 An ideal gas at pressure $P$ is enclosed in a container that is placed in a reservoir at constant temperature $T$ . If the volume of the gas is increased to two times its original value, then the new pressure $P^{'} =$ P

1

\frac{1}{2}

2 2

3 1

4 Cannot be determined

Explanation:

A Given that, $V_{1} = V, V^{'} = 2 V, P_{1} = P, P^{'} =$ ?
According to Boyle's law -
$P \propto \frac{1}{V}$
$P_{1} V_{1} = P^{'} V^{'}$
$P^{'} = \frac{P_{1} V_{1}}{V^{'}} = \frac{PV}{2 V}$
$P^{'} = P / 2$
$P^{'} = 1 / 2 P$

Shift-II

Kinetic Theory of Gases

138981 For a monoatomic ideal gas following the cyclic process ABCA shown in the $U$ vs $ρ$ plot, identify the incorrect option.

1 Molar heat capacity of the process

AB

is

\frac{R}{2}

2 Heat is rejected by the system in path

BC

3 Molar heat capacity for the process

BC

is

\frac{2 R}{3}

4 Work done by the system in the process CA is

\frac{2 U_{\circ}}{3} \ln 4

Explanation:

C For process $AB$
$TV =$ constant
${PV}^{2} =$ constant
Comparing with polytropic equation ${PV}^{m} =$ constant $m = 2$
for option (a)
molar heat capacity $C = \frac{R}{γ - 1} - \frac{R}{m - 1}, γ = \frac{5}{3}$
$C = \frac{3 R}{2} - R = \frac{R}{2}$
Hence option (a) is correct.
For option (b)
The process is isochoric and temperature is decreasing. Hence (b) is correct.
For option (c)
$C_{V} = \frac{3}{2} R$
Hence (c) is incorrect.
For option (d).
The process is isothermal-
$W = \frac{2}{3} (\frac{3}{2} nRT) \ln (4) = \frac{2}{3} U_{0} \ln (4)$

Shift-I]

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: