QBManager

PHXI12:THERMODYNAMICS

371288 For a certain mass of gas, the isothermal curves between $p$ and $V$ at $T_{1}$ and $T_{2}$ temperatures are 1 and 2 as shown in figure. Then,
supporting img

1

T_{1} = T_{2}

2

T_{1} < T_{2}

3

T_{1} > T_{2}

4 Nothing can be predicted

Explanation:

In isothermal process, curves between $p$ and $V$ is such that $T_{2}$ is farther from the origin than $T_{1}$ . Therefore, $T_{2} > T_{1}$ .

PHXI12:THERMODYNAMICS

371289 $P - V$ plots for two gases during adiabatic process are shown in the figure. Plots 1 and 2 should correspond respectively, to
supporting img

1

H e

and

N_{2}

2

N_{2}

and

H e

3

H e

and

A r

4

O_{2}

and

N_{2}

Explanation:

For an adiabatic process, $P V^{γ} =$ constant,
Differentiating, we have
$γ P V^{γ - 1} + \frac{d P}{d V} V^{γ} = 0$ or
$\frac{d P}{d V} = - \frac{γ P}{V}$
Thus, the slope of $P - V$ curve is proportional to $γ$ . Now, for a diatomic gas, $γ (= 7 / 5)$ is less then that for a monoatomic gas for which $γ = 5 / 3$ . Therefore, the slope of the $P - V$ curve is less for a diatomic gas than for a monoatomic gas. Hence curve 1 corresponds to diatomic gas ans curve 2 to monoatomic gas. Thus the correct choice is (2).

PHXI12:THERMODYNAMICS

371290 The PV diagram of $4 g m$ of $H e$ gas for a certain process $A$ to $B$ is shown in the figure. What is the heat given to the gas during the process $A$ to $B$ ?
supporting img

1

4 P_{0} V_{0}

2

6 P_{0} V_{0}

3

4.5 P_{0} V_{0}

4

2 P_{0} V_{0}

Explanation:

Number of moles $n = \frac{4 g}{4 g} = 1$
$\begin{aligned} W_{A B} = P_{0} V_{0} + \frac{1}{2} P_{0} V_{0} = \frac{3 P_{0} V_{0}}{2} \\ Δ U_{A B} = n c_{v} Δ T = \frac{3}{2} (2 P_{0} 2 V_{0} - P_{0} V_{0}) \\ Δ U_{A B} = \frac{9}{2} P_{0} V_{0} \\ Δ Q = W + Δ U_{A B} = 6 P_{0} V_{0} \end{aligned}$

PHXI12:THERMODYNAMICS

371291 Consider a process shown in the figure. During this process the work done by the system
supporting img

1 Continuously decreases

2 Continuously increases

3 First increases, then decreases

4 first decreases, then increases

PHXI12:THERMODYNAMICS

371292 Choose the correct statement for processes $A$ and $B$ shown in figure.
supporting img

1

P V^{γ} = k

for process

B

and

P V = k

for process

A

.

2

\frac{T^{γ}}{P^{γ - 1}} = k

for process

A

and

P V = k

for process

B

.

3

P V = k

for process

B

and

A

.

4

\frac{P^{γ - 1}}{T^{γ}} = k

for process

B

and

T = k

for process

A

.

Explanation:

Slope of adiabatic curve is greater than that of an isothermal curve
So, process $A$ is isothermal
$∴ P V = k$ for process $A$ and process $B$ is adiabatic
$∴ P V^{γ} = k$ for process $B$ .
supporting img

JEE - 2024

PHXI12:THERMODYNAMICS

371288 For a certain mass of gas, the isothermal curves between $p$ and $V$ at $T_{1}$ and $T_{2}$ temperatures are 1 and 2 as shown in figure. Then,
supporting img

1

T_{1} = T_{2}

2

T_{1} < T_{2}

3

T_{1} > T_{2}

4 Nothing can be predicted

Explanation:

In isothermal process, curves between $p$ and $V$ is such that $T_{2}$ is farther from the origin than $T_{1}$ . Therefore, $T_{2} > T_{1}$ .

PHXI12:THERMODYNAMICS

371289 $P - V$ plots for two gases during adiabatic process are shown in the figure. Plots 1 and 2 should correspond respectively, to
supporting img

1

H e

and

N_{2}

2

N_{2}

and

H e

3

H e

and

A r

4

O_{2}

and

N_{2}

Explanation:

For an adiabatic process, $P V^{γ} =$ constant,
Differentiating, we have
$γ P V^{γ - 1} + \frac{d P}{d V} V^{γ} = 0$ or
$\frac{d P}{d V} = - \frac{γ P}{V}$
Thus, the slope of $P - V$ curve is proportional to $γ$ . Now, for a diatomic gas, $γ (= 7 / 5)$ is less then that for a monoatomic gas for which $γ = 5 / 3$ . Therefore, the slope of the $P - V$ curve is less for a diatomic gas than for a monoatomic gas. Hence curve 1 corresponds to diatomic gas ans curve 2 to monoatomic gas. Thus the correct choice is (2).

PHXI12:THERMODYNAMICS

371290 The PV diagram of $4 g m$ of $H e$ gas for a certain process $A$ to $B$ is shown in the figure. What is the heat given to the gas during the process $A$ to $B$ ?
supporting img

1

4 P_{0} V_{0}

2

6 P_{0} V_{0}

3

4.5 P_{0} V_{0}

4

2 P_{0} V_{0}

Explanation:

Number of moles $n = \frac{4 g}{4 g} = 1$
$\begin{aligned} W_{A B} = P_{0} V_{0} + \frac{1}{2} P_{0} V_{0} = \frac{3 P_{0} V_{0}}{2} \\ Δ U_{A B} = n c_{v} Δ T = \frac{3}{2} (2 P_{0} 2 V_{0} - P_{0} V_{0}) \\ Δ U_{A B} = \frac{9}{2} P_{0} V_{0} \\ Δ Q = W + Δ U_{A B} = 6 P_{0} V_{0} \end{aligned}$

PHXI12:THERMODYNAMICS

371291 Consider a process shown in the figure. During this process the work done by the system
supporting img

1 Continuously decreases

2 Continuously increases

3 First increases, then decreases

4 first decreases, then increases

PHXI12:THERMODYNAMICS

371292 Choose the correct statement for processes $A$ and $B$ shown in figure.
supporting img

1

P V^{γ} = k

for process

B

and

P V = k

for process

A

.

2

\frac{T^{γ}}{P^{γ - 1}} = k

for process

A

and

P V = k

for process

B

.

3

P V = k

for process

B

and

A

.

4

\frac{P^{γ - 1}}{T^{γ}} = k

for process

B

and

T = k

for process

A

.

Explanation:

Slope of adiabatic curve is greater than that of an isothermal curve
So, process $A$ is isothermal
$∴ P V = k$ for process $A$ and process $B$ is adiabatic
$∴ P V^{γ} = k$ for process $B$ .
supporting img

JEE - 2024

PHXI12:THERMODYNAMICS

371288 For a certain mass of gas, the isothermal curves between $p$ and $V$ at $T_{1}$ and $T_{2}$ temperatures are 1 and 2 as shown in figure. Then,
supporting img

1

T_{1} = T_{2}

2

T_{1} < T_{2}

3

T_{1} > T_{2}

4 Nothing can be predicted

Explanation:

In isothermal process, curves between $p$ and $V$ is such that $T_{2}$ is farther from the origin than $T_{1}$ . Therefore, $T_{2} > T_{1}$ .

PHXI12:THERMODYNAMICS

371289 $P - V$ plots for two gases during adiabatic process are shown in the figure. Plots 1 and 2 should correspond respectively, to
supporting img

1

H e

and

N_{2}

2

N_{2}

and

H e

3

H e

and

A r

4

O_{2}

and

N_{2}

Explanation:

For an adiabatic process, $P V^{γ} =$ constant,
Differentiating, we have
$γ P V^{γ - 1} + \frac{d P}{d V} V^{γ} = 0$ or
$\frac{d P}{d V} = - \frac{γ P}{V}$
Thus, the slope of $P - V$ curve is proportional to $γ$ . Now, for a diatomic gas, $γ (= 7 / 5)$ is less then that for a monoatomic gas for which $γ = 5 / 3$ . Therefore, the slope of the $P - V$ curve is less for a diatomic gas than for a monoatomic gas. Hence curve 1 corresponds to diatomic gas ans curve 2 to monoatomic gas. Thus the correct choice is (2).

PHXI12:THERMODYNAMICS

371290 The PV diagram of $4 g m$ of $H e$ gas for a certain process $A$ to $B$ is shown in the figure. What is the heat given to the gas during the process $A$ to $B$ ?
supporting img

1

4 P_{0} V_{0}

2

6 P_{0} V_{0}

3

4.5 P_{0} V_{0}

4

2 P_{0} V_{0}

Explanation:

Number of moles $n = \frac{4 g}{4 g} = 1$
$\begin{aligned} W_{A B} = P_{0} V_{0} + \frac{1}{2} P_{0} V_{0} = \frac{3 P_{0} V_{0}}{2} \\ Δ U_{A B} = n c_{v} Δ T = \frac{3}{2} (2 P_{0} 2 V_{0} - P_{0} V_{0}) \\ Δ U_{A B} = \frac{9}{2} P_{0} V_{0} \\ Δ Q = W + Δ U_{A B} = 6 P_{0} V_{0} \end{aligned}$

PHXI12:THERMODYNAMICS

371291 Consider a process shown in the figure. During this process the work done by the system
supporting img

1 Continuously decreases

2 Continuously increases

3 First increases, then decreases

4 first decreases, then increases

PHXI12:THERMODYNAMICS

371292 Choose the correct statement for processes $A$ and $B$ shown in figure.
supporting img

1

P V^{γ} = k

for process

B

and

P V = k

for process

A

.

2

\frac{T^{γ}}{P^{γ - 1}} = k

for process

A

and

P V = k

for process

B

.

3

P V = k

for process

B

and

A

.

4

\frac{P^{γ - 1}}{T^{γ}} = k

for process

B

and

T = k

for process

A

.

Explanation:

Slope of adiabatic curve is greater than that of an isothermal curve
So, process $A$ is isothermal
$∴ P V = k$ for process $A$ and process $B$ is adiabatic
$∴ P V^{γ} = k$ for process $B$ .
supporting img

JEE - 2024

PHXI12:THERMODYNAMICS

371288 For a certain mass of gas, the isothermal curves between $p$ and $V$ at $T_{1}$ and $T_{2}$ temperatures are 1 and 2 as shown in figure. Then,
supporting img

1

T_{1} = T_{2}

2

T_{1} < T_{2}

3

T_{1} > T_{2}

4 Nothing can be predicted

Explanation:

In isothermal process, curves between $p$ and $V$ is such that $T_{2}$ is farther from the origin than $T_{1}$ . Therefore, $T_{2} > T_{1}$ .

PHXI12:THERMODYNAMICS

371289 $P - V$ plots for two gases during adiabatic process are shown in the figure. Plots 1 and 2 should correspond respectively, to
supporting img

1

H e

and

N_{2}

2

N_{2}

and

H e

3

H e

and

A r

4

O_{2}

and

N_{2}

Explanation:

For an adiabatic process, $P V^{γ} =$ constant,
Differentiating, we have
$γ P V^{γ - 1} + \frac{d P}{d V} V^{γ} = 0$ or
$\frac{d P}{d V} = - \frac{γ P}{V}$
Thus, the slope of $P - V$ curve is proportional to $γ$ . Now, for a diatomic gas, $γ (= 7 / 5)$ is less then that for a monoatomic gas for which $γ = 5 / 3$ . Therefore, the slope of the $P - V$ curve is less for a diatomic gas than for a monoatomic gas. Hence curve 1 corresponds to diatomic gas ans curve 2 to monoatomic gas. Thus the correct choice is (2).

PHXI12:THERMODYNAMICS

371290 The PV diagram of $4 g m$ of $H e$ gas for a certain process $A$ to $B$ is shown in the figure. What is the heat given to the gas during the process $A$ to $B$ ?
supporting img

1

4 P_{0} V_{0}

2

6 P_{0} V_{0}

3

4.5 P_{0} V_{0}

4

2 P_{0} V_{0}

Explanation:

Number of moles $n = \frac{4 g}{4 g} = 1$
$\begin{aligned} W_{A B} = P_{0} V_{0} + \frac{1}{2} P_{0} V_{0} = \frac{3 P_{0} V_{0}}{2} \\ Δ U_{A B} = n c_{v} Δ T = \frac{3}{2} (2 P_{0} 2 V_{0} - P_{0} V_{0}) \\ Δ U_{A B} = \frac{9}{2} P_{0} V_{0} \\ Δ Q = W + Δ U_{A B} = 6 P_{0} V_{0} \end{aligned}$

PHXI12:THERMODYNAMICS

371291 Consider a process shown in the figure. During this process the work done by the system
supporting img

1 Continuously decreases

2 Continuously increases

3 First increases, then decreases

4 first decreases, then increases

PHXI12:THERMODYNAMICS

371292 Choose the correct statement for processes $A$ and $B$ shown in figure.
supporting img

1

P V^{γ} = k

for process

B

and

P V = k

for process

A

.

2

\frac{T^{γ}}{P^{γ - 1}} = k

for process

A

and

P V = k

for process

B

.

3

P V = k

for process

B

and

A

.

4

\frac{P^{γ - 1}}{T^{γ}} = k

for process

B

and

T = k

for process

A

.

Explanation:

Slope of adiabatic curve is greater than that of an isothermal curve
So, process $A$ is isothermal
$∴ P V = k$ for process $A$ and process $B$ is adiabatic
$∴ P V^{γ} = k$ for process $B$ .
supporting img

JEE - 2024

PHXI12:THERMODYNAMICS

371288 For a certain mass of gas, the isothermal curves between $p$ and $V$ at $T_{1}$ and $T_{2}$ temperatures are 1 and 2 as shown in figure. Then,
supporting img

1

T_{1} = T_{2}

2

T_{1} < T_{2}

3

T_{1} > T_{2}

4 Nothing can be predicted

Explanation:

In isothermal process, curves between $p$ and $V$ is such that $T_{2}$ is farther from the origin than $T_{1}$ . Therefore, $T_{2} > T_{1}$ .

PHXI12:THERMODYNAMICS

371289 $P - V$ plots for two gases during adiabatic process are shown in the figure. Plots 1 and 2 should correspond respectively, to
supporting img

1

H e

and

N_{2}

2

N_{2}

and

H e

3

H e

and

A r

4

O_{2}

and

N_{2}

Explanation:

For an adiabatic process, $P V^{γ} =$ constant,
Differentiating, we have
$γ P V^{γ - 1} + \frac{d P}{d V} V^{γ} = 0$ or
$\frac{d P}{d V} = - \frac{γ P}{V}$
Thus, the slope of $P - V$ curve is proportional to $γ$ . Now, for a diatomic gas, $γ (= 7 / 5)$ is less then that for a monoatomic gas for which $γ = 5 / 3$ . Therefore, the slope of the $P - V$ curve is less for a diatomic gas than for a monoatomic gas. Hence curve 1 corresponds to diatomic gas ans curve 2 to monoatomic gas. Thus the correct choice is (2).

PHXI12:THERMODYNAMICS

371290 The PV diagram of $4 g m$ of $H e$ gas for a certain process $A$ to $B$ is shown in the figure. What is the heat given to the gas during the process $A$ to $B$ ?
supporting img

1

4 P_{0} V_{0}

2

6 P_{0} V_{0}

3

4.5 P_{0} V_{0}

4

2 P_{0} V_{0}

Explanation:

Number of moles $n = \frac{4 g}{4 g} = 1$
$\begin{aligned} W_{A B} = P_{0} V_{0} + \frac{1}{2} P_{0} V_{0} = \frac{3 P_{0} V_{0}}{2} \\ Δ U_{A B} = n c_{v} Δ T = \frac{3}{2} (2 P_{0} 2 V_{0} - P_{0} V_{0}) \\ Δ U_{A B} = \frac{9}{2} P_{0} V_{0} \\ Δ Q = W + Δ U_{A B} = 6 P_{0} V_{0} \end{aligned}$

PHXI12:THERMODYNAMICS

371291 Consider a process shown in the figure. During this process the work done by the system
supporting img

1 Continuously decreases

2 Continuously increases

3 First increases, then decreases

4 first decreases, then increases

PHXI12:THERMODYNAMICS

371292 Choose the correct statement for processes $A$ and $B$ shown in figure.
supporting img

1

P V^{γ} = k

for process

B

and

P V = k

for process

A

.

2

\frac{T^{γ}}{P^{γ - 1}} = k

for process

A

and

P V = k

for process

B

.

3

P V = k

for process

B

and

A

.

4

\frac{P^{γ - 1}}{T^{γ}} = k

for process

B

and

T = k

for process

A

.

Explanation:

Slope of adiabatic curve is greater than that of an isothermal curve
So, process $A$ is isothermal
$∴ P V = k$ for process $A$ and process $B$ is adiabatic
$∴ P V^{γ} = k$ for process $B$ .
supporting img

JEE - 2024