QBManager

Thermodynamics

148334 When a system is taken from state $i$ to state $f$ along path ia $f$ in the figure, the heat absorbed $Q=50$ cal and the work done $W=20$ cal. If $W$ $=-13$ cal for the return path $f i, Q$ for this path is

1 $17 \mathrm{cal}$

2 $-17 \mathrm{cal}$

3 $43 \mathrm{cal}$

4 $-43 \mathrm{cal}$

AMU-2008

Thermodynamics

148335 The ratio of the slopes of isothermal and adiabatic curves is

1 1

2 $\gamma$

3 $\frac{1}{\gamma}$

4 $\frac{3}{2}$

AP EAMCET (17.09.2020) Shift-I

Thermodynamics

148336 One mole of an diatomic ideal gas undergoes a process shown in P-V diagram. The total heat given to the gas $(\ell \mathbf{n} 2=0.7)$ is

1 $2.5 \mathrm{P}_{0} \mathrm{~V}_{0}$

2 $3.9 \mathrm{P}_{0} \mathrm{~V}_{0}$

3 $1.1 \mathrm{P}_{0} \mathrm{~V}_{0}$

4 $1.4 \mathrm{P}_{0} \mathrm{~V}_{0}$

Explanation:

B
We know that ideal gas equation,
$\mathrm{PV}=\mathrm{nRT}$
$\mathrm{P}=\left(\frac{\mathrm{nR}}{\mathrm{V}}\right) \mathrm{T}$
$\mathrm{P} \propto \mathrm{T}$
$\mathrm{P}_{0} \rightarrow \mathrm{T}_{0}$ (A)
$2 \mathrm{P}_{0} \rightarrow 2 \mathrm{~T}_{0}$ (B)
Heat supplied,
$\mathrm{Q}_{\mathrm{AB}}=\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{P} \Delta \mathrm{V}$
for (constant volume) $\Delta \mathrm{V}=0$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{R}\left[\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right]$ For diatomic $\mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{RT}_{0}$
Again $\mathrm{PV}=\mathrm{nRT}$
$\mathrm{PV}=\mathrm{RT}$
(For one mole gas)
$\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{RT}_{0}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
Heat supplied $\rightarrow$ B to C
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=0+(1) \mathrm{R}\left(2 \mathrm{~T}_{0}\right) \ln \left(\frac{2 \mathrm{~V}_{0}}{\mathrm{~V}_{0}}\right)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{RT}_{0} \ln (2)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{P}_{0} \mathrm{~V}_{0} \ln (2)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)$
Total heat
$\mathrm{Q}=\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}$
$\mathrm{Q}=\frac{5}{2} \quad \mathrm{P}_{0} \mathrm{~V}_{0}+2 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)$
$\mathrm{Q}=\frac{5 \mathrm{P}_{0} \mathrm{~V}_{0}+4 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)}{2}$
$\mathrm{Q}=\frac{5 \mathrm{P}_{0} \mathrm{~V}_{0}+2.8 \mathrm{P}_{0} \mathrm{~V}_{0}}{2}$
$\mathrm{Q}=\frac{7.8 \mathrm{P}_{0} \mathrm{~V}_{0}}{2}$
$\mathrm{Q}=3.9 \mathrm{P}_{0} \mathrm{~V}_{0}$

WB JEE 2022

Thermodynamics

148338 A gas system is taken through the thermodynamic cyclic process $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$ as shown below. The amount of heat transfer

1 $-\mathrm{P} \frac{\mathrm{V}}{2}$

2 PV

3 $\frac{\mathrm{PV}}{2}$

4 $\frac{-3 \mathrm{PV}}{2}$

TS EAMCET 20.07.2022

Thermodynamics

148334 When a system is taken from state $i$ to state $f$ along path ia $f$ in the figure, the heat absorbed $Q=50$ cal and the work done $W=20$ cal. If $W$ $=-13$ cal for the return path $f i, Q$ for this path is

1 $17 \mathrm{cal}$

2 $-17 \mathrm{cal}$

3 $43 \mathrm{cal}$

4 $-43 \mathrm{cal}$

AMU-2008

Thermodynamics

148335 The ratio of the slopes of isothermal and adiabatic curves is

1 1

2 $\gamma$

3 $\frac{1}{\gamma}$

4 $\frac{3}{2}$

AP EAMCET (17.09.2020) Shift-I

Thermodynamics

148336 One mole of an diatomic ideal gas undergoes a process shown in P-V diagram. The total heat given to the gas $(\ell \mathbf{n} 2=0.7)$ is

1 $2.5 \mathrm{P}_{0} \mathrm{~V}_{0}$

2 $3.9 \mathrm{P}_{0} \mathrm{~V}_{0}$

3 $1.1 \mathrm{P}_{0} \mathrm{~V}_{0}$

4 $1.4 \mathrm{P}_{0} \mathrm{~V}_{0}$

Explanation:

B
We know that ideal gas equation,
$\mathrm{PV}=\mathrm{nRT}$
$\mathrm{P}=\left(\frac{\mathrm{nR}}{\mathrm{V}}\right) \mathrm{T}$
$\mathrm{P} \propto \mathrm{T}$
$\mathrm{P}_{0} \rightarrow \mathrm{T}_{0}$ (A)
$2 \mathrm{P}_{0} \rightarrow 2 \mathrm{~T}_{0}$ (B)
Heat supplied,
$\mathrm{Q}_{\mathrm{AB}}=\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{P} \Delta \mathrm{V}$
for (constant volume) $\Delta \mathrm{V}=0$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{R}\left[\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right]$ For diatomic $\mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{RT}_{0}$
Again $\mathrm{PV}=\mathrm{nRT}$
$\mathrm{PV}=\mathrm{RT}$
(For one mole gas)
$\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{RT}_{0}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
Heat supplied $\rightarrow$ B to C
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=0+(1) \mathrm{R}\left(2 \mathrm{~T}_{0}\right) \ln \left(\frac{2 \mathrm{~V}_{0}}{\mathrm{~V}_{0}}\right)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{RT}_{0} \ln (2)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{P}_{0} \mathrm{~V}_{0} \ln (2)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)$
Total heat
$\mathrm{Q}=\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}$
$\mathrm{Q}=\frac{5}{2} \quad \mathrm{P}_{0} \mathrm{~V}_{0}+2 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)$
$\mathrm{Q}=\frac{5 \mathrm{P}_{0} \mathrm{~V}_{0}+4 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)}{2}$
$\mathrm{Q}=\frac{5 \mathrm{P}_{0} \mathrm{~V}_{0}+2.8 \mathrm{P}_{0} \mathrm{~V}_{0}}{2}$
$\mathrm{Q}=\frac{7.8 \mathrm{P}_{0} \mathrm{~V}_{0}}{2}$
$\mathrm{Q}=3.9 \mathrm{P}_{0} \mathrm{~V}_{0}$

WB JEE 2022

Thermodynamics

148338 A gas system is taken through the thermodynamic cyclic process $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$ as shown below. The amount of heat transfer

1 $-\mathrm{P} \frac{\mathrm{V}}{2}$

2 PV

3 $\frac{\mathrm{PV}}{2}$

4 $\frac{-3 \mathrm{PV}}{2}$

TS EAMCET 20.07.2022

Thermodynamics

148334 When a system is taken from state $i$ to state $f$ along path ia $f$ in the figure, the heat absorbed $Q=50$ cal and the work done $W=20$ cal. If $W$ $=-13$ cal for the return path $f i, Q$ for this path is

1 $17 \mathrm{cal}$

2 $-17 \mathrm{cal}$

3 $43 \mathrm{cal}$

4 $-43 \mathrm{cal}$

AMU-2008

Thermodynamics

148335 The ratio of the slopes of isothermal and adiabatic curves is

1 1

2 $\gamma$

3 $\frac{1}{\gamma}$

4 $\frac{3}{2}$

AP EAMCET (17.09.2020) Shift-I

Thermodynamics

148336 One mole of an diatomic ideal gas undergoes a process shown in P-V diagram. The total heat given to the gas $(\ell \mathbf{n} 2=0.7)$ is

1 $2.5 \mathrm{P}_{0} \mathrm{~V}_{0}$

2 $3.9 \mathrm{P}_{0} \mathrm{~V}_{0}$

3 $1.1 \mathrm{P}_{0} \mathrm{~V}_{0}$

4 $1.4 \mathrm{P}_{0} \mathrm{~V}_{0}$

Explanation:

B
We know that ideal gas equation,
$\mathrm{PV}=\mathrm{nRT}$
$\mathrm{P}=\left(\frac{\mathrm{nR}}{\mathrm{V}}\right) \mathrm{T}$
$\mathrm{P} \propto \mathrm{T}$
$\mathrm{P}_{0} \rightarrow \mathrm{T}_{0}$ (A)
$2 \mathrm{P}_{0} \rightarrow 2 \mathrm{~T}_{0}$ (B)
Heat supplied,
$\mathrm{Q}_{\mathrm{AB}}=\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{P} \Delta \mathrm{V}$
for (constant volume) $\Delta \mathrm{V}=0$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{R}\left[\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right]$ For diatomic $\mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{RT}_{0}$
Again $\mathrm{PV}=\mathrm{nRT}$
$\mathrm{PV}=\mathrm{RT}$
(For one mole gas)
$\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{RT}_{0}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
Heat supplied $\rightarrow$ B to C
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=0+(1) \mathrm{R}\left(2 \mathrm{~T}_{0}\right) \ln \left(\frac{2 \mathrm{~V}_{0}}{\mathrm{~V}_{0}}\right)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{RT}_{0} \ln (2)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{P}_{0} \mathrm{~V}_{0} \ln (2)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)$
Total heat
$\mathrm{Q}=\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}$
$\mathrm{Q}=\frac{5}{2} \quad \mathrm{P}_{0} \mathrm{~V}_{0}+2 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)$
$\mathrm{Q}=\frac{5 \mathrm{P}_{0} \mathrm{~V}_{0}+4 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)}{2}$
$\mathrm{Q}=\frac{5 \mathrm{P}_{0} \mathrm{~V}_{0}+2.8 \mathrm{P}_{0} \mathrm{~V}_{0}}{2}$
$\mathrm{Q}=\frac{7.8 \mathrm{P}_{0} \mathrm{~V}_{0}}{2}$
$\mathrm{Q}=3.9 \mathrm{P}_{0} \mathrm{~V}_{0}$

WB JEE 2022

Thermodynamics

148338 A gas system is taken through the thermodynamic cyclic process $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$ as shown below. The amount of heat transfer

1 $-\mathrm{P} \frac{\mathrm{V}}{2}$

2 PV

3 $\frac{\mathrm{PV}}{2}$

4 $\frac{-3 \mathrm{PV}}{2}$

TS EAMCET 20.07.2022

Thermodynamics

148334 When a system is taken from state $i$ to state $f$ along path ia $f$ in the figure, the heat absorbed $Q=50$ cal and the work done $W=20$ cal. If $W$ $=-13$ cal for the return path $f i, Q$ for this path is

1 $17 \mathrm{cal}$

2 $-17 \mathrm{cal}$

3 $43 \mathrm{cal}$

4 $-43 \mathrm{cal}$

AMU-2008

Thermodynamics

148335 The ratio of the slopes of isothermal and adiabatic curves is

1 1

2 $\gamma$

3 $\frac{1}{\gamma}$

4 $\frac{3}{2}$

AP EAMCET (17.09.2020) Shift-I

Thermodynamics

148336 One mole of an diatomic ideal gas undergoes a process shown in P-V diagram. The total heat given to the gas $(\ell \mathbf{n} 2=0.7)$ is

1 $2.5 \mathrm{P}_{0} \mathrm{~V}_{0}$

2 $3.9 \mathrm{P}_{0} \mathrm{~V}_{0}$

3 $1.1 \mathrm{P}_{0} \mathrm{~V}_{0}$

4 $1.4 \mathrm{P}_{0} \mathrm{~V}_{0}$

Explanation:

B
We know that ideal gas equation,
$\mathrm{PV}=\mathrm{nRT}$
$\mathrm{P}=\left(\frac{\mathrm{nR}}{\mathrm{V}}\right) \mathrm{T}$
$\mathrm{P} \propto \mathrm{T}$
$\mathrm{P}_{0} \rightarrow \mathrm{T}_{0}$ (A)
$2 \mathrm{P}_{0} \rightarrow 2 \mathrm{~T}_{0}$ (B)
Heat supplied,
$\mathrm{Q}_{\mathrm{AB}}=\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{P} \Delta \mathrm{V}$
for (constant volume) $\Delta \mathrm{V}=0$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{R}\left[\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right]$ For diatomic $\mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{RT}_{0}$
Again $\mathrm{PV}=\mathrm{nRT}$
$\mathrm{PV}=\mathrm{RT}$
(For one mole gas)
$\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{RT}_{0}$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{5}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
Heat supplied $\rightarrow$ B to C
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=0+(1) \mathrm{R}\left(2 \mathrm{~T}_{0}\right) \ln \left(\frac{2 \mathrm{~V}_{0}}{\mathrm{~V}_{0}}\right)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{RT}_{0} \ln (2)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{P}_{0} \mathrm{~V}_{0} \ln (2)$
$\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}=2 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)$
Total heat
$\mathrm{Q}=\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}$
$\mathrm{Q}=\frac{5}{2} \quad \mathrm{P}_{0} \mathrm{~V}_{0}+2 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)$
$\mathrm{Q}=\frac{5 \mathrm{P}_{0} \mathrm{~V}_{0}+4 \mathrm{P}_{0} \mathrm{~V}_{0}(0.7)}{2}$
$\mathrm{Q}=\frac{5 \mathrm{P}_{0} \mathrm{~V}_{0}+2.8 \mathrm{P}_{0} \mathrm{~V}_{0}}{2}$
$\mathrm{Q}=\frac{7.8 \mathrm{P}_{0} \mathrm{~V}_{0}}{2}$
$\mathrm{Q}=3.9 \mathrm{P}_{0} \mathrm{~V}_{0}$

WB JEE 2022

Thermodynamics

148338 A gas system is taken through the thermodynamic cyclic process $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$ as shown below. The amount of heat transfer

1 $-\mathrm{P} \frac{\mathrm{V}}{2}$

2 PV

3 $\frac{\mathrm{PV}}{2}$

4 $\frac{-3 \mathrm{PV}}{2}$

TS EAMCET 20.07.2022