148126
Heat supplied in a process is used completely to do work against the external surroundings. Then, the process may be identified as which one of the following?
1 Adiabatic
2 Isochoric
3 Isothermal
4 None of the above
Explanation:
C In a thermodynamic process, heat supplied in a process is used completely in doing work against the external surroundings. Hence, this process may be designated as isothermal. \(\Delta \mathrm{U}=0=\mathrm{Q}-\mathrm{W}\) \(\therefore \mathrm{Q}=\mathrm{W}\) Heat added to the system to do work \(\mathrm{W}=\mathrm{nRT} \ln \left[\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right]\)
SCRA-2012
Thermodynamics
148127
What is the external work done (approximately) when $1 \mathrm{~g}$ of helium is heated through $1^{\circ} \mathrm{C}$ at constant pressure?
1 $1 \mathrm{~J}$
2 $2 \mathrm{~J}$
3 $4 \mathrm{~J}$
4 $8 \mathrm{~J}$
Explanation:
B Given, Temperature, $\mathrm{T}=1^{\circ} \mathrm{C}$ Mass, $\mathrm{m}=1 \mathrm{~g}$ No. of moles in $1 \mathrm{~g}$ of Helium, $\mathrm{n}=0.25$ mole $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ $\mathrm{W} =\int \mathrm{Pdv}$ $=\mathrm{nR} \times 1=\mathrm{nR}$ $=0.25 \times 8.314$ $\mathrm{W} \approx 2 \mathrm{~J}$
SCRA-2012
Thermodynamics
148129
What is the change in internal energy of one mole of a gas, when volume changes from $V$ to $2 \mathrm{~V}$ at constant pressure $P$ ?
1 $\frac{\mathrm{R}}{(\gamma-1)}$
2 PV
3 $\frac{\mathrm{PV}}{(\gamma-1)}$
4 $\frac{\gamma \mathrm{PV}}{(\gamma-1)}$ Where $\gamma$ is the ratio of specific heat of the gas at constant pressure to that at constant volume
Explanation:
C Given that, Mole of gas, $n=1$ Initial volume $=\mathrm{V}$ Final volume $=2 \mathrm{~V}$ Pressure, $\mathrm{P}=$ Constant According to ideal gas equation, $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{P}(2 \mathrm{~V}-\mathrm{V})=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ As we know that, $\gamma=\frac{C_{p}}{C_{v}}$ Substract one in both side, $\gamma-1=\frac{C_{p}}{C_{v}}-1$ $\gamma-1=\frac{C_{p}-C_{v}}{C_{v}}$ $=\frac{R}{C_{v}}$ $C_{v}=\frac{R}{\gamma-1}$ From equation (i) and (ii), $\Delta \mathrm{u}=\frac{\mathrm{PV}}{\gamma-1}$
SCRA-2009
Thermodynamics
148130
The volume of one mole of an ideal gas changes from $V$ to $2 V$ at temperature $300 \mathrm{~K}$. If $R$ is universal gas constant, then work done in this process is
1 $300 \mathrm{R} \ln 2$
2 $600 \mathrm{R} \ln 2$
3 $300 \ln 2$
4 $600 \ln 2$
Explanation:
A Given, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=2 \mathrm{~V}$ We know that-: Work done in an isothermal process $\mathrm{W}=\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $\mathrm{W}=1 \times \mathrm{R} \times 300 \ln$ $\mathrm{W}=300 \mathrm{R} \ln 2$ $\therefore \quad \mathrm{W}=1 \times \mathrm{R} \times 300 \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$
J and K CET- 2010
Thermodynamics
148131
A gas for which $\gamma=4 / 3$ is heated at constant pressure. Fraction of heat supplied for doing external work is
148126
Heat supplied in a process is used completely to do work against the external surroundings. Then, the process may be identified as which one of the following?
1 Adiabatic
2 Isochoric
3 Isothermal
4 None of the above
Explanation:
C In a thermodynamic process, heat supplied in a process is used completely in doing work against the external surroundings. Hence, this process may be designated as isothermal. \(\Delta \mathrm{U}=0=\mathrm{Q}-\mathrm{W}\) \(\therefore \mathrm{Q}=\mathrm{W}\) Heat added to the system to do work \(\mathrm{W}=\mathrm{nRT} \ln \left[\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right]\)
SCRA-2012
Thermodynamics
148127
What is the external work done (approximately) when $1 \mathrm{~g}$ of helium is heated through $1^{\circ} \mathrm{C}$ at constant pressure?
1 $1 \mathrm{~J}$
2 $2 \mathrm{~J}$
3 $4 \mathrm{~J}$
4 $8 \mathrm{~J}$
Explanation:
B Given, Temperature, $\mathrm{T}=1^{\circ} \mathrm{C}$ Mass, $\mathrm{m}=1 \mathrm{~g}$ No. of moles in $1 \mathrm{~g}$ of Helium, $\mathrm{n}=0.25$ mole $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ $\mathrm{W} =\int \mathrm{Pdv}$ $=\mathrm{nR} \times 1=\mathrm{nR}$ $=0.25 \times 8.314$ $\mathrm{W} \approx 2 \mathrm{~J}$
SCRA-2012
Thermodynamics
148129
What is the change in internal energy of one mole of a gas, when volume changes from $V$ to $2 \mathrm{~V}$ at constant pressure $P$ ?
1 $\frac{\mathrm{R}}{(\gamma-1)}$
2 PV
3 $\frac{\mathrm{PV}}{(\gamma-1)}$
4 $\frac{\gamma \mathrm{PV}}{(\gamma-1)}$ Where $\gamma$ is the ratio of specific heat of the gas at constant pressure to that at constant volume
Explanation:
C Given that, Mole of gas, $n=1$ Initial volume $=\mathrm{V}$ Final volume $=2 \mathrm{~V}$ Pressure, $\mathrm{P}=$ Constant According to ideal gas equation, $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{P}(2 \mathrm{~V}-\mathrm{V})=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ As we know that, $\gamma=\frac{C_{p}}{C_{v}}$ Substract one in both side, $\gamma-1=\frac{C_{p}}{C_{v}}-1$ $\gamma-1=\frac{C_{p}-C_{v}}{C_{v}}$ $=\frac{R}{C_{v}}$ $C_{v}=\frac{R}{\gamma-1}$ From equation (i) and (ii), $\Delta \mathrm{u}=\frac{\mathrm{PV}}{\gamma-1}$
SCRA-2009
Thermodynamics
148130
The volume of one mole of an ideal gas changes from $V$ to $2 V$ at temperature $300 \mathrm{~K}$. If $R$ is universal gas constant, then work done in this process is
1 $300 \mathrm{R} \ln 2$
2 $600 \mathrm{R} \ln 2$
3 $300 \ln 2$
4 $600 \ln 2$
Explanation:
A Given, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=2 \mathrm{~V}$ We know that-: Work done in an isothermal process $\mathrm{W}=\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $\mathrm{W}=1 \times \mathrm{R} \times 300 \ln$ $\mathrm{W}=300 \mathrm{R} \ln 2$ $\therefore \quad \mathrm{W}=1 \times \mathrm{R} \times 300 \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$
J and K CET- 2010
Thermodynamics
148131
A gas for which $\gamma=4 / 3$ is heated at constant pressure. Fraction of heat supplied for doing external work is
148126
Heat supplied in a process is used completely to do work against the external surroundings. Then, the process may be identified as which one of the following?
1 Adiabatic
2 Isochoric
3 Isothermal
4 None of the above
Explanation:
C In a thermodynamic process, heat supplied in a process is used completely in doing work against the external surroundings. Hence, this process may be designated as isothermal. \(\Delta \mathrm{U}=0=\mathrm{Q}-\mathrm{W}\) \(\therefore \mathrm{Q}=\mathrm{W}\) Heat added to the system to do work \(\mathrm{W}=\mathrm{nRT} \ln \left[\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right]\)
SCRA-2012
Thermodynamics
148127
What is the external work done (approximately) when $1 \mathrm{~g}$ of helium is heated through $1^{\circ} \mathrm{C}$ at constant pressure?
1 $1 \mathrm{~J}$
2 $2 \mathrm{~J}$
3 $4 \mathrm{~J}$
4 $8 \mathrm{~J}$
Explanation:
B Given, Temperature, $\mathrm{T}=1^{\circ} \mathrm{C}$ Mass, $\mathrm{m}=1 \mathrm{~g}$ No. of moles in $1 \mathrm{~g}$ of Helium, $\mathrm{n}=0.25$ mole $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ $\mathrm{W} =\int \mathrm{Pdv}$ $=\mathrm{nR} \times 1=\mathrm{nR}$ $=0.25 \times 8.314$ $\mathrm{W} \approx 2 \mathrm{~J}$
SCRA-2012
Thermodynamics
148129
What is the change in internal energy of one mole of a gas, when volume changes from $V$ to $2 \mathrm{~V}$ at constant pressure $P$ ?
1 $\frac{\mathrm{R}}{(\gamma-1)}$
2 PV
3 $\frac{\mathrm{PV}}{(\gamma-1)}$
4 $\frac{\gamma \mathrm{PV}}{(\gamma-1)}$ Where $\gamma$ is the ratio of specific heat of the gas at constant pressure to that at constant volume
Explanation:
C Given that, Mole of gas, $n=1$ Initial volume $=\mathrm{V}$ Final volume $=2 \mathrm{~V}$ Pressure, $\mathrm{P}=$ Constant According to ideal gas equation, $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{P}(2 \mathrm{~V}-\mathrm{V})=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ As we know that, $\gamma=\frac{C_{p}}{C_{v}}$ Substract one in both side, $\gamma-1=\frac{C_{p}}{C_{v}}-1$ $\gamma-1=\frac{C_{p}-C_{v}}{C_{v}}$ $=\frac{R}{C_{v}}$ $C_{v}=\frac{R}{\gamma-1}$ From equation (i) and (ii), $\Delta \mathrm{u}=\frac{\mathrm{PV}}{\gamma-1}$
SCRA-2009
Thermodynamics
148130
The volume of one mole of an ideal gas changes from $V$ to $2 V$ at temperature $300 \mathrm{~K}$. If $R$ is universal gas constant, then work done in this process is
1 $300 \mathrm{R} \ln 2$
2 $600 \mathrm{R} \ln 2$
3 $300 \ln 2$
4 $600 \ln 2$
Explanation:
A Given, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=2 \mathrm{~V}$ We know that-: Work done in an isothermal process $\mathrm{W}=\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $\mathrm{W}=1 \times \mathrm{R} \times 300 \ln$ $\mathrm{W}=300 \mathrm{R} \ln 2$ $\therefore \quad \mathrm{W}=1 \times \mathrm{R} \times 300 \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$
J and K CET- 2010
Thermodynamics
148131
A gas for which $\gamma=4 / 3$ is heated at constant pressure. Fraction of heat supplied for doing external work is
148126
Heat supplied in a process is used completely to do work against the external surroundings. Then, the process may be identified as which one of the following?
1 Adiabatic
2 Isochoric
3 Isothermal
4 None of the above
Explanation:
C In a thermodynamic process, heat supplied in a process is used completely in doing work against the external surroundings. Hence, this process may be designated as isothermal. \(\Delta \mathrm{U}=0=\mathrm{Q}-\mathrm{W}\) \(\therefore \mathrm{Q}=\mathrm{W}\) Heat added to the system to do work \(\mathrm{W}=\mathrm{nRT} \ln \left[\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right]\)
SCRA-2012
Thermodynamics
148127
What is the external work done (approximately) when $1 \mathrm{~g}$ of helium is heated through $1^{\circ} \mathrm{C}$ at constant pressure?
1 $1 \mathrm{~J}$
2 $2 \mathrm{~J}$
3 $4 \mathrm{~J}$
4 $8 \mathrm{~J}$
Explanation:
B Given, Temperature, $\mathrm{T}=1^{\circ} \mathrm{C}$ Mass, $\mathrm{m}=1 \mathrm{~g}$ No. of moles in $1 \mathrm{~g}$ of Helium, $\mathrm{n}=0.25$ mole $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ $\mathrm{W} =\int \mathrm{Pdv}$ $=\mathrm{nR} \times 1=\mathrm{nR}$ $=0.25 \times 8.314$ $\mathrm{W} \approx 2 \mathrm{~J}$
SCRA-2012
Thermodynamics
148129
What is the change in internal energy of one mole of a gas, when volume changes from $V$ to $2 \mathrm{~V}$ at constant pressure $P$ ?
1 $\frac{\mathrm{R}}{(\gamma-1)}$
2 PV
3 $\frac{\mathrm{PV}}{(\gamma-1)}$
4 $\frac{\gamma \mathrm{PV}}{(\gamma-1)}$ Where $\gamma$ is the ratio of specific heat of the gas at constant pressure to that at constant volume
Explanation:
C Given that, Mole of gas, $n=1$ Initial volume $=\mathrm{V}$ Final volume $=2 \mathrm{~V}$ Pressure, $\mathrm{P}=$ Constant According to ideal gas equation, $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{P}(2 \mathrm{~V}-\mathrm{V})=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ As we know that, $\gamma=\frac{C_{p}}{C_{v}}$ Substract one in both side, $\gamma-1=\frac{C_{p}}{C_{v}}-1$ $\gamma-1=\frac{C_{p}-C_{v}}{C_{v}}$ $=\frac{R}{C_{v}}$ $C_{v}=\frac{R}{\gamma-1}$ From equation (i) and (ii), $\Delta \mathrm{u}=\frac{\mathrm{PV}}{\gamma-1}$
SCRA-2009
Thermodynamics
148130
The volume of one mole of an ideal gas changes from $V$ to $2 V$ at temperature $300 \mathrm{~K}$. If $R$ is universal gas constant, then work done in this process is
1 $300 \mathrm{R} \ln 2$
2 $600 \mathrm{R} \ln 2$
3 $300 \ln 2$
4 $600 \ln 2$
Explanation:
A Given, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=2 \mathrm{~V}$ We know that-: Work done in an isothermal process $\mathrm{W}=\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $\mathrm{W}=1 \times \mathrm{R} \times 300 \ln$ $\mathrm{W}=300 \mathrm{R} \ln 2$ $\therefore \quad \mathrm{W}=1 \times \mathrm{R} \times 300 \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$
J and K CET- 2010
Thermodynamics
148131
A gas for which $\gamma=4 / 3$ is heated at constant pressure. Fraction of heat supplied for doing external work is
148126
Heat supplied in a process is used completely to do work against the external surroundings. Then, the process may be identified as which one of the following?
1 Adiabatic
2 Isochoric
3 Isothermal
4 None of the above
Explanation:
C In a thermodynamic process, heat supplied in a process is used completely in doing work against the external surroundings. Hence, this process may be designated as isothermal. \(\Delta \mathrm{U}=0=\mathrm{Q}-\mathrm{W}\) \(\therefore \mathrm{Q}=\mathrm{W}\) Heat added to the system to do work \(\mathrm{W}=\mathrm{nRT} \ln \left[\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right]\)
SCRA-2012
Thermodynamics
148127
What is the external work done (approximately) when $1 \mathrm{~g}$ of helium is heated through $1^{\circ} \mathrm{C}$ at constant pressure?
1 $1 \mathrm{~J}$
2 $2 \mathrm{~J}$
3 $4 \mathrm{~J}$
4 $8 \mathrm{~J}$
Explanation:
B Given, Temperature, $\mathrm{T}=1^{\circ} \mathrm{C}$ Mass, $\mathrm{m}=1 \mathrm{~g}$ No. of moles in $1 \mathrm{~g}$ of Helium, $\mathrm{n}=0.25$ mole $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ $\mathrm{W} =\int \mathrm{Pdv}$ $=\mathrm{nR} \times 1=\mathrm{nR}$ $=0.25 \times 8.314$ $\mathrm{W} \approx 2 \mathrm{~J}$
SCRA-2012
Thermodynamics
148129
What is the change in internal energy of one mole of a gas, when volume changes from $V$ to $2 \mathrm{~V}$ at constant pressure $P$ ?
1 $\frac{\mathrm{R}}{(\gamma-1)}$
2 PV
3 $\frac{\mathrm{PV}}{(\gamma-1)}$
4 $\frac{\gamma \mathrm{PV}}{(\gamma-1)}$ Where $\gamma$ is the ratio of specific heat of the gas at constant pressure to that at constant volume
Explanation:
C Given that, Mole of gas, $n=1$ Initial volume $=\mathrm{V}$ Final volume $=2 \mathrm{~V}$ Pressure, $\mathrm{P}=$ Constant According to ideal gas equation, $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{P}(2 \mathrm{~V}-\mathrm{V})=\mathrm{nR} \Delta \mathrm{T}$ $\mathrm{PV}=\mathrm{nR} \Delta \mathrm{T}$ As we know that, $\gamma=\frac{C_{p}}{C_{v}}$ Substract one in both side, $\gamma-1=\frac{C_{p}}{C_{v}}-1$ $\gamma-1=\frac{C_{p}-C_{v}}{C_{v}}$ $=\frac{R}{C_{v}}$ $C_{v}=\frac{R}{\gamma-1}$ From equation (i) and (ii), $\Delta \mathrm{u}=\frac{\mathrm{PV}}{\gamma-1}$
SCRA-2009
Thermodynamics
148130
The volume of one mole of an ideal gas changes from $V$ to $2 V$ at temperature $300 \mathrm{~K}$. If $R$ is universal gas constant, then work done in this process is
1 $300 \mathrm{R} \ln 2$
2 $600 \mathrm{R} \ln 2$
3 $300 \ln 2$
4 $600 \ln 2$
Explanation:
A Given, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=2 \mathrm{~V}$ We know that-: Work done in an isothermal process $\mathrm{W}=\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $\mathrm{W}=1 \times \mathrm{R} \times 300 \ln$ $\mathrm{W}=300 \mathrm{R} \ln 2$ $\therefore \quad \mathrm{W}=1 \times \mathrm{R} \times 300 \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$
J and K CET- 2010
Thermodynamics
148131
A gas for which $\gamma=4 / 3$ is heated at constant pressure. Fraction of heat supplied for doing external work is
