A For an adiabatic process, $\Delta \mathrm{Q}=0$ $\because \quad \mathrm{PV}^{\gamma}=$ constant From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ or $\quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$ Putting the value of $\mathrm{P}$ in equation (i), we get $\frac{\mathrm{nRT}}{\mathrm{V}} \mathrm{V}^{\gamma}=\text { constant }$ or $\quad \mathrm{TV}^{\gamma-1}=\mathrm{constant}$
Manipal UGET -2020
Thermodynamics
148253
A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is
1 $20 \mathrm{~J}$
2 $-20 \mathrm{~J}$
3 $400 \mathrm{~J}$
4 $-374 \mathrm{~J}$
Explanation:
B Net work done by the system $=$ Area of PQRS $=(300-100) \times(200-100)$ $=200 \times 100$ $=2 \times 10^{4} \mathrm{kPa}-\mathrm{cm}^{3}$ $=2 \times 10^{4} \times 10^{3} \times 10^{-6} \mathrm{~Pa}-\mathrm{m}^{3}$ $\mathrm{~W}_{\text {net }} =20 \mathrm{~J}$ Here, direction of process is anticlockwise so work done will be negative i.e. $\Delta \mathrm{W}=-20 \mathrm{~J}$
Manipal UGET-2019
Thermodynamics
148254
A gas at the temperature $250 \mathrm{~K}$ is contained in a closed vessel. If the gas is heated through $1 \mathrm{~K}$, then percentage increase in its pressure will be
1 $0.4 \%$
2 $0.2 \%$
3 $0.1 \%$
4 $0.8 \%$
Explanation:
A Given, $\mathrm{T}_{1}=250 \mathrm{~K}, \mathrm{~T}_{2}=251 \mathrm{~K}$ $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=251-250=1 \mathrm{~K}$ If mass and volume (v) are constant. $\mathrm{P} \propto T$ $\frac{P}{T}=\text { constant }$ $\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$ $\frac{P_{2}}{P_{1}}=\frac{T_{2}}{T_{1}}$ $\frac{P_{2}-P_{1}}{P_{1}}=\frac{T_{2}-T_{1}}{T_{1}}$ $\frac{P_{2}-P_{1}}{P_{1}}=\frac{1}{250}$ So, $\%$ increase in pressure $=\frac{P_{2}-P_{1}}{P_{1}} \times 100$ $=\frac{1}{250} \times 100$ $=0.4 \%$
Manipal UGET-2018
Thermodynamics
148255
Work done by air when it expands from $50 \mathrm{~L}$ to $150 \mathrm{~L}$ at a constant pressure of 2 atmosphere is
A For an adiabatic process, $\Delta \mathrm{Q}=0$ $\because \quad \mathrm{PV}^{\gamma}=$ constant From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ or $\quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$ Putting the value of $\mathrm{P}$ in equation (i), we get $\frac{\mathrm{nRT}}{\mathrm{V}} \mathrm{V}^{\gamma}=\text { constant }$ or $\quad \mathrm{TV}^{\gamma-1}=\mathrm{constant}$
Manipal UGET -2020
Thermodynamics
148253
A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is
1 $20 \mathrm{~J}$
2 $-20 \mathrm{~J}$
3 $400 \mathrm{~J}$
4 $-374 \mathrm{~J}$
Explanation:
B Net work done by the system $=$ Area of PQRS $=(300-100) \times(200-100)$ $=200 \times 100$ $=2 \times 10^{4} \mathrm{kPa}-\mathrm{cm}^{3}$ $=2 \times 10^{4} \times 10^{3} \times 10^{-6} \mathrm{~Pa}-\mathrm{m}^{3}$ $\mathrm{~W}_{\text {net }} =20 \mathrm{~J}$ Here, direction of process is anticlockwise so work done will be negative i.e. $\Delta \mathrm{W}=-20 \mathrm{~J}$
Manipal UGET-2019
Thermodynamics
148254
A gas at the temperature $250 \mathrm{~K}$ is contained in a closed vessel. If the gas is heated through $1 \mathrm{~K}$, then percentage increase in its pressure will be
1 $0.4 \%$
2 $0.2 \%$
3 $0.1 \%$
4 $0.8 \%$
Explanation:
A Given, $\mathrm{T}_{1}=250 \mathrm{~K}, \mathrm{~T}_{2}=251 \mathrm{~K}$ $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=251-250=1 \mathrm{~K}$ If mass and volume (v) are constant. $\mathrm{P} \propto T$ $\frac{P}{T}=\text { constant }$ $\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$ $\frac{P_{2}}{P_{1}}=\frac{T_{2}}{T_{1}}$ $\frac{P_{2}-P_{1}}{P_{1}}=\frac{T_{2}-T_{1}}{T_{1}}$ $\frac{P_{2}-P_{1}}{P_{1}}=\frac{1}{250}$ So, $\%$ increase in pressure $=\frac{P_{2}-P_{1}}{P_{1}} \times 100$ $=\frac{1}{250} \times 100$ $=0.4 \%$
Manipal UGET-2018
Thermodynamics
148255
Work done by air when it expands from $50 \mathrm{~L}$ to $150 \mathrm{~L}$ at a constant pressure of 2 atmosphere is
A For an adiabatic process, $\Delta \mathrm{Q}=0$ $\because \quad \mathrm{PV}^{\gamma}=$ constant From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ or $\quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$ Putting the value of $\mathrm{P}$ in equation (i), we get $\frac{\mathrm{nRT}}{\mathrm{V}} \mathrm{V}^{\gamma}=\text { constant }$ or $\quad \mathrm{TV}^{\gamma-1}=\mathrm{constant}$
Manipal UGET -2020
Thermodynamics
148253
A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is
1 $20 \mathrm{~J}$
2 $-20 \mathrm{~J}$
3 $400 \mathrm{~J}$
4 $-374 \mathrm{~J}$
Explanation:
B Net work done by the system $=$ Area of PQRS $=(300-100) \times(200-100)$ $=200 \times 100$ $=2 \times 10^{4} \mathrm{kPa}-\mathrm{cm}^{3}$ $=2 \times 10^{4} \times 10^{3} \times 10^{-6} \mathrm{~Pa}-\mathrm{m}^{3}$ $\mathrm{~W}_{\text {net }} =20 \mathrm{~J}$ Here, direction of process is anticlockwise so work done will be negative i.e. $\Delta \mathrm{W}=-20 \mathrm{~J}$
Manipal UGET-2019
Thermodynamics
148254
A gas at the temperature $250 \mathrm{~K}$ is contained in a closed vessel. If the gas is heated through $1 \mathrm{~K}$, then percentage increase in its pressure will be
1 $0.4 \%$
2 $0.2 \%$
3 $0.1 \%$
4 $0.8 \%$
Explanation:
A Given, $\mathrm{T}_{1}=250 \mathrm{~K}, \mathrm{~T}_{2}=251 \mathrm{~K}$ $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=251-250=1 \mathrm{~K}$ If mass and volume (v) are constant. $\mathrm{P} \propto T$ $\frac{P}{T}=\text { constant }$ $\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$ $\frac{P_{2}}{P_{1}}=\frac{T_{2}}{T_{1}}$ $\frac{P_{2}-P_{1}}{P_{1}}=\frac{T_{2}-T_{1}}{T_{1}}$ $\frac{P_{2}-P_{1}}{P_{1}}=\frac{1}{250}$ So, $\%$ increase in pressure $=\frac{P_{2}-P_{1}}{P_{1}} \times 100$ $=\frac{1}{250} \times 100$ $=0.4 \%$
Manipal UGET-2018
Thermodynamics
148255
Work done by air when it expands from $50 \mathrm{~L}$ to $150 \mathrm{~L}$ at a constant pressure of 2 atmosphere is
A For an adiabatic process, $\Delta \mathrm{Q}=0$ $\because \quad \mathrm{PV}^{\gamma}=$ constant From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ or $\quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$ Putting the value of $\mathrm{P}$ in equation (i), we get $\frac{\mathrm{nRT}}{\mathrm{V}} \mathrm{V}^{\gamma}=\text { constant }$ or $\quad \mathrm{TV}^{\gamma-1}=\mathrm{constant}$
Manipal UGET -2020
Thermodynamics
148253
A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is
1 $20 \mathrm{~J}$
2 $-20 \mathrm{~J}$
3 $400 \mathrm{~J}$
4 $-374 \mathrm{~J}$
Explanation:
B Net work done by the system $=$ Area of PQRS $=(300-100) \times(200-100)$ $=200 \times 100$ $=2 \times 10^{4} \mathrm{kPa}-\mathrm{cm}^{3}$ $=2 \times 10^{4} \times 10^{3} \times 10^{-6} \mathrm{~Pa}-\mathrm{m}^{3}$ $\mathrm{~W}_{\text {net }} =20 \mathrm{~J}$ Here, direction of process is anticlockwise so work done will be negative i.e. $\Delta \mathrm{W}=-20 \mathrm{~J}$
Manipal UGET-2019
Thermodynamics
148254
A gas at the temperature $250 \mathrm{~K}$ is contained in a closed vessel. If the gas is heated through $1 \mathrm{~K}$, then percentage increase in its pressure will be
1 $0.4 \%$
2 $0.2 \%$
3 $0.1 \%$
4 $0.8 \%$
Explanation:
A Given, $\mathrm{T}_{1}=250 \mathrm{~K}, \mathrm{~T}_{2}=251 \mathrm{~K}$ $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=251-250=1 \mathrm{~K}$ If mass and volume (v) are constant. $\mathrm{P} \propto T$ $\frac{P}{T}=\text { constant }$ $\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$ $\frac{P_{2}}{P_{1}}=\frac{T_{2}}{T_{1}}$ $\frac{P_{2}-P_{1}}{P_{1}}=\frac{T_{2}-T_{1}}{T_{1}}$ $\frac{P_{2}-P_{1}}{P_{1}}=\frac{1}{250}$ So, $\%$ increase in pressure $=\frac{P_{2}-P_{1}}{P_{1}} \times 100$ $=\frac{1}{250} \times 100$ $=0.4 \%$
Manipal UGET-2018
Thermodynamics
148255
Work done by air when it expands from $50 \mathrm{~L}$ to $150 \mathrm{~L}$ at a constant pressure of 2 atmosphere is
