148590
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?
1 $600 \mathrm{~K}, 400 \mathrm{~K}$
2 $400 \mathrm{~K}, 200 \mathrm{~K}$
3 $500 \mathrm{~K}, 300 \mathrm{~K}$
4 $300 \mathrm{~K}, 100 \mathrm{~K}$
Explanation:
D Efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}$ (a) $\eta_{1}=1-\frac{400}{600}=0.33$ (b) $\eta_{2}=1-\frac{200}{400}=0.50$ (c) $\eta_{3}=1-\frac{300}{500}=0.40$ (d) $\eta_{4}=1-\frac{100}{300}=0.66$ So the efficiency of carnot's heat engine is maximum for $\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}$ and $\mathrm{T}_{\mathrm{H}}=100 \mathrm{~K}$ Hence, Option (d) is correct.
Karnataka CET-2013
Thermodynamics
148592
If $\gamma$ is the ratio of specific heats and $R$ is the universal gas constant, then the molar specific heat at constant volume $C_{V}$ is given by:
1 $\gamma \mathrm{R}$
2 $\frac{(\gamma-1) R}{\gamma}$
3 $\frac{\mathrm{R}}{\gamma-1}$
4 $\frac{\gamma \mathrm{R}}{\gamma-1}$
Explanation:
C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale. $\mathrm{R}=\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{v}}$ Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume. $\gamma=\frac{C_{p}}{C_{v}}$ $\mathrm{C}_{\mathrm{P}}=\gamma \mathrm{C}_{\mathrm{V}}$ Put the value of $C_{p}$ in equation (i) $\gamma \mathrm{C}_{\mathrm{v}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}(\gamma-1)=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}$
Karnataka CET-2008
Thermodynamics
148593
A Carnot engine taken heat from a reservoir at $627^{\circ} \mathrm{C}$ and rejects heat to a sink at $27^{\circ} \mathrm{C}$. Its efficiency will be :
1 $3 / 5$
2 $1 / 3$
3 $2 / 3$
4 $200 / 209$
Explanation:
C Given that, $\mathrm{T}_{1}=627^{\circ} \mathrm{C}=627+273=900 \mathrm{~K}$ $\mathrm{~T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ We know, efficiency of carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{300}{900}=1-\frac{1}{3}$ $\eta=\frac{2}{3}$
Karnataka CET-2006
Thermodynamics
148594
A monoatomic gas is suddenly compressed to $(1 / 8)^{\text {th }}$ of its initial volume adiabatically. The ratio of its final pressure to the initial pressure is (given the ratio of the specific heat of the given gas to be $5 / 3$ )
1 32
2 $40 / 3$
3 $24 / 5$
4 8
Explanation:
A Given that, Initial volume $=\mathrm{V}$ Final volume $=\frac{\mathrm{V}}{8}$ $\text { Adiabatic index, } \gamma=\frac{5}{3}$ We know for adiabatic process $\mathrm{PV}^{\gamma}=$ constant $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}}{\frac{\mathrm{V}}{8}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=(8)^{5 / 3}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(2^{3}\right)^{\frac{5}{3}} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=2^{5}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=32$
Karnataka CET-2006
Thermodynamics
148595
A Carnot's engine is made to work between $200^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ first and then between $0^{\circ} \mathrm{C}$ and $-200^{\circ} \mathrm{C}$. The ratio of efficiencies of the engine in the two cases is:
1 $1: 1.73$
2 $1.73: 1$
3 $1: 2$
4 $1: 1$
Explanation:
A As we know that the efficiency of Carnot's engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ In $1^{\text {st }}$ case when temperature is between $200^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$. $\mathrm{T}_{2}=0^{\circ} \mathrm{C}=(0+273) \mathrm{K}=273 \mathrm{~K}$ $\mathrm{~T}_{1}=200^{\circ} \mathrm{C}=(200+273) \mathrm{K}=473 \mathrm{~K}$ $\eta_{1}=1-\frac{273}{473}=0.423$ For $2^{\text {nd }}$ case when temperature is between $0^{\circ} \mathrm{C}$ to $200^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=\left(0^{\circ}+273\right)=273 \mathrm{~K}$ $\mathrm{~T}_{2}=(-200+273)=73 \mathrm{~K}$ $\eta_{2}=1-\frac{73}{273}=0.733$ $\therefore \quad \eta_{1}: \eta_{2}=\frac{0.423}{0.733}=\frac{1}{1.73}=1: 1.73$
148590
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?
1 $600 \mathrm{~K}, 400 \mathrm{~K}$
2 $400 \mathrm{~K}, 200 \mathrm{~K}$
3 $500 \mathrm{~K}, 300 \mathrm{~K}$
4 $300 \mathrm{~K}, 100 \mathrm{~K}$
Explanation:
D Efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}$ (a) $\eta_{1}=1-\frac{400}{600}=0.33$ (b) $\eta_{2}=1-\frac{200}{400}=0.50$ (c) $\eta_{3}=1-\frac{300}{500}=0.40$ (d) $\eta_{4}=1-\frac{100}{300}=0.66$ So the efficiency of carnot's heat engine is maximum for $\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}$ and $\mathrm{T}_{\mathrm{H}}=100 \mathrm{~K}$ Hence, Option (d) is correct.
Karnataka CET-2013
Thermodynamics
148592
If $\gamma$ is the ratio of specific heats and $R$ is the universal gas constant, then the molar specific heat at constant volume $C_{V}$ is given by:
1 $\gamma \mathrm{R}$
2 $\frac{(\gamma-1) R}{\gamma}$
3 $\frac{\mathrm{R}}{\gamma-1}$
4 $\frac{\gamma \mathrm{R}}{\gamma-1}$
Explanation:
C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale. $\mathrm{R}=\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{v}}$ Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume. $\gamma=\frac{C_{p}}{C_{v}}$ $\mathrm{C}_{\mathrm{P}}=\gamma \mathrm{C}_{\mathrm{V}}$ Put the value of $C_{p}$ in equation (i) $\gamma \mathrm{C}_{\mathrm{v}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}(\gamma-1)=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}$
Karnataka CET-2008
Thermodynamics
148593
A Carnot engine taken heat from a reservoir at $627^{\circ} \mathrm{C}$ and rejects heat to a sink at $27^{\circ} \mathrm{C}$. Its efficiency will be :
1 $3 / 5$
2 $1 / 3$
3 $2 / 3$
4 $200 / 209$
Explanation:
C Given that, $\mathrm{T}_{1}=627^{\circ} \mathrm{C}=627+273=900 \mathrm{~K}$ $\mathrm{~T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ We know, efficiency of carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{300}{900}=1-\frac{1}{3}$ $\eta=\frac{2}{3}$
Karnataka CET-2006
Thermodynamics
148594
A monoatomic gas is suddenly compressed to $(1 / 8)^{\text {th }}$ of its initial volume adiabatically. The ratio of its final pressure to the initial pressure is (given the ratio of the specific heat of the given gas to be $5 / 3$ )
1 32
2 $40 / 3$
3 $24 / 5$
4 8
Explanation:
A Given that, Initial volume $=\mathrm{V}$ Final volume $=\frac{\mathrm{V}}{8}$ $\text { Adiabatic index, } \gamma=\frac{5}{3}$ We know for adiabatic process $\mathrm{PV}^{\gamma}=$ constant $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}}{\frac{\mathrm{V}}{8}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=(8)^{5 / 3}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(2^{3}\right)^{\frac{5}{3}} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=2^{5}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=32$
Karnataka CET-2006
Thermodynamics
148595
A Carnot's engine is made to work between $200^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ first and then between $0^{\circ} \mathrm{C}$ and $-200^{\circ} \mathrm{C}$. The ratio of efficiencies of the engine in the two cases is:
1 $1: 1.73$
2 $1.73: 1$
3 $1: 2$
4 $1: 1$
Explanation:
A As we know that the efficiency of Carnot's engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ In $1^{\text {st }}$ case when temperature is between $200^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$. $\mathrm{T}_{2}=0^{\circ} \mathrm{C}=(0+273) \mathrm{K}=273 \mathrm{~K}$ $\mathrm{~T}_{1}=200^{\circ} \mathrm{C}=(200+273) \mathrm{K}=473 \mathrm{~K}$ $\eta_{1}=1-\frac{273}{473}=0.423$ For $2^{\text {nd }}$ case when temperature is between $0^{\circ} \mathrm{C}$ to $200^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=\left(0^{\circ}+273\right)=273 \mathrm{~K}$ $\mathrm{~T}_{2}=(-200+273)=73 \mathrm{~K}$ $\eta_{2}=1-\frac{73}{273}=0.733$ $\therefore \quad \eta_{1}: \eta_{2}=\frac{0.423}{0.733}=\frac{1}{1.73}=1: 1.73$
148590
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?
1 $600 \mathrm{~K}, 400 \mathrm{~K}$
2 $400 \mathrm{~K}, 200 \mathrm{~K}$
3 $500 \mathrm{~K}, 300 \mathrm{~K}$
4 $300 \mathrm{~K}, 100 \mathrm{~K}$
Explanation:
D Efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}$ (a) $\eta_{1}=1-\frac{400}{600}=0.33$ (b) $\eta_{2}=1-\frac{200}{400}=0.50$ (c) $\eta_{3}=1-\frac{300}{500}=0.40$ (d) $\eta_{4}=1-\frac{100}{300}=0.66$ So the efficiency of carnot's heat engine is maximum for $\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}$ and $\mathrm{T}_{\mathrm{H}}=100 \mathrm{~K}$ Hence, Option (d) is correct.
Karnataka CET-2013
Thermodynamics
148592
If $\gamma$ is the ratio of specific heats and $R$ is the universal gas constant, then the molar specific heat at constant volume $C_{V}$ is given by:
1 $\gamma \mathrm{R}$
2 $\frac{(\gamma-1) R}{\gamma}$
3 $\frac{\mathrm{R}}{\gamma-1}$
4 $\frac{\gamma \mathrm{R}}{\gamma-1}$
Explanation:
C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale. $\mathrm{R}=\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{v}}$ Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume. $\gamma=\frac{C_{p}}{C_{v}}$ $\mathrm{C}_{\mathrm{P}}=\gamma \mathrm{C}_{\mathrm{V}}$ Put the value of $C_{p}$ in equation (i) $\gamma \mathrm{C}_{\mathrm{v}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}(\gamma-1)=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}$
Karnataka CET-2008
Thermodynamics
148593
A Carnot engine taken heat from a reservoir at $627^{\circ} \mathrm{C}$ and rejects heat to a sink at $27^{\circ} \mathrm{C}$. Its efficiency will be :
1 $3 / 5$
2 $1 / 3$
3 $2 / 3$
4 $200 / 209$
Explanation:
C Given that, $\mathrm{T}_{1}=627^{\circ} \mathrm{C}=627+273=900 \mathrm{~K}$ $\mathrm{~T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ We know, efficiency of carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{300}{900}=1-\frac{1}{3}$ $\eta=\frac{2}{3}$
Karnataka CET-2006
Thermodynamics
148594
A monoatomic gas is suddenly compressed to $(1 / 8)^{\text {th }}$ of its initial volume adiabatically. The ratio of its final pressure to the initial pressure is (given the ratio of the specific heat of the given gas to be $5 / 3$ )
1 32
2 $40 / 3$
3 $24 / 5$
4 8
Explanation:
A Given that, Initial volume $=\mathrm{V}$ Final volume $=\frac{\mathrm{V}}{8}$ $\text { Adiabatic index, } \gamma=\frac{5}{3}$ We know for adiabatic process $\mathrm{PV}^{\gamma}=$ constant $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}}{\frac{\mathrm{V}}{8}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=(8)^{5 / 3}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(2^{3}\right)^{\frac{5}{3}} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=2^{5}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=32$
Karnataka CET-2006
Thermodynamics
148595
A Carnot's engine is made to work between $200^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ first and then between $0^{\circ} \mathrm{C}$ and $-200^{\circ} \mathrm{C}$. The ratio of efficiencies of the engine in the two cases is:
1 $1: 1.73$
2 $1.73: 1$
3 $1: 2$
4 $1: 1$
Explanation:
A As we know that the efficiency of Carnot's engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ In $1^{\text {st }}$ case when temperature is between $200^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$. $\mathrm{T}_{2}=0^{\circ} \mathrm{C}=(0+273) \mathrm{K}=273 \mathrm{~K}$ $\mathrm{~T}_{1}=200^{\circ} \mathrm{C}=(200+273) \mathrm{K}=473 \mathrm{~K}$ $\eta_{1}=1-\frac{273}{473}=0.423$ For $2^{\text {nd }}$ case when temperature is between $0^{\circ} \mathrm{C}$ to $200^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=\left(0^{\circ}+273\right)=273 \mathrm{~K}$ $\mathrm{~T}_{2}=(-200+273)=73 \mathrm{~K}$ $\eta_{2}=1-\frac{73}{273}=0.733$ $\therefore \quad \eta_{1}: \eta_{2}=\frac{0.423}{0.733}=\frac{1}{1.73}=1: 1.73$
148590
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?
1 $600 \mathrm{~K}, 400 \mathrm{~K}$
2 $400 \mathrm{~K}, 200 \mathrm{~K}$
3 $500 \mathrm{~K}, 300 \mathrm{~K}$
4 $300 \mathrm{~K}, 100 \mathrm{~K}$
Explanation:
D Efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}$ (a) $\eta_{1}=1-\frac{400}{600}=0.33$ (b) $\eta_{2}=1-\frac{200}{400}=0.50$ (c) $\eta_{3}=1-\frac{300}{500}=0.40$ (d) $\eta_{4}=1-\frac{100}{300}=0.66$ So the efficiency of carnot's heat engine is maximum for $\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}$ and $\mathrm{T}_{\mathrm{H}}=100 \mathrm{~K}$ Hence, Option (d) is correct.
Karnataka CET-2013
Thermodynamics
148592
If $\gamma$ is the ratio of specific heats and $R$ is the universal gas constant, then the molar specific heat at constant volume $C_{V}$ is given by:
1 $\gamma \mathrm{R}$
2 $\frac{(\gamma-1) R}{\gamma}$
3 $\frac{\mathrm{R}}{\gamma-1}$
4 $\frac{\gamma \mathrm{R}}{\gamma-1}$
Explanation:
C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale. $\mathrm{R}=\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{v}}$ Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume. $\gamma=\frac{C_{p}}{C_{v}}$ $\mathrm{C}_{\mathrm{P}}=\gamma \mathrm{C}_{\mathrm{V}}$ Put the value of $C_{p}$ in equation (i) $\gamma \mathrm{C}_{\mathrm{v}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}(\gamma-1)=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}$
Karnataka CET-2008
Thermodynamics
148593
A Carnot engine taken heat from a reservoir at $627^{\circ} \mathrm{C}$ and rejects heat to a sink at $27^{\circ} \mathrm{C}$. Its efficiency will be :
1 $3 / 5$
2 $1 / 3$
3 $2 / 3$
4 $200 / 209$
Explanation:
C Given that, $\mathrm{T}_{1}=627^{\circ} \mathrm{C}=627+273=900 \mathrm{~K}$ $\mathrm{~T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ We know, efficiency of carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{300}{900}=1-\frac{1}{3}$ $\eta=\frac{2}{3}$
Karnataka CET-2006
Thermodynamics
148594
A monoatomic gas is suddenly compressed to $(1 / 8)^{\text {th }}$ of its initial volume adiabatically. The ratio of its final pressure to the initial pressure is (given the ratio of the specific heat of the given gas to be $5 / 3$ )
1 32
2 $40 / 3$
3 $24 / 5$
4 8
Explanation:
A Given that, Initial volume $=\mathrm{V}$ Final volume $=\frac{\mathrm{V}}{8}$ $\text { Adiabatic index, } \gamma=\frac{5}{3}$ We know for adiabatic process $\mathrm{PV}^{\gamma}=$ constant $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}}{\frac{\mathrm{V}}{8}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=(8)^{5 / 3}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(2^{3}\right)^{\frac{5}{3}} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=2^{5}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=32$
Karnataka CET-2006
Thermodynamics
148595
A Carnot's engine is made to work between $200^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ first and then between $0^{\circ} \mathrm{C}$ and $-200^{\circ} \mathrm{C}$. The ratio of efficiencies of the engine in the two cases is:
1 $1: 1.73$
2 $1.73: 1$
3 $1: 2$
4 $1: 1$
Explanation:
A As we know that the efficiency of Carnot's engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ In $1^{\text {st }}$ case when temperature is between $200^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$. $\mathrm{T}_{2}=0^{\circ} \mathrm{C}=(0+273) \mathrm{K}=273 \mathrm{~K}$ $\mathrm{~T}_{1}=200^{\circ} \mathrm{C}=(200+273) \mathrm{K}=473 \mathrm{~K}$ $\eta_{1}=1-\frac{273}{473}=0.423$ For $2^{\text {nd }}$ case when temperature is between $0^{\circ} \mathrm{C}$ to $200^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=\left(0^{\circ}+273\right)=273 \mathrm{~K}$ $\mathrm{~T}_{2}=(-200+273)=73 \mathrm{~K}$ $\eta_{2}=1-\frac{73}{273}=0.733$ $\therefore \quad \eta_{1}: \eta_{2}=\frac{0.423}{0.733}=\frac{1}{1.73}=1: 1.73$
148590
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?
1 $600 \mathrm{~K}, 400 \mathrm{~K}$
2 $400 \mathrm{~K}, 200 \mathrm{~K}$
3 $500 \mathrm{~K}, 300 \mathrm{~K}$
4 $300 \mathrm{~K}, 100 \mathrm{~K}$
Explanation:
D Efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}$ (a) $\eta_{1}=1-\frac{400}{600}=0.33$ (b) $\eta_{2}=1-\frac{200}{400}=0.50$ (c) $\eta_{3}=1-\frac{300}{500}=0.40$ (d) $\eta_{4}=1-\frac{100}{300}=0.66$ So the efficiency of carnot's heat engine is maximum for $\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}$ and $\mathrm{T}_{\mathrm{H}}=100 \mathrm{~K}$ Hence, Option (d) is correct.
Karnataka CET-2013
Thermodynamics
148592
If $\gamma$ is the ratio of specific heats and $R$ is the universal gas constant, then the molar specific heat at constant volume $C_{V}$ is given by:
1 $\gamma \mathrm{R}$
2 $\frac{(\gamma-1) R}{\gamma}$
3 $\frac{\mathrm{R}}{\gamma-1}$
4 $\frac{\gamma \mathrm{R}}{\gamma-1}$
Explanation:
C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale. $\mathrm{R}=\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{v}}$ Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume. $\gamma=\frac{C_{p}}{C_{v}}$ $\mathrm{C}_{\mathrm{P}}=\gamma \mathrm{C}_{\mathrm{V}}$ Put the value of $C_{p}$ in equation (i) $\gamma \mathrm{C}_{\mathrm{v}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}(\gamma-1)=\mathrm{R}$ $\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}$
Karnataka CET-2008
Thermodynamics
148593
A Carnot engine taken heat from a reservoir at $627^{\circ} \mathrm{C}$ and rejects heat to a sink at $27^{\circ} \mathrm{C}$. Its efficiency will be :
1 $3 / 5$
2 $1 / 3$
3 $2 / 3$
4 $200 / 209$
Explanation:
C Given that, $\mathrm{T}_{1}=627^{\circ} \mathrm{C}=627+273=900 \mathrm{~K}$ $\mathrm{~T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ We know, efficiency of carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{300}{900}=1-\frac{1}{3}$ $\eta=\frac{2}{3}$
Karnataka CET-2006
Thermodynamics
148594
A monoatomic gas is suddenly compressed to $(1 / 8)^{\text {th }}$ of its initial volume adiabatically. The ratio of its final pressure to the initial pressure is (given the ratio of the specific heat of the given gas to be $5 / 3$ )
1 32
2 $40 / 3$
3 $24 / 5$
4 8
Explanation:
A Given that, Initial volume $=\mathrm{V}$ Final volume $=\frac{\mathrm{V}}{8}$ $\text { Adiabatic index, } \gamma=\frac{5}{3}$ We know for adiabatic process $\mathrm{PV}^{\gamma}=$ constant $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}}{\frac{\mathrm{V}}{8}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=(8)^{5 / 3}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(2^{3}\right)^{\frac{5}{3}} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=2^{5}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=32$
Karnataka CET-2006
Thermodynamics
148595
A Carnot's engine is made to work between $200^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ first and then between $0^{\circ} \mathrm{C}$ and $-200^{\circ} \mathrm{C}$. The ratio of efficiencies of the engine in the two cases is:
1 $1: 1.73$
2 $1.73: 1$
3 $1: 2$
4 $1: 1$
Explanation:
A As we know that the efficiency of Carnot's engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ In $1^{\text {st }}$ case when temperature is between $200^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$. $\mathrm{T}_{2}=0^{\circ} \mathrm{C}=(0+273) \mathrm{K}=273 \mathrm{~K}$ $\mathrm{~T}_{1}=200^{\circ} \mathrm{C}=(200+273) \mathrm{K}=473 \mathrm{~K}$ $\eta_{1}=1-\frac{273}{473}=0.423$ For $2^{\text {nd }}$ case when temperature is between $0^{\circ} \mathrm{C}$ to $200^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=\left(0^{\circ}+273\right)=273 \mathrm{~K}$ $\mathrm{~T}_{2}=(-200+273)=73 \mathrm{~K}$ $\eta_{2}=1-\frac{73}{273}=0.733$ $\therefore \quad \eta_{1}: \eta_{2}=\frac{0.423}{0.733}=\frac{1}{1.73}=1: 1.73$
