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Thermodynamics

148590 For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?

1

600 K, 400 K

2

400 K, 200 K

3

500 K, 300 K

4

300 K, 100 K

Explanation:

D Efficiency of carnot engine $η = 1 - \frac{T_{L}}{T_{H}}$
(a) $η_{1} = 1 - \frac{400}{600} = 0.33$
(b) $η_{2} = 1 - \frac{200}{400} = 0.50$
(c) $η_{3} = 1 - \frac{300}{500} = 0.40$
(d) $η_{4} = 1 - \frac{100}{300} = 0.66$
So the efficiency of carnot's heat engine is maximum for $T_{L} = 300 K$ and $T_{H} = 100 K$
Hence, Option (d) is correct.

Karnataka CET-2013

Thermodynamics

148592 If $γ$ 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

γ R

2

\frac{(γ - 1) R}{γ}

3

\frac{R}{γ - 1}

4

\frac{γ R}{γ - 1}

Explanation:

C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale.
$R = C_{P} - C_{v}$
Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume.
$γ = \frac{C_{p}}{C_{v}}$
$C_{P} = γ C_{V}$
Put the value of $C_{p}$ in equation (i)
$γ C_{v} - C_{v} = R$
$C_{v} (γ - 1) = R$
$C_{v} = \frac{R}{γ - 1}$

Karnataka CET-2008

Thermodynamics

148593 A Carnot engine taken heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be :

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

C Given that,
$T_{1} = 627^{\circ} C = 627 + 273 = 900 K$
$T_{2} = 27^{\circ} C = 27 + 273 = 300 K$
We know, efficiency of carnot engine,
$η = 1 - \frac{T_{2}}{T_{1}}$
$η = 1 - \frac{300}{900} = 1 - \frac{1}{3}$
$η = \frac{2}{3}$

Karnataka CET-2006

Thermodynamics

148594 A monoatomic gas is suddenly compressed to $(1 / 8)^{th}$ 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 $= V$
Final volume $= \frac{V}{8}$
$Adiabatic index, γ = \frac{5}{3}$
We know for adiabatic process ${PV}^{γ} =$ constant
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V}{\frac{V}{8}})}^{γ}$
$\frac{P_{2}}{P_{1}} = (8)^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{\frac{5}{3}} \Rightarrow \frac{P_{2}}{P_{1}} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$

Karnataka CET-2006

Thermodynamics

148595 A Carnot's engine is made to work between $200^{\circ} C$ and $0^{\circ} C$ first and then between $0^{\circ} C$ and $- 200^{\circ} C$ . 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 $(η) = 1 - \frac{T_{2}}{T_{1}}$
In $1^{st}$ case when temperature is between $200^{\circ} C$ to $0^{\circ} C$ .
$T_{2} = 0^{\circ} C = (0 + 273) K = 273 K$
$T_{1} = 200^{\circ} C = (200 + 273) K = 473 K$
$η_{1} = 1 - \frac{273}{473} = 0.423$
For $2^{nd}$ case when temperature is between $0^{\circ} C$ to $200^{\circ} C$
$T_{1} = (0^{\circ} + 273) = 273 K$
$T_{2} = (- 200 + 273) = 73 K$
$η_{2} = 1 - \frac{73}{273} = 0.733$
$∴ η_{1} : η_{2} = \frac{0.423}{0.733} = \frac{1}{1.73} = 1 : 1.73$

Karnataka CET-2002

Thermodynamics

148590 For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?

1

600 K, 400 K

2

400 K, 200 K

3

500 K, 300 K

4

300 K, 100 K

Explanation:

D Efficiency of carnot engine $η = 1 - \frac{T_{L}}{T_{H}}$
(a) $η_{1} = 1 - \frac{400}{600} = 0.33$
(b) $η_{2} = 1 - \frac{200}{400} = 0.50$
(c) $η_{3} = 1 - \frac{300}{500} = 0.40$
(d) $η_{4} = 1 - \frac{100}{300} = 0.66$
So the efficiency of carnot's heat engine is maximum for $T_{L} = 300 K$ and $T_{H} = 100 K$
Hence, Option (d) is correct.

Karnataka CET-2013

Thermodynamics

148592 If $γ$ 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

γ R

2

\frac{(γ - 1) R}{γ}

3

\frac{R}{γ - 1}

4

\frac{γ R}{γ - 1}

Explanation:

C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale.
$R = C_{P} - C_{v}$
Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume.
$γ = \frac{C_{p}}{C_{v}}$
$C_{P} = γ C_{V}$
Put the value of $C_{p}$ in equation (i)
$γ C_{v} - C_{v} = R$
$C_{v} (γ - 1) = R$
$C_{v} = \frac{R}{γ - 1}$

Karnataka CET-2008

Thermodynamics

148593 A Carnot engine taken heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be :

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

C Given that,
$T_{1} = 627^{\circ} C = 627 + 273 = 900 K$
$T_{2} = 27^{\circ} C = 27 + 273 = 300 K$
We know, efficiency of carnot engine,
$η = 1 - \frac{T_{2}}{T_{1}}$
$η = 1 - \frac{300}{900} = 1 - \frac{1}{3}$
$η = \frac{2}{3}$

Karnataka CET-2006

Thermodynamics

148594 A monoatomic gas is suddenly compressed to $(1 / 8)^{th}$ 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 $= V$
Final volume $= \frac{V}{8}$
$Adiabatic index, γ = \frac{5}{3}$
We know for adiabatic process ${PV}^{γ} =$ constant
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V}{\frac{V}{8}})}^{γ}$
$\frac{P_{2}}{P_{1}} = (8)^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{\frac{5}{3}} \Rightarrow \frac{P_{2}}{P_{1}} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$

Karnataka CET-2006

Thermodynamics

148595 A Carnot's engine is made to work between $200^{\circ} C$ and $0^{\circ} C$ first and then between $0^{\circ} C$ and $- 200^{\circ} C$ . 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 $(η) = 1 - \frac{T_{2}}{T_{1}}$
In $1^{st}$ case when temperature is between $200^{\circ} C$ to $0^{\circ} C$ .
$T_{2} = 0^{\circ} C = (0 + 273) K = 273 K$
$T_{1} = 200^{\circ} C = (200 + 273) K = 473 K$
$η_{1} = 1 - \frac{273}{473} = 0.423$
For $2^{nd}$ case when temperature is between $0^{\circ} C$ to $200^{\circ} C$
$T_{1} = (0^{\circ} + 273) = 273 K$
$T_{2} = (- 200 + 273) = 73 K$
$η_{2} = 1 - \frac{73}{273} = 0.733$
$∴ η_{1} : η_{2} = \frac{0.423}{0.733} = \frac{1}{1.73} = 1 : 1.73$

Karnataka CET-2002

Thermodynamics

148590 For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?

1

600 K, 400 K

2

400 K, 200 K

3

500 K, 300 K

4

300 K, 100 K

Explanation:

D Efficiency of carnot engine $η = 1 - \frac{T_{L}}{T_{H}}$
(a) $η_{1} = 1 - \frac{400}{600} = 0.33$
(b) $η_{2} = 1 - \frac{200}{400} = 0.50$
(c) $η_{3} = 1 - \frac{300}{500} = 0.40$
(d) $η_{4} = 1 - \frac{100}{300} = 0.66$
So the efficiency of carnot's heat engine is maximum for $T_{L} = 300 K$ and $T_{H} = 100 K$
Hence, Option (d) is correct.

Karnataka CET-2013

Thermodynamics

148592 If $γ$ 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

γ R

2

\frac{(γ - 1) R}{γ}

3

\frac{R}{γ - 1}

4

\frac{γ R}{γ - 1}

Explanation:

C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale.
$R = C_{P} - C_{v}$
Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume.
$γ = \frac{C_{p}}{C_{v}}$
$C_{P} = γ C_{V}$
Put the value of $C_{p}$ in equation (i)
$γ C_{v} - C_{v} = R$
$C_{v} (γ - 1) = R$
$C_{v} = \frac{R}{γ - 1}$

Karnataka CET-2008

Thermodynamics

148593 A Carnot engine taken heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be :

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

C Given that,
$T_{1} = 627^{\circ} C = 627 + 273 = 900 K$
$T_{2} = 27^{\circ} C = 27 + 273 = 300 K$
We know, efficiency of carnot engine,
$η = 1 - \frac{T_{2}}{T_{1}}$
$η = 1 - \frac{300}{900} = 1 - \frac{1}{3}$
$η = \frac{2}{3}$

Karnataka CET-2006

Thermodynamics

148594 A monoatomic gas is suddenly compressed to $(1 / 8)^{th}$ 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 $= V$
Final volume $= \frac{V}{8}$
$Adiabatic index, γ = \frac{5}{3}$
We know for adiabatic process ${PV}^{γ} =$ constant
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V}{\frac{V}{8}})}^{γ}$
$\frac{P_{2}}{P_{1}} = (8)^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{\frac{5}{3}} \Rightarrow \frac{P_{2}}{P_{1}} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$

Karnataka CET-2006

Thermodynamics

148595 A Carnot's engine is made to work between $200^{\circ} C$ and $0^{\circ} C$ first and then between $0^{\circ} C$ and $- 200^{\circ} C$ . 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 $(η) = 1 - \frac{T_{2}}{T_{1}}$
In $1^{st}$ case when temperature is between $200^{\circ} C$ to $0^{\circ} C$ .
$T_{2} = 0^{\circ} C = (0 + 273) K = 273 K$
$T_{1} = 200^{\circ} C = (200 + 273) K = 473 K$
$η_{1} = 1 - \frac{273}{473} = 0.423$
For $2^{nd}$ case when temperature is between $0^{\circ} C$ to $200^{\circ} C$
$T_{1} = (0^{\circ} + 273) = 273 K$
$T_{2} = (- 200 + 273) = 73 K$
$η_{2} = 1 - \frac{73}{273} = 0.733$
$∴ η_{1} : η_{2} = \frac{0.423}{0.733} = \frac{1}{1.73} = 1 : 1.73$

Karnataka CET-2002

Thermodynamics

148590 For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?

1

600 K, 400 K

2

400 K, 200 K

3

500 K, 300 K

4

300 K, 100 K

Explanation:

D Efficiency of carnot engine $η = 1 - \frac{T_{L}}{T_{H}}$
(a) $η_{1} = 1 - \frac{400}{600} = 0.33$
(b) $η_{2} = 1 - \frac{200}{400} = 0.50$
(c) $η_{3} = 1 - \frac{300}{500} = 0.40$
(d) $η_{4} = 1 - \frac{100}{300} = 0.66$
So the efficiency of carnot's heat engine is maximum for $T_{L} = 300 K$ and $T_{H} = 100 K$
Hence, Option (d) is correct.

Karnataka CET-2013

Thermodynamics

148592 If $γ$ 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

γ R

2

\frac{(γ - 1) R}{γ}

3

\frac{R}{γ - 1}

4

\frac{γ R}{γ - 1}

Explanation:

C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale.
$R = C_{P} - C_{v}$
Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume.
$γ = \frac{C_{p}}{C_{v}}$
$C_{P} = γ C_{V}$
Put the value of $C_{p}$ in equation (i)
$γ C_{v} - C_{v} = R$
$C_{v} (γ - 1) = R$
$C_{v} = \frac{R}{γ - 1}$

Karnataka CET-2008

Thermodynamics

148593 A Carnot engine taken heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be :

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

C Given that,
$T_{1} = 627^{\circ} C = 627 + 273 = 900 K$
$T_{2} = 27^{\circ} C = 27 + 273 = 300 K$
We know, efficiency of carnot engine,
$η = 1 - \frac{T_{2}}{T_{1}}$
$η = 1 - \frac{300}{900} = 1 - \frac{1}{3}$
$η = \frac{2}{3}$

Karnataka CET-2006

Thermodynamics

148594 A monoatomic gas is suddenly compressed to $(1 / 8)^{th}$ 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 $= V$
Final volume $= \frac{V}{8}$
$Adiabatic index, γ = \frac{5}{3}$
We know for adiabatic process ${PV}^{γ} =$ constant
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V}{\frac{V}{8}})}^{γ}$
$\frac{P_{2}}{P_{1}} = (8)^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{\frac{5}{3}} \Rightarrow \frac{P_{2}}{P_{1}} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$

Karnataka CET-2006

Thermodynamics

148595 A Carnot's engine is made to work between $200^{\circ} C$ and $0^{\circ} C$ first and then between $0^{\circ} C$ and $- 200^{\circ} C$ . 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 $(η) = 1 - \frac{T_{2}}{T_{1}}$
In $1^{st}$ case when temperature is between $200^{\circ} C$ to $0^{\circ} C$ .
$T_{2} = 0^{\circ} C = (0 + 273) K = 273 K$
$T_{1} = 200^{\circ} C = (200 + 273) K = 473 K$
$η_{1} = 1 - \frac{273}{473} = 0.423$
For $2^{nd}$ case when temperature is between $0^{\circ} C$ to $200^{\circ} C$
$T_{1} = (0^{\circ} + 273) = 273 K$
$T_{2} = (- 200 + 273) = 73 K$
$η_{2} = 1 - \frac{73}{273} = 0.733$
$∴ η_{1} : η_{2} = \frac{0.423}{0.733} = \frac{1}{1.73} = 1 : 1.73$

Karnataka CET-2002

Thermodynamics

148590 For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum ?

1

600 K, 400 K

2

400 K, 200 K

3

500 K, 300 K

4

300 K, 100 K

Explanation:

D Efficiency of carnot engine $η = 1 - \frac{T_{L}}{T_{H}}$
(a) $η_{1} = 1 - \frac{400}{600} = 0.33$
(b) $η_{2} = 1 - \frac{200}{400} = 0.50$
(c) $η_{3} = 1 - \frac{300}{500} = 0.40$
(d) $η_{4} = 1 - \frac{100}{300} = 0.66$
So the efficiency of carnot's heat engine is maximum for $T_{L} = 300 K$ and $T_{H} = 100 K$
Hence, Option (d) is correct.

Karnataka CET-2013

Thermodynamics

148592 If $γ$ 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

γ R

2

\frac{(γ - 1) R}{γ}

3

\frac{R}{γ - 1}

4

\frac{γ R}{γ - 1}

Explanation:

C Gas constant (R): It is the proportionality constant used to relate the energy scale to the temperature scale.
$R = C_{P} - C_{v}$
Heat capacity ratio is the ratio of specific heat capacity at constant pressure to that of constant volume.
$γ = \frac{C_{p}}{C_{v}}$
$C_{P} = γ C_{V}$
Put the value of $C_{p}$ in equation (i)
$γ C_{v} - C_{v} = R$
$C_{v} (γ - 1) = R$
$C_{v} = \frac{R}{γ - 1}$

Karnataka CET-2008

Thermodynamics

148593 A Carnot engine taken heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be :

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

C Given that,
$T_{1} = 627^{\circ} C = 627 + 273 = 900 K$
$T_{2} = 27^{\circ} C = 27 + 273 = 300 K$
We know, efficiency of carnot engine,
$η = 1 - \frac{T_{2}}{T_{1}}$
$η = 1 - \frac{300}{900} = 1 - \frac{1}{3}$
$η = \frac{2}{3}$

Karnataka CET-2006

Thermodynamics

148594 A monoatomic gas is suddenly compressed to $(1 / 8)^{th}$ 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 $= V$
Final volume $= \frac{V}{8}$
$Adiabatic index, γ = \frac{5}{3}$
We know for adiabatic process ${PV}^{γ} =$ constant
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V}{\frac{V}{8}})}^{γ}$
$\frac{P_{2}}{P_{1}} = (8)^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{\frac{5}{3}} \Rightarrow \frac{P_{2}}{P_{1}} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$

Karnataka CET-2006

Thermodynamics

148595 A Carnot's engine is made to work between $200^{\circ} C$ and $0^{\circ} C$ first and then between $0^{\circ} C$ and $- 200^{\circ} C$ . 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 $(η) = 1 - \frac{T_{2}}{T_{1}}$
In $1^{st}$ case when temperature is between $200^{\circ} C$ to $0^{\circ} C$ .
$T_{2} = 0^{\circ} C = (0 + 273) K = 273 K$
$T_{1} = 200^{\circ} C = (200 + 273) K = 473 K$
$η_{1} = 1 - \frac{273}{473} = 0.423$
For $2^{nd}$ case when temperature is between $0^{\circ} C$ to $200^{\circ} C$
$T_{1} = (0^{\circ} + 273) = 273 K$
$T_{2} = (- 200 + 273) = 73 K$
$η_{2} = 1 - \frac{73}{273} = 0.733$
$∴ η_{1} : η_{2} = \frac{0.423}{0.733} = \frac{1}{1.73} = 1 : 1.73$

Karnataka CET-2002