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Thermodynamics

148549 A diatomic ideal gas is used in Carnot's engine as working substance. During adiabatic expansion of the cycle, if the volume of the gas increases from $V$ to $32 V$ , then the efficiency of the engine is

1 0.25

2 0.5

3 0.67

4 0.75

Explanation:

D Given

For an adiabatic process,
${TV}^{γ - 1} =$ constant
$∴ T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
Or
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V}{32 V})}^{γ - 1} = \frac{1}{(32)^{γ - 1}}$
For diatomic gas, $γ = \frac{7}{5}$
$∴ \frac{T_{2}}{T_{1}} = \frac{1}{(32)^{7 / 3 - 1}} = \frac{1}{{(2^{5})}^{7 / 3 - 1}} = \frac{1}{{(2^{5})}^{2 / 5}} = \frac{1}{4}$
Now efficiency of Carnot cycle,
$η = 1 - \frac{T_{2}}{T_{1}}$
$= 1 - \frac{1}{4}$
$η = \frac{3}{4} = 0.75$

AP EAMCET (22.04.2018) Shift-II

Thermodynamics

148550 A reversible Carnot heat engine converts $\frac{1}{4}$ th of its input heat into work. When the temperature of the sink is reduced by $50 K$ , its efficiency becomes $33 \frac{1}{3} %$ . The initial temperatures of the source and the sink respectively are

1

600 K, 550 K

2

600 K, 450 K

3

300 K, 150 K

4

450 K, 350 K

Explanation:

B Given,
Work done, $W = \frac{Q}{4}$
Where,
$Q = Heat input$
Let, the source temperature $= T_{1}$ and sink Temperature $= T_{2}$
$Efficiency, η = \frac{W}{Q} = \frac{Q / 4}{Q}$
Sink Temperature reduced by $50 K$
Then, $T_{2} - 50 K$ and $η = \frac{1}{4}$
$η = 1 - \frac{T_{2}}{T_{1}} = \frac{1}{4}$
When temperature of sink is reduced by $50 K$ , efficiency become $\frac{100}{3} %$
$η = \frac{100}{3 \times 100} = \frac{1}{3}$
So,
$1 - \frac{T_{2} - 50}{T_{1}} = \frac{1}{3}$
$1 - \frac{T_{2}}{T_{1}} + \frac{50}{T_{1}} = \frac{1}{3}$
Put the value of equation (i) in equation (ii),
$\frac{1}{4} + \frac{50}{T_{1}} = \frac{1}{3}$
$\frac{50}{T_{1}} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$
$T_{1} = 600 K$
From equation (i),
$1 - \frac{T_{2}}{T_{1}} = \frac{1}{4}$
$Then, \frac{T_{2}}{T_{1}} = \frac{3}{4}$
$T_{2} = \frac{3}{4} \times 600$
$= 450 K$

AP EAMCET (20.04.2019) Shift-1

Thermodynamics

148551 Freezing compartment of a refrigerator is at $0^{\circ} C$ and room temperature is ${27.3}^{\circ} C$ . Work done by the refrigerator to freeze $1 g$ of water at $0^{\circ} C$ is $(L_{ice} = 80 cal g^{- 1})$

1

336 J

2

33.6 J

3

3.36 J

4

40 J

Explanation:

B Coefficient of performance of refrigerator is
$α = \frac{T_{2}}{T_{1} - T_{2}}$
$T_{2} = 0^{\circ} C = 0 + 273 = 273 K$
$T_{1} = {27.3}^{\circ} C + 273 = 300.3 ≃ 300 K$
$α = \frac{273}{300 - 273} = 10.11 ≃ 10$
Now, if $Q_{2}$ is heat extracted and $W$ is the work done,
$α = \frac{Q_{2}}{W} or W = \frac{Q}{α}$
Where $Q_{2} =$ heat extracted form $1 g$ of water to make it ice.
$Q_{2} = Q = L_{ice} = 80 cal = 80 \times 4.2 = 336 J$
So,
$W = 336 / 10$
$= 33.6 J$

AP EAMCET (23.04.2018) Shift-1

Thermodynamics

148552 A Carnot engine works first between $200^{\circ} C$ and $0^{\circ} C$ and then between $0^{\circ} C$ and $- 200^{\circ} C$ . The ratio of its efficiency in the two cases is

1 1.0

2 0.577

3 0.34

4 0.68

Explanation:

B For case $I : T_{1} = 200^{\circ} C = 273 + 200 = 473 K$ $T_{2} = 0^{\circ} C + 273 = 273 K$
$η_{1} = 1 - \frac{T_{2}}{T_{1}}$
$η_{1} = 1 - \frac{273}{473} = \frac{200}{473}$
For case II : $T_{1}^{'} = 0^{\circ} C + 273 = 273 K$
$T_{2}^{'} = - 200^{\circ} C + 273 = 73 K$
$η_{2} = 1 - \frac{T_{2}^{'}}{T_{1}^{'}}$
$η_{2} = 1 - \frac{73}{273} = \frac{200}{273}$
$\frac{η_{1}}{η_{2}} = \frac{200 / 473}{200 / 273} = \frac{273}{473} = 0.577$
$\frac{η_{1}}{η_{2}} = 0.577$

BITSAT-2007

Thermodynamics

148549 A diatomic ideal gas is used in Carnot's engine as working substance. During adiabatic expansion of the cycle, if the volume of the gas increases from $V$ to $32 V$ , then the efficiency of the engine is

1 0.25

2 0.5

3 0.67

4 0.75

Explanation:

D Given

For an adiabatic process,
${TV}^{γ - 1} =$ constant
$∴ T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
Or
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V}{32 V})}^{γ - 1} = \frac{1}{(32)^{γ - 1}}$
For diatomic gas, $γ = \frac{7}{5}$
$∴ \frac{T_{2}}{T_{1}} = \frac{1}{(32)^{7 / 3 - 1}} = \frac{1}{{(2^{5})}^{7 / 3 - 1}} = \frac{1}{{(2^{5})}^{2 / 5}} = \frac{1}{4}$
Now efficiency of Carnot cycle,
$η = 1 - \frac{T_{2}}{T_{1}}$
$= 1 - \frac{1}{4}$
$η = \frac{3}{4} = 0.75$

AP EAMCET (22.04.2018) Shift-II

Thermodynamics

148550 A reversible Carnot heat engine converts $\frac{1}{4}$ th of its input heat into work. When the temperature of the sink is reduced by $50 K$ , its efficiency becomes $33 \frac{1}{3} %$ . The initial temperatures of the source and the sink respectively are

1

600 K, 550 K

2

600 K, 450 K

3

300 K, 150 K

4

450 K, 350 K

Explanation:

B Given,
Work done, $W = \frac{Q}{4}$
Where,
$Q = Heat input$
Let, the source temperature $= T_{1}$ and sink Temperature $= T_{2}$
$Efficiency, η = \frac{W}{Q} = \frac{Q / 4}{Q}$
Sink Temperature reduced by $50 K$
Then, $T_{2} - 50 K$ and $η = \frac{1}{4}$
$η = 1 - \frac{T_{2}}{T_{1}} = \frac{1}{4}$
When temperature of sink is reduced by $50 K$ , efficiency become $\frac{100}{3} %$
$η = \frac{100}{3 \times 100} = \frac{1}{3}$
So,
$1 - \frac{T_{2} - 50}{T_{1}} = \frac{1}{3}$
$1 - \frac{T_{2}}{T_{1}} + \frac{50}{T_{1}} = \frac{1}{3}$
Put the value of equation (i) in equation (ii),
$\frac{1}{4} + \frac{50}{T_{1}} = \frac{1}{3}$
$\frac{50}{T_{1}} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$
$T_{1} = 600 K$
From equation (i),
$1 - \frac{T_{2}}{T_{1}} = \frac{1}{4}$
$Then, \frac{T_{2}}{T_{1}} = \frac{3}{4}$
$T_{2} = \frac{3}{4} \times 600$
$= 450 K$

AP EAMCET (20.04.2019) Shift-1

Thermodynamics

148551 Freezing compartment of a refrigerator is at $0^{\circ} C$ and room temperature is ${27.3}^{\circ} C$ . Work done by the refrigerator to freeze $1 g$ of water at $0^{\circ} C$ is $(L_{ice} = 80 cal g^{- 1})$

1

336 J

2

33.6 J

3

3.36 J

4

40 J

Explanation:

B Coefficient of performance of refrigerator is
$α = \frac{T_{2}}{T_{1} - T_{2}}$
$T_{2} = 0^{\circ} C = 0 + 273 = 273 K$
$T_{1} = {27.3}^{\circ} C + 273 = 300.3 ≃ 300 K$
$α = \frac{273}{300 - 273} = 10.11 ≃ 10$
Now, if $Q_{2}$ is heat extracted and $W$ is the work done,
$α = \frac{Q_{2}}{W} or W = \frac{Q}{α}$
Where $Q_{2} =$ heat extracted form $1 g$ of water to make it ice.
$Q_{2} = Q = L_{ice} = 80 cal = 80 \times 4.2 = 336 J$
So,
$W = 336 / 10$
$= 33.6 J$

AP EAMCET (23.04.2018) Shift-1

Thermodynamics

148552 A Carnot engine works first between $200^{\circ} C$ and $0^{\circ} C$ and then between $0^{\circ} C$ and $- 200^{\circ} C$ . The ratio of its efficiency in the two cases is

1 1.0

2 0.577

3 0.34

4 0.68

Explanation:

B For case $I : T_{1} = 200^{\circ} C = 273 + 200 = 473 K$ $T_{2} = 0^{\circ} C + 273 = 273 K$
$η_{1} = 1 - \frac{T_{2}}{T_{1}}$
$η_{1} = 1 - \frac{273}{473} = \frac{200}{473}$
For case II : $T_{1}^{'} = 0^{\circ} C + 273 = 273 K$
$T_{2}^{'} = - 200^{\circ} C + 273 = 73 K$
$η_{2} = 1 - \frac{T_{2}^{'}}{T_{1}^{'}}$
$η_{2} = 1 - \frac{73}{273} = \frac{200}{273}$
$\frac{η_{1}}{η_{2}} = \frac{200 / 473}{200 / 273} = \frac{273}{473} = 0.577$
$\frac{η_{1}}{η_{2}} = 0.577$

BITSAT-2007

Thermodynamics

148549 A diatomic ideal gas is used in Carnot's engine as working substance. During adiabatic expansion of the cycle, if the volume of the gas increases from $V$ to $32 V$ , then the efficiency of the engine is

1 0.25

2 0.5

3 0.67

4 0.75

Explanation:

D Given

For an adiabatic process,
${TV}^{γ - 1} =$ constant
$∴ T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
Or
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V}{32 V})}^{γ - 1} = \frac{1}{(32)^{γ - 1}}$
For diatomic gas, $γ = \frac{7}{5}$
$∴ \frac{T_{2}}{T_{1}} = \frac{1}{(32)^{7 / 3 - 1}} = \frac{1}{{(2^{5})}^{7 / 3 - 1}} = \frac{1}{{(2^{5})}^{2 / 5}} = \frac{1}{4}$
Now efficiency of Carnot cycle,
$η = 1 - \frac{T_{2}}{T_{1}}$
$= 1 - \frac{1}{4}$
$η = \frac{3}{4} = 0.75$

AP EAMCET (22.04.2018) Shift-II

Thermodynamics

148550 A reversible Carnot heat engine converts $\frac{1}{4}$ th of its input heat into work. When the temperature of the sink is reduced by $50 K$ , its efficiency becomes $33 \frac{1}{3} %$ . The initial temperatures of the source and the sink respectively are

1

600 K, 550 K

2

600 K, 450 K

3

300 K, 150 K

4

450 K, 350 K

Explanation:

B Given,
Work done, $W = \frac{Q}{4}$
Where,
$Q = Heat input$
Let, the source temperature $= T_{1}$ and sink Temperature $= T_{2}$
$Efficiency, η = \frac{W}{Q} = \frac{Q / 4}{Q}$
Sink Temperature reduced by $50 K$
Then, $T_{2} - 50 K$ and $η = \frac{1}{4}$
$η = 1 - \frac{T_{2}}{T_{1}} = \frac{1}{4}$
When temperature of sink is reduced by $50 K$ , efficiency become $\frac{100}{3} %$
$η = \frac{100}{3 \times 100} = \frac{1}{3}$
So,
$1 - \frac{T_{2} - 50}{T_{1}} = \frac{1}{3}$
$1 - \frac{T_{2}}{T_{1}} + \frac{50}{T_{1}} = \frac{1}{3}$
Put the value of equation (i) in equation (ii),
$\frac{1}{4} + \frac{50}{T_{1}} = \frac{1}{3}$
$\frac{50}{T_{1}} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$
$T_{1} = 600 K$
From equation (i),
$1 - \frac{T_{2}}{T_{1}} = \frac{1}{4}$
$Then, \frac{T_{2}}{T_{1}} = \frac{3}{4}$
$T_{2} = \frac{3}{4} \times 600$
$= 450 K$

AP EAMCET (20.04.2019) Shift-1

Thermodynamics

148551 Freezing compartment of a refrigerator is at $0^{\circ} C$ and room temperature is ${27.3}^{\circ} C$ . Work done by the refrigerator to freeze $1 g$ of water at $0^{\circ} C$ is $(L_{ice} = 80 cal g^{- 1})$

1

336 J

2

33.6 J

3

3.36 J

4

40 J

Explanation:

B Coefficient of performance of refrigerator is
$α = \frac{T_{2}}{T_{1} - T_{2}}$
$T_{2} = 0^{\circ} C = 0 + 273 = 273 K$
$T_{1} = {27.3}^{\circ} C + 273 = 300.3 ≃ 300 K$
$α = \frac{273}{300 - 273} = 10.11 ≃ 10$
Now, if $Q_{2}$ is heat extracted and $W$ is the work done,
$α = \frac{Q_{2}}{W} or W = \frac{Q}{α}$
Where $Q_{2} =$ heat extracted form $1 g$ of water to make it ice.
$Q_{2} = Q = L_{ice} = 80 cal = 80 \times 4.2 = 336 J$
So,
$W = 336 / 10$
$= 33.6 J$

AP EAMCET (23.04.2018) Shift-1

Thermodynamics

148552 A Carnot engine works first between $200^{\circ} C$ and $0^{\circ} C$ and then between $0^{\circ} C$ and $- 200^{\circ} C$ . The ratio of its efficiency in the two cases is

1 1.0

2 0.577

3 0.34

4 0.68

Explanation:

B For case $I : T_{1} = 200^{\circ} C = 273 + 200 = 473 K$ $T_{2} = 0^{\circ} C + 273 = 273 K$
$η_{1} = 1 - \frac{T_{2}}{T_{1}}$
$η_{1} = 1 - \frac{273}{473} = \frac{200}{473}$
For case II : $T_{1}^{'} = 0^{\circ} C + 273 = 273 K$
$T_{2}^{'} = - 200^{\circ} C + 273 = 73 K$
$η_{2} = 1 - \frac{T_{2}^{'}}{T_{1}^{'}}$
$η_{2} = 1 - \frac{73}{273} = \frac{200}{273}$
$\frac{η_{1}}{η_{2}} = \frac{200 / 473}{200 / 273} = \frac{273}{473} = 0.577$
$\frac{η_{1}}{η_{2}} = 0.577$

BITSAT-2007

Thermodynamics

148549 A diatomic ideal gas is used in Carnot's engine as working substance. During adiabatic expansion of the cycle, if the volume of the gas increases from $V$ to $32 V$ , then the efficiency of the engine is

1 0.25

2 0.5

3 0.67

4 0.75

Explanation:

D Given

For an adiabatic process,
${TV}^{γ - 1} =$ constant
$∴ T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
Or
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V}{32 V})}^{γ - 1} = \frac{1}{(32)^{γ - 1}}$
For diatomic gas, $γ = \frac{7}{5}$
$∴ \frac{T_{2}}{T_{1}} = \frac{1}{(32)^{7 / 3 - 1}} = \frac{1}{{(2^{5})}^{7 / 3 - 1}} = \frac{1}{{(2^{5})}^{2 / 5}} = \frac{1}{4}$
Now efficiency of Carnot cycle,
$η = 1 - \frac{T_{2}}{T_{1}}$
$= 1 - \frac{1}{4}$
$η = \frac{3}{4} = 0.75$

AP EAMCET (22.04.2018) Shift-II

Thermodynamics

148550 A reversible Carnot heat engine converts $\frac{1}{4}$ th of its input heat into work. When the temperature of the sink is reduced by $50 K$ , its efficiency becomes $33 \frac{1}{3} %$ . The initial temperatures of the source and the sink respectively are

1

600 K, 550 K

2

600 K, 450 K

3

300 K, 150 K

4

450 K, 350 K

Explanation:

B Given,
Work done, $W = \frac{Q}{4}$
Where,
$Q = Heat input$
Let, the source temperature $= T_{1}$ and sink Temperature $= T_{2}$
$Efficiency, η = \frac{W}{Q} = \frac{Q / 4}{Q}$
Sink Temperature reduced by $50 K$
Then, $T_{2} - 50 K$ and $η = \frac{1}{4}$
$η = 1 - \frac{T_{2}}{T_{1}} = \frac{1}{4}$
When temperature of sink is reduced by $50 K$ , efficiency become $\frac{100}{3} %$
$η = \frac{100}{3 \times 100} = \frac{1}{3}$
So,
$1 - \frac{T_{2} - 50}{T_{1}} = \frac{1}{3}$
$1 - \frac{T_{2}}{T_{1}} + \frac{50}{T_{1}} = \frac{1}{3}$
Put the value of equation (i) in equation (ii),
$\frac{1}{4} + \frac{50}{T_{1}} = \frac{1}{3}$
$\frac{50}{T_{1}} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$
$T_{1} = 600 K$
From equation (i),
$1 - \frac{T_{2}}{T_{1}} = \frac{1}{4}$
$Then, \frac{T_{2}}{T_{1}} = \frac{3}{4}$
$T_{2} = \frac{3}{4} \times 600$
$= 450 K$

AP EAMCET (20.04.2019) Shift-1

Thermodynamics

148551 Freezing compartment of a refrigerator is at $0^{\circ} C$ and room temperature is ${27.3}^{\circ} C$ . Work done by the refrigerator to freeze $1 g$ of water at $0^{\circ} C$ is $(L_{ice} = 80 cal g^{- 1})$

1

336 J

2

33.6 J

3

3.36 J

4

40 J

Explanation:

B Coefficient of performance of refrigerator is
$α = \frac{T_{2}}{T_{1} - T_{2}}$
$T_{2} = 0^{\circ} C = 0 + 273 = 273 K$
$T_{1} = {27.3}^{\circ} C + 273 = 300.3 ≃ 300 K$
$α = \frac{273}{300 - 273} = 10.11 ≃ 10$
Now, if $Q_{2}$ is heat extracted and $W$ is the work done,
$α = \frac{Q_{2}}{W} or W = \frac{Q}{α}$
Where $Q_{2} =$ heat extracted form $1 g$ of water to make it ice.
$Q_{2} = Q = L_{ice} = 80 cal = 80 \times 4.2 = 336 J$
So,
$W = 336 / 10$
$= 33.6 J$

AP EAMCET (23.04.2018) Shift-1

Thermodynamics

148552 A Carnot engine works first between $200^{\circ} C$ and $0^{\circ} C$ and then between $0^{\circ} C$ and $- 200^{\circ} C$ . The ratio of its efficiency in the two cases is

1 1.0

2 0.577

3 0.34

4 0.68

Explanation:

B For case $I : T_{1} = 200^{\circ} C = 273 + 200 = 473 K$ $T_{2} = 0^{\circ} C + 273 = 273 K$
$η_{1} = 1 - \frac{T_{2}}{T_{1}}$
$η_{1} = 1 - \frac{273}{473} = \frac{200}{473}$
For case II : $T_{1}^{'} = 0^{\circ} C + 273 = 273 K$
$T_{2}^{'} = - 200^{\circ} C + 273 = 73 K$
$η_{2} = 1 - \frac{T_{2}^{'}}{T_{1}^{'}}$
$η_{2} = 1 - \frac{73}{273} = \frac{200}{273}$
$\frac{η_{1}}{η_{2}} = \frac{200 / 473}{200 / 273} = \frac{273}{473} = 0.577$
$\frac{η_{1}}{η_{2}} = 0.577$

BITSAT-2007

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