QBManager

Thermodynamics

148097 An ideal gas is taken through the cycle $A \to B \to C \to A$ , as shown in the figure below. If the net heat supplied to the gas is $5 J$ , then the work done by the gas in the process $C \to A$ is

1

- 5 J

2

- 10 J

3

- 15 J

4

- 20 J

Explanation:

A Given, $Q_{net} = 5 J$
$W_{BC} = PdV = 0 (∵ constant volume)$
$W_{AB} = 10 \times (2 - 1) = 10 J$
$\oint dQ = \oint dW$
$5 = W_{AB} + W_{BC} + W_{CA}$
$W_{CA} = 5 - W_{AB} - W_{BC}$
$W_{CA} = 5 - 10 - 0 = - 5 J$
$W_{CA} = - 5 J$

CG PET -2018

Thermodynamics

148098 Calculate the heat required to increases the temperature of 1 mole of one atomic gas from $0^{\circ} C$ to $150^{\circ} C$ , when no work is done. $[C_{p} = 2.5 R$ and $R = 83 J {mol}^{- 1} K^{- 1}$ ]

1

867.5 J

2

186.5 J

3

1867.5 J

4

86.7 J

Explanation:

C Given
$n = 1$ mole
$Δ T = T_{2} - T_{1} = (150 - 0) = 150^{\circ} C$
$R = 8.3 J / K . mol, C_{P} = 2.5 R$
$C_{P} - C_{V} = R \Rightarrow C_{V} = C_{P} - R = 2.5 R - R = 1.5 R$
$dQ = dU + dW \Rightarrow dQ = dU (∵ dW = 0)$
$dQ = {nC}_{V} Δ T$
$dQ = n 1.5 R Δ T$
$dQ = 1 \times 1.5 \times 8.3 \times 150 = 1867.5 J$
$dQ = 1867.5 J$

CG PET- 2014

Thermodynamics

148100 Ideal gas is contained in a thermally insulated and rigid container and it is heated through a resistance $100 Ω$ by passing a current of $1 A$ for five minutes, then change in internal energy of the gas is

1

0 kJ

2

30 kJ

3

10 kJ

4

20 kJ

Explanation:

B Given, $R = 100 Ω, i = 1 A$
$t = 5 \min . = 5 \times 60 \sec = 300 \sec$
When current is flowing through a wire then heat produces,
$Q = i^{2} Rt = 1 \times 100 \times 300 = 3 \times 10^{4} J$
$Q = 30 kJ$
Therefore change in internal energy is $30 kJ$ .

Manipal UGET-2014

Thermodynamics

148101 $1 {cm}^{3}$ of water at its boiling point absorbs 540 cal of heat to becomes steam with a volume of $1671 {cm}^{3}$ . If the atmospheric pressure $=$ $1.013 \times 10^{5} N / m^{2}$ and the mechanical equivalent of heat $= 4.19 J / cal$ , the energy spent in this process in overcoming intermolecular process in overcoming intermolecular forces is

1

540 cal

2

40 cal

3

500 cal

4 zero

Explanation:

C Given, $V_{1} = 1 {cm}^{3}, V_{2} = 1671 {cm}^{3}$
and $Q = 540 cal$
Atmospheric pressure $= 1.013 \times 10^{5} N / m^{2}$
Mechanical equivalent $(J) = 4.19 J / cal$
According to first law of thermodynamics
$Δ Δ = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$Δ U = 540 - \frac{P (V_{2} - V_{1})}{J}$
$Δ U = 540 - \frac{1.013 \times 10^{5} \times [(1671 - 1) \times 10^{- 6}]}{4.2}$
$Δ U = 540 - 40 = 500 cal$

Manipal UGET-2010

Thermodynamics

148097 An ideal gas is taken through the cycle $A \to B \to C \to A$ , as shown in the figure below. If the net heat supplied to the gas is $5 J$ , then the work done by the gas in the process $C \to A$ is

1

- 5 J

2

- 10 J

3

- 15 J

4

- 20 J

Explanation:

A Given, $Q_{net} = 5 J$
$W_{BC} = PdV = 0 (∵ constant volume)$
$W_{AB} = 10 \times (2 - 1) = 10 J$
$\oint dQ = \oint dW$
$5 = W_{AB} + W_{BC} + W_{CA}$
$W_{CA} = 5 - W_{AB} - W_{BC}$
$W_{CA} = 5 - 10 - 0 = - 5 J$
$W_{CA} = - 5 J$

CG PET -2018

Thermodynamics

148098 Calculate the heat required to increases the temperature of 1 mole of one atomic gas from $0^{\circ} C$ to $150^{\circ} C$ , when no work is done. $[C_{p} = 2.5 R$ and $R = 83 J {mol}^{- 1} K^{- 1}$ ]

1

867.5 J

2

186.5 J

3

1867.5 J

4

86.7 J

Explanation:

C Given
$n = 1$ mole
$Δ T = T_{2} - T_{1} = (150 - 0) = 150^{\circ} C$
$R = 8.3 J / K . mol, C_{P} = 2.5 R$
$C_{P} - C_{V} = R \Rightarrow C_{V} = C_{P} - R = 2.5 R - R = 1.5 R$
$dQ = dU + dW \Rightarrow dQ = dU (∵ dW = 0)$
$dQ = {nC}_{V} Δ T$
$dQ = n 1.5 R Δ T$
$dQ = 1 \times 1.5 \times 8.3 \times 150 = 1867.5 J$
$dQ = 1867.5 J$

CG PET- 2014

Thermodynamics

148100 Ideal gas is contained in a thermally insulated and rigid container and it is heated through a resistance $100 Ω$ by passing a current of $1 A$ for five minutes, then change in internal energy of the gas is

1

0 kJ

2

30 kJ

3

10 kJ

4

20 kJ

Explanation:

B Given, $R = 100 Ω, i = 1 A$
$t = 5 \min . = 5 \times 60 \sec = 300 \sec$
When current is flowing through a wire then heat produces,
$Q = i^{2} Rt = 1 \times 100 \times 300 = 3 \times 10^{4} J$
$Q = 30 kJ$
Therefore change in internal energy is $30 kJ$ .

Manipal UGET-2014

Thermodynamics

148101 $1 {cm}^{3}$ of water at its boiling point absorbs 540 cal of heat to becomes steam with a volume of $1671 {cm}^{3}$ . If the atmospheric pressure $=$ $1.013 \times 10^{5} N / m^{2}$ and the mechanical equivalent of heat $= 4.19 J / cal$ , the energy spent in this process in overcoming intermolecular process in overcoming intermolecular forces is

1

540 cal

2

40 cal

3

500 cal

4 zero

Explanation:

C Given, $V_{1} = 1 {cm}^{3}, V_{2} = 1671 {cm}^{3}$
and $Q = 540 cal$
Atmospheric pressure $= 1.013 \times 10^{5} N / m^{2}$
Mechanical equivalent $(J) = 4.19 J / cal$
According to first law of thermodynamics
$Δ Δ = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$Δ U = 540 - \frac{P (V_{2} - V_{1})}{J}$
$Δ U = 540 - \frac{1.013 \times 10^{5} \times [(1671 - 1) \times 10^{- 6}]}{4.2}$
$Δ U = 540 - 40 = 500 cal$

Manipal UGET-2010

Thermodynamics

148097 An ideal gas is taken through the cycle $A \to B \to C \to A$ , as shown in the figure below. If the net heat supplied to the gas is $5 J$ , then the work done by the gas in the process $C \to A$ is

1

- 5 J

2

- 10 J

3

- 15 J

4

- 20 J

Explanation:

A Given, $Q_{net} = 5 J$
$W_{BC} = PdV = 0 (∵ constant volume)$
$W_{AB} = 10 \times (2 - 1) = 10 J$
$\oint dQ = \oint dW$
$5 = W_{AB} + W_{BC} + W_{CA}$
$W_{CA} = 5 - W_{AB} - W_{BC}$
$W_{CA} = 5 - 10 - 0 = - 5 J$
$W_{CA} = - 5 J$

CG PET -2018

Thermodynamics

148098 Calculate the heat required to increases the temperature of 1 mole of one atomic gas from $0^{\circ} C$ to $150^{\circ} C$ , when no work is done. $[C_{p} = 2.5 R$ and $R = 83 J {mol}^{- 1} K^{- 1}$ ]

1

867.5 J

2

186.5 J

3

1867.5 J

4

86.7 J

Explanation:

C Given
$n = 1$ mole
$Δ T = T_{2} - T_{1} = (150 - 0) = 150^{\circ} C$
$R = 8.3 J / K . mol, C_{P} = 2.5 R$
$C_{P} - C_{V} = R \Rightarrow C_{V} = C_{P} - R = 2.5 R - R = 1.5 R$
$dQ = dU + dW \Rightarrow dQ = dU (∵ dW = 0)$
$dQ = {nC}_{V} Δ T$
$dQ = n 1.5 R Δ T$
$dQ = 1 \times 1.5 \times 8.3 \times 150 = 1867.5 J$
$dQ = 1867.5 J$

CG PET- 2014

Thermodynamics

148100 Ideal gas is contained in a thermally insulated and rigid container and it is heated through a resistance $100 Ω$ by passing a current of $1 A$ for five minutes, then change in internal energy of the gas is

1

0 kJ

2

30 kJ

3

10 kJ

4

20 kJ

Explanation:

B Given, $R = 100 Ω, i = 1 A$
$t = 5 \min . = 5 \times 60 \sec = 300 \sec$
When current is flowing through a wire then heat produces,
$Q = i^{2} Rt = 1 \times 100 \times 300 = 3 \times 10^{4} J$
$Q = 30 kJ$
Therefore change in internal energy is $30 kJ$ .

Manipal UGET-2014

Thermodynamics

148101 $1 {cm}^{3}$ of water at its boiling point absorbs 540 cal of heat to becomes steam with a volume of $1671 {cm}^{3}$ . If the atmospheric pressure $=$ $1.013 \times 10^{5} N / m^{2}$ and the mechanical equivalent of heat $= 4.19 J / cal$ , the energy spent in this process in overcoming intermolecular process in overcoming intermolecular forces is

1

540 cal

2

40 cal

3

500 cal

4 zero

Explanation:

C Given, $V_{1} = 1 {cm}^{3}, V_{2} = 1671 {cm}^{3}$
and $Q = 540 cal$
Atmospheric pressure $= 1.013 \times 10^{5} N / m^{2}$
Mechanical equivalent $(J) = 4.19 J / cal$
According to first law of thermodynamics
$Δ Δ = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$Δ U = 540 - \frac{P (V_{2} - V_{1})}{J}$
$Δ U = 540 - \frac{1.013 \times 10^{5} \times [(1671 - 1) \times 10^{- 6}]}{4.2}$
$Δ U = 540 - 40 = 500 cal$

Manipal UGET-2010

Thermodynamics

148097 An ideal gas is taken through the cycle $A \to B \to C \to A$ , as shown in the figure below. If the net heat supplied to the gas is $5 J$ , then the work done by the gas in the process $C \to A$ is

1

- 5 J

2

- 10 J

3

- 15 J

4

- 20 J

Explanation:

A Given, $Q_{net} = 5 J$
$W_{BC} = PdV = 0 (∵ constant volume)$
$W_{AB} = 10 \times (2 - 1) = 10 J$
$\oint dQ = \oint dW$
$5 = W_{AB} + W_{BC} + W_{CA}$
$W_{CA} = 5 - W_{AB} - W_{BC}$
$W_{CA} = 5 - 10 - 0 = - 5 J$
$W_{CA} = - 5 J$

CG PET -2018

Thermodynamics

148098 Calculate the heat required to increases the temperature of 1 mole of one atomic gas from $0^{\circ} C$ to $150^{\circ} C$ , when no work is done. $[C_{p} = 2.5 R$ and $R = 83 J {mol}^{- 1} K^{- 1}$ ]

1

867.5 J

2

186.5 J

3

1867.5 J

4

86.7 J

Explanation:

C Given
$n = 1$ mole
$Δ T = T_{2} - T_{1} = (150 - 0) = 150^{\circ} C$
$R = 8.3 J / K . mol, C_{P} = 2.5 R$
$C_{P} - C_{V} = R \Rightarrow C_{V} = C_{P} - R = 2.5 R - R = 1.5 R$
$dQ = dU + dW \Rightarrow dQ = dU (∵ dW = 0)$
$dQ = {nC}_{V} Δ T$
$dQ = n 1.5 R Δ T$
$dQ = 1 \times 1.5 \times 8.3 \times 150 = 1867.5 J$
$dQ = 1867.5 J$

CG PET- 2014

Thermodynamics

148100 Ideal gas is contained in a thermally insulated and rigid container and it is heated through a resistance $100 Ω$ by passing a current of $1 A$ for five minutes, then change in internal energy of the gas is

1

0 kJ

2

30 kJ

3

10 kJ

4

20 kJ

Explanation:

B Given, $R = 100 Ω, i = 1 A$
$t = 5 \min . = 5 \times 60 \sec = 300 \sec$
When current is flowing through a wire then heat produces,
$Q = i^{2} Rt = 1 \times 100 \times 300 = 3 \times 10^{4} J$
$Q = 30 kJ$
Therefore change in internal energy is $30 kJ$ .

Manipal UGET-2014

Thermodynamics

148101 $1 {cm}^{3}$ of water at its boiling point absorbs 540 cal of heat to becomes steam with a volume of $1671 {cm}^{3}$ . If the atmospheric pressure $=$ $1.013 \times 10^{5} N / m^{2}$ and the mechanical equivalent of heat $= 4.19 J / cal$ , the energy spent in this process in overcoming intermolecular process in overcoming intermolecular forces is

1

540 cal

2

40 cal

3

500 cal

4 zero

Explanation:

C Given, $V_{1} = 1 {cm}^{3}, V_{2} = 1671 {cm}^{3}$
and $Q = 540 cal$
Atmospheric pressure $= 1.013 \times 10^{5} N / m^{2}$
Mechanical equivalent $(J) = 4.19 J / cal$
According to first law of thermodynamics
$Δ Δ = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$Δ U = 540 - \frac{P (V_{2} - V_{1})}{J}$
$Δ U = 540 - \frac{1.013 \times 10^{5} \times [(1671 - 1) \times 10^{- 6}]}{4.2}$
$Δ U = 540 - 40 = 500 cal$

Manipal UGET-2010