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Thermodynamics

148339 A gas undergoes a thermodynamic cycle as shown in pressure-volume diagram below. The heat transferred to the gas is $300 J$ and $140 J$ along path $1 \to 2$ and $2 \to 3$ respectively. The internal energy is changed by $- 260 J$ along path $3 \to 1$ . The work done by gas along path $1 \to 2 \to 3$ is

1

- 80 J

2

140 J

3

160 J

4

180 J

Explanation:

D Given that,
Heat transfer to gas along path $1 \to 2$ is $Q_{1} = 300 J$
Heat transfer to gas along path $2 \to 3$ is $Q_{2} = 140 J$
The change in internal energy along $3 \to 1$ is $Δ U = - 260 J$ According to first law of thermodynamic,
$W = Q_{1} + Q_{2} - Δ U$
$= 300 + 140 - 260$
$= 440 - 260$
$= 180 J$

TS EAMCET 08.05.2019

Thermodynamics

148340 A gas system is taken through the thermodynamic cycle (abcda) as shown in the figure. The amount of heat change experienced by the gas during the process is

1

40 J

of heat is rejected

2

40 J

of heat is absorbed

3

100 j

of heat is rejected

4

100 J

of heat is absorbed

Explanation:

A
For process ab (constant pressure process) $W_{a b} = P d V$
$= 150 \times 10^{3} (400 - 200) \times 10^{- 6}$
$= 150 \times 10^{3} \times 200 \times 10^{- 6}$
$= 15 \times 2 \times 10^{6} \times 10^{- 6}$
$= 30 \times 10^{6} \times 10^{- 6}$
$= 30 Joule$
For process bc (constant volume process)
$Δ V = 0$
$Therefore, W_{bc} = 0$
For process cd (constant pressure process)
$W_{cd} = PdV$
$= 350 \times 10^{3} [200 - 400] \times 10^{- 6}$
$= 350 \times 10^{3} (- 200) \times 10^{- 6}$
$= - 70 Joule$
For process da (volume at constant process)
$W_{da} = 0$
Total work done $= W_{ab} + W_{bc} + W_{cd} + W_{da}$
$30 + 0 - 70 + 0$
$W_{total} = - 40$ Joule
For process abcda
Internal energy $(Δ U) = 0$ (Because it is a cyclic process) The amount of heat
$dQ = dW + dU$
$dQ = - 40$ Joule (-ve sign indicate heat is rejected)

TS EAMCET 03.05.2018

Thermodynamics

148341 7 mole of certain monoatomic ideal gas undergoes a temperature increase of $40 K$ at constant pressure. The increase in the internal energy of the gas in this process is
(Given $R = 8.3 {JK}^{- 1} {mol}^{- 1}$ )

1

5810 J

2

3486 J

3

11620 J

4

6972 J

Explanation:

B Given that,
$n = 7 mole, Δ P = 0, Δ T = 40 K$
$Δ U = ?$
For monoatomic internal energy,
$Δ U = \frac{3}{2} \times n \times R \times Δ T$
$= \frac{3}{2} \times 7 \times 8.3 \times 40$
$3 \times 7 \times 8.3 \times 20$
$Δ U = 3486$ Joule

JEE Main-26.07.2022

Thermodynamics

148342 Carbon monoxide $(γ = \frac{7}{5})$ is carried around a
closed cyclic process abc, in which 'bc' is an isothermal process as shown in figure. The gas absorbs $6000 J$ of heat as its temperature is increased from $200 K$ to $800 K$ in going from 'a' to ' $b$ '. The quantity of heat ejected by the gas during the process 'ca' is

1

6000 J

2

2400 J

3

8400 J

4

4800 J

Explanation:

C Given, Heat absorbed $a \to b$
$(Δ Q)_{ab} = + 6000$ Jule $= ($ at constant volume $)$
$+ 6000 = {mC}_{V} (Δ T)$
$6000 = {mC}_{V} (800 - 200)$
$6000 = {mC}_{V} \times 600$
${mC}_{v} = 10$
For the process $c$ to a
$T_{a} = 200 K T_{b} = T_{c} = 800 K$
$(Δ Q)_{ca} = {mC}_{P} (Δ T)$
$(Δ Q)_{ca} = {mC}_{P} (200 - 800)$
$(Δ Q)_{ca} = - {mC}_{P} \times 600$
$= - m (C_{V} + R) \times 600$
For carbon monoxide
$γ = 7 / 5$
$C_{v} = \frac{R}{γ - 1} = \frac{R}{7 / 5 - 1} = \frac{5 R}{2}$
From equation (i) and (iii)
$6000 = m \frac{.5 R}{2} \times 600$
$10 = \frac{m 5 R}{2}$
$mR = 4$
$(Q)_{ca} = - m (C_{V} + R) \times 600$
$(Q)_{ca} = (- {mC}_{V} - mR) \times 600$
$= (- 10 - 4) \times 600$
$= - 14 \times 600$
$(Q)_{ca} = - 8400 (- ve$ sign indicates heat ejected) $(Q)_{ca} = 8400 Joule$

AP EAMCET-24.04.2019

Thermodynamics

148339 A gas undergoes a thermodynamic cycle as shown in pressure-volume diagram below. The heat transferred to the gas is $300 J$ and $140 J$ along path $1 \to 2$ and $2 \to 3$ respectively. The internal energy is changed by $- 260 J$ along path $3 \to 1$ . The work done by gas along path $1 \to 2 \to 3$ is

1

- 80 J

2

140 J

3

160 J

4

180 J

Explanation:

D Given that,
Heat transfer to gas along path $1 \to 2$ is $Q_{1} = 300 J$
Heat transfer to gas along path $2 \to 3$ is $Q_{2} = 140 J$
The change in internal energy along $3 \to 1$ is $Δ U = - 260 J$ According to first law of thermodynamic,
$W = Q_{1} + Q_{2} - Δ U$
$= 300 + 140 - 260$
$= 440 - 260$
$= 180 J$

TS EAMCET 08.05.2019

Thermodynamics

148340 A gas system is taken through the thermodynamic cycle (abcda) as shown in the figure. The amount of heat change experienced by the gas during the process is

1

40 J

of heat is rejected

2

40 J

of heat is absorbed

3

100 j

of heat is rejected

4

100 J

of heat is absorbed

Explanation:

A
For process ab (constant pressure process) $W_{a b} = P d V$
$= 150 \times 10^{3} (400 - 200) \times 10^{- 6}$
$= 150 \times 10^{3} \times 200 \times 10^{- 6}$
$= 15 \times 2 \times 10^{6} \times 10^{- 6}$
$= 30 \times 10^{6} \times 10^{- 6}$
$= 30 Joule$
For process bc (constant volume process)
$Δ V = 0$
$Therefore, W_{bc} = 0$
For process cd (constant pressure process)
$W_{cd} = PdV$
$= 350 \times 10^{3} [200 - 400] \times 10^{- 6}$
$= 350 \times 10^{3} (- 200) \times 10^{- 6}$
$= - 70 Joule$
For process da (volume at constant process)
$W_{da} = 0$
Total work done $= W_{ab} + W_{bc} + W_{cd} + W_{da}$
$30 + 0 - 70 + 0$
$W_{total} = - 40$ Joule
For process abcda
Internal energy $(Δ U) = 0$ (Because it is a cyclic process) The amount of heat
$dQ = dW + dU$
$dQ = - 40$ Joule (-ve sign indicate heat is rejected)

TS EAMCET 03.05.2018

Thermodynamics

148341 7 mole of certain monoatomic ideal gas undergoes a temperature increase of $40 K$ at constant pressure. The increase in the internal energy of the gas in this process is
(Given $R = 8.3 {JK}^{- 1} {mol}^{- 1}$ )

1

5810 J

2

3486 J

3

11620 J

4

6972 J

Explanation:

B Given that,
$n = 7 mole, Δ P = 0, Δ T = 40 K$
$Δ U = ?$
For monoatomic internal energy,
$Δ U = \frac{3}{2} \times n \times R \times Δ T$
$= \frac{3}{2} \times 7 \times 8.3 \times 40$
$3 \times 7 \times 8.3 \times 20$
$Δ U = 3486$ Joule

JEE Main-26.07.2022

Thermodynamics

148342 Carbon monoxide $(γ = \frac{7}{5})$ is carried around a
closed cyclic process abc, in which 'bc' is an isothermal process as shown in figure. The gas absorbs $6000 J$ of heat as its temperature is increased from $200 K$ to $800 K$ in going from 'a' to ' $b$ '. The quantity of heat ejected by the gas during the process 'ca' is

1

6000 J

2

2400 J

3

8400 J

4

4800 J

Explanation:

C Given, Heat absorbed $a \to b$
$(Δ Q)_{ab} = + 6000$ Jule $= ($ at constant volume $)$
$+ 6000 = {mC}_{V} (Δ T)$
$6000 = {mC}_{V} (800 - 200)$
$6000 = {mC}_{V} \times 600$
${mC}_{v} = 10$
For the process $c$ to a
$T_{a} = 200 K T_{b} = T_{c} = 800 K$
$(Δ Q)_{ca} = {mC}_{P} (Δ T)$
$(Δ Q)_{ca} = {mC}_{P} (200 - 800)$
$(Δ Q)_{ca} = - {mC}_{P} \times 600$
$= - m (C_{V} + R) \times 600$
For carbon monoxide
$γ = 7 / 5$
$C_{v} = \frac{R}{γ - 1} = \frac{R}{7 / 5 - 1} = \frac{5 R}{2}$
From equation (i) and (iii)
$6000 = m \frac{.5 R}{2} \times 600$
$10 = \frac{m 5 R}{2}$
$mR = 4$
$(Q)_{ca} = - m (C_{V} + R) \times 600$
$(Q)_{ca} = (- {mC}_{V} - mR) \times 600$
$= (- 10 - 4) \times 600$
$= - 14 \times 600$
$(Q)_{ca} = - 8400 (- ve$ sign indicates heat ejected) $(Q)_{ca} = 8400 Joule$

AP EAMCET-24.04.2019

Thermodynamics

148339 A gas undergoes a thermodynamic cycle as shown in pressure-volume diagram below. The heat transferred to the gas is $300 J$ and $140 J$ along path $1 \to 2$ and $2 \to 3$ respectively. The internal energy is changed by $- 260 J$ along path $3 \to 1$ . The work done by gas along path $1 \to 2 \to 3$ is

1

- 80 J

2

140 J

3

160 J

4

180 J

Explanation:

D Given that,
Heat transfer to gas along path $1 \to 2$ is $Q_{1} = 300 J$
Heat transfer to gas along path $2 \to 3$ is $Q_{2} = 140 J$
The change in internal energy along $3 \to 1$ is $Δ U = - 260 J$ According to first law of thermodynamic,
$W = Q_{1} + Q_{2} - Δ U$
$= 300 + 140 - 260$
$= 440 - 260$
$= 180 J$

TS EAMCET 08.05.2019

Thermodynamics

148340 A gas system is taken through the thermodynamic cycle (abcda) as shown in the figure. The amount of heat change experienced by the gas during the process is

1

40 J

of heat is rejected

2

40 J

of heat is absorbed

3

100 j

of heat is rejected

4

100 J

of heat is absorbed

Explanation:

A
For process ab (constant pressure process) $W_{a b} = P d V$
$= 150 \times 10^{3} (400 - 200) \times 10^{- 6}$
$= 150 \times 10^{3} \times 200 \times 10^{- 6}$
$= 15 \times 2 \times 10^{6} \times 10^{- 6}$
$= 30 \times 10^{6} \times 10^{- 6}$
$= 30 Joule$
For process bc (constant volume process)
$Δ V = 0$
$Therefore, W_{bc} = 0$
For process cd (constant pressure process)
$W_{cd} = PdV$
$= 350 \times 10^{3} [200 - 400] \times 10^{- 6}$
$= 350 \times 10^{3} (- 200) \times 10^{- 6}$
$= - 70 Joule$
For process da (volume at constant process)
$W_{da} = 0$
Total work done $= W_{ab} + W_{bc} + W_{cd} + W_{da}$
$30 + 0 - 70 + 0$
$W_{total} = - 40$ Joule
For process abcda
Internal energy $(Δ U) = 0$ (Because it is a cyclic process) The amount of heat
$dQ = dW + dU$
$dQ = - 40$ Joule (-ve sign indicate heat is rejected)

TS EAMCET 03.05.2018

Thermodynamics

148341 7 mole of certain monoatomic ideal gas undergoes a temperature increase of $40 K$ at constant pressure. The increase in the internal energy of the gas in this process is
(Given $R = 8.3 {JK}^{- 1} {mol}^{- 1}$ )

1

5810 J

2

3486 J

3

11620 J

4

6972 J

Explanation:

B Given that,
$n = 7 mole, Δ P = 0, Δ T = 40 K$
$Δ U = ?$
For monoatomic internal energy,
$Δ U = \frac{3}{2} \times n \times R \times Δ T$
$= \frac{3}{2} \times 7 \times 8.3 \times 40$
$3 \times 7 \times 8.3 \times 20$
$Δ U = 3486$ Joule

JEE Main-26.07.2022

Thermodynamics

148342 Carbon monoxide $(γ = \frac{7}{5})$ is carried around a
closed cyclic process abc, in which 'bc' is an isothermal process as shown in figure. The gas absorbs $6000 J$ of heat as its temperature is increased from $200 K$ to $800 K$ in going from 'a' to ' $b$ '. The quantity of heat ejected by the gas during the process 'ca' is

1

6000 J

2

2400 J

3

8400 J

4

4800 J

Explanation:

C Given, Heat absorbed $a \to b$
$(Δ Q)_{ab} = + 6000$ Jule $= ($ at constant volume $)$
$+ 6000 = {mC}_{V} (Δ T)$
$6000 = {mC}_{V} (800 - 200)$
$6000 = {mC}_{V} \times 600$
${mC}_{v} = 10$
For the process $c$ to a
$T_{a} = 200 K T_{b} = T_{c} = 800 K$
$(Δ Q)_{ca} = {mC}_{P} (Δ T)$
$(Δ Q)_{ca} = {mC}_{P} (200 - 800)$
$(Δ Q)_{ca} = - {mC}_{P} \times 600$
$= - m (C_{V} + R) \times 600$
For carbon monoxide
$γ = 7 / 5$
$C_{v} = \frac{R}{γ - 1} = \frac{R}{7 / 5 - 1} = \frac{5 R}{2}$
From equation (i) and (iii)
$6000 = m \frac{.5 R}{2} \times 600$
$10 = \frac{m 5 R}{2}$
$mR = 4$
$(Q)_{ca} = - m (C_{V} + R) \times 600$
$(Q)_{ca} = (- {mC}_{V} - mR) \times 600$
$= (- 10 - 4) \times 600$
$= - 14 \times 600$
$(Q)_{ca} = - 8400 (- ve$ sign indicates heat ejected) $(Q)_{ca} = 8400 Joule$

AP EAMCET-24.04.2019

Thermodynamics

148339 A gas undergoes a thermodynamic cycle as shown in pressure-volume diagram below. The heat transferred to the gas is $300 J$ and $140 J$ along path $1 \to 2$ and $2 \to 3$ respectively. The internal energy is changed by $- 260 J$ along path $3 \to 1$ . The work done by gas along path $1 \to 2 \to 3$ is

1

- 80 J

2

140 J

3

160 J

4

180 J

Explanation:

D Given that,
Heat transfer to gas along path $1 \to 2$ is $Q_{1} = 300 J$
Heat transfer to gas along path $2 \to 3$ is $Q_{2} = 140 J$
The change in internal energy along $3 \to 1$ is $Δ U = - 260 J$ According to first law of thermodynamic,
$W = Q_{1} + Q_{2} - Δ U$
$= 300 + 140 - 260$
$= 440 - 260$
$= 180 J$

TS EAMCET 08.05.2019

Thermodynamics

148340 A gas system is taken through the thermodynamic cycle (abcda) as shown in the figure. The amount of heat change experienced by the gas during the process is

1

40 J

of heat is rejected

2

40 J

of heat is absorbed

3

100 j

of heat is rejected

4

100 J

of heat is absorbed

Explanation:

A
For process ab (constant pressure process) $W_{a b} = P d V$
$= 150 \times 10^{3} (400 - 200) \times 10^{- 6}$
$= 150 \times 10^{3} \times 200 \times 10^{- 6}$
$= 15 \times 2 \times 10^{6} \times 10^{- 6}$
$= 30 \times 10^{6} \times 10^{- 6}$
$= 30 Joule$
For process bc (constant volume process)
$Δ V = 0$
$Therefore, W_{bc} = 0$
For process cd (constant pressure process)
$W_{cd} = PdV$
$= 350 \times 10^{3} [200 - 400] \times 10^{- 6}$
$= 350 \times 10^{3} (- 200) \times 10^{- 6}$
$= - 70 Joule$
For process da (volume at constant process)
$W_{da} = 0$
Total work done $= W_{ab} + W_{bc} + W_{cd} + W_{da}$
$30 + 0 - 70 + 0$
$W_{total} = - 40$ Joule
For process abcda
Internal energy $(Δ U) = 0$ (Because it is a cyclic process) The amount of heat
$dQ = dW + dU$
$dQ = - 40$ Joule (-ve sign indicate heat is rejected)

TS EAMCET 03.05.2018

Thermodynamics

148341 7 mole of certain monoatomic ideal gas undergoes a temperature increase of $40 K$ at constant pressure. The increase in the internal energy of the gas in this process is
(Given $R = 8.3 {JK}^{- 1} {mol}^{- 1}$ )

1

5810 J

2

3486 J

3

11620 J

4

6972 J

Explanation:

B Given that,
$n = 7 mole, Δ P = 0, Δ T = 40 K$
$Δ U = ?$
For monoatomic internal energy,
$Δ U = \frac{3}{2} \times n \times R \times Δ T$
$= \frac{3}{2} \times 7 \times 8.3 \times 40$
$3 \times 7 \times 8.3 \times 20$
$Δ U = 3486$ Joule

JEE Main-26.07.2022

Thermodynamics

148342 Carbon monoxide $(γ = \frac{7}{5})$ is carried around a
closed cyclic process abc, in which 'bc' is an isothermal process as shown in figure. The gas absorbs $6000 J$ of heat as its temperature is increased from $200 K$ to $800 K$ in going from 'a' to ' $b$ '. The quantity of heat ejected by the gas during the process 'ca' is

1

6000 J

2

2400 J

3

8400 J

4

4800 J

Explanation:

C Given, Heat absorbed $a \to b$
$(Δ Q)_{ab} = + 6000$ Jule $= ($ at constant volume $)$
$+ 6000 = {mC}_{V} (Δ T)$
$6000 = {mC}_{V} (800 - 200)$
$6000 = {mC}_{V} \times 600$
${mC}_{v} = 10$
For the process $c$ to a
$T_{a} = 200 K T_{b} = T_{c} = 800 K$
$(Δ Q)_{ca} = {mC}_{P} (Δ T)$
$(Δ Q)_{ca} = {mC}_{P} (200 - 800)$
$(Δ Q)_{ca} = - {mC}_{P} \times 600$
$= - m (C_{V} + R) \times 600$
For carbon monoxide
$γ = 7 / 5$
$C_{v} = \frac{R}{γ - 1} = \frac{R}{7 / 5 - 1} = \frac{5 R}{2}$
From equation (i) and (iii)
$6000 = m \frac{.5 R}{2} \times 600$
$10 = \frac{m 5 R}{2}$
$mR = 4$
$(Q)_{ca} = - m (C_{V} + R) \times 600$
$(Q)_{ca} = (- {mC}_{V} - mR) \times 600$
$= (- 10 - 4) \times 600$
$= - 14 \times 600$
$(Q)_{ca} = - 8400 (- ve$ sign indicates heat ejected) $(Q)_{ca} = 8400 Joule$

AP EAMCET-24.04.2019

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