QBManager

Thermodynamics

272423 The Gibbs free energy for the decomposition of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is $482.5 kJ / mol$ . The potential difference needed for electrolytic reduction of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is at least

1

2.5 V

2

5.0 V

3

1.25 V

4

4.5 V

Explanation:

Reaction formed,
$\frac{2}{3} {Al}_{2} O_{3} ⟶ \frac{4}{3} Al + O_{2}$
Where, $Δ_{r} G = + 482.5 kJ {mol}^{- 1}$
(i) $2 O^{2 -} + 4 e^{-} ⟶ \underset{(A + 2 mode)}{O_{2}] \times 3}$
(ii) ${Al}^{3 +} ⟶ Al + 3 e^{-}] \times 4$
Accurate reaction,
$\frac{4}{3} A 1 + 2 O^{2 -} ⟶ O_{2} + \frac{4}{3} Al$
So, number of electron transferred $= \frac{12}{3} = 4$
From formula,
$Δ G^{\circ} = - {nFE}^{\circ}$
$(482.5 \times 1000) J / mol = - 4 \times 96500 \times E^{\circ}$
$E^{\circ} = \frac{(482.5 \times 1000) J / mol}{4 \times 96500} = 1.25$

AMU-2019

Thermodynamics

272425 An ideal gas initially at temperature, pressure and volume, $27^{\circ} C, 1.00$ bar and $10 L$ , respectively is heated at constant volume until pressure is 10.0 bar; it then undergoes a reversible isothermal expansion until pressure is 1.00 bar. What is the total work $W$ , during this process?

1

- 23.02 \times 10^{3} J

2

- 14.0 \times 10^{3} J

3

14.0 \times 10^{3} J

4 Zero

Explanation:

For process 1-2: $V = C$
$W_{1} = - P Δ V$
$∴ Δ V = 0$
$∴ W_{1} = 0$
For process 2-3: $T = C$
$W_{2} = - 2.303 nRT \log \frac{P_{2}}{P_{3}}$
We know that, $P V = n R T$
$W_{2} = - 2.303 \times P \times V \times \log \frac{10}{1}$
$= - 2.303 \times 1 \times 10^{5} \times 10^{- 3} \log 10$
$= - 2.303 \times 10^{2} J$
$= - 23.03 \times 10^{3} J \approx 23.02 \times 10^{3} J$

AMU-2018

Thermodynamics

272426 For the given PV isotherms, which of the following is correct for $T_{1}, T_{2}, T_{3}$ ?

1

T_{1} < T_{2} < T_{3}

2

T_{3} < T_{2} < T_{1}

3

T_{2} < T_{3} < T_{1}

4

T_{1} > T_{3} > T_{2}

AMU-2018

Thermodynamics

272432 Calculate the entropy change for
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
${JK}^{- 1}$ mole $^{- 1}$ are $186.2, 188.7, 130.6, 197.6$ respectively. The entropy change is

1

- 46 {JK}^{- 1} {mole}^{- 1}

2

+ 46 {JK}^{- 1} {mole}^{- 1}

3

- 214.8 {JK}^{- 1} {mole}^{- 1}

4

+ 214.8 {JK}^{- 1} {mole}^{- 1}

Explanation:

Given,
${CH}_{4} (g) = 186.2 {JK}^{- 1} {mol}^{- 1}$
$H_{2} O (g) = 188.7 {JK}^{- 1} {mol}^{- 1}$
$H_{2} (g) = 130.6 {JK}^{- 1} {mol}^{- 1}$
$CO (g) = 197.6 {JK}^{- 1} {mol}^{- 1}$
The entropy change $(Δ S^{\circ})$ for the reaction,
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
$Entropy of a reaction = Entropy of products - Entropy$
$of reactants$
$Entropy change (Δ S^{\circ}) = (197.6 + 3 \times 130.6) - (186.2 + 188.7)$
$= 589.4 - 374.9$
$= 214.5 {JK}^{- 1} mole$

AMU-2005

Thermodynamics

272438 For the gaseous reaction $N_{2} O_{4} (g) \to 2 {NO}_{2} (g)$

1

Δ H = 0

2

Δ H = Δ U

3

Δ H < Δ U

4

Δ H > Δ U

Explanation:

$Δ H = Δ U + Δ n_{g} RT$
$Δ n_{g} = n_{p} - n_{R} = 2 - 1 = 1$
$∴ Δ H = Δ U + (1) RT$
$∴ Δ H > Δ U$

Assam CEE-2014

Thermodynamics

272423 The Gibbs free energy for the decomposition of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is $482.5 kJ / mol$ . The potential difference needed for electrolytic reduction of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is at least

1

2.5 V

2

5.0 V

3

1.25 V

4

4.5 V

Explanation:

Reaction formed,
$\frac{2}{3} {Al}_{2} O_{3} ⟶ \frac{4}{3} Al + O_{2}$
Where, $Δ_{r} G = + 482.5 kJ {mol}^{- 1}$
(i) $2 O^{2 -} + 4 e^{-} ⟶ \underset{(A + 2 mode)}{O_{2}] \times 3}$
(ii) ${Al}^{3 +} ⟶ Al + 3 e^{-}] \times 4$
Accurate reaction,
$\frac{4}{3} A 1 + 2 O^{2 -} ⟶ O_{2} + \frac{4}{3} Al$
So, number of electron transferred $= \frac{12}{3} = 4$
From formula,
$Δ G^{\circ} = - {nFE}^{\circ}$
$(482.5 \times 1000) J / mol = - 4 \times 96500 \times E^{\circ}$
$E^{\circ} = \frac{(482.5 \times 1000) J / mol}{4 \times 96500} = 1.25$

AMU-2019

Thermodynamics

272425 An ideal gas initially at temperature, pressure and volume, $27^{\circ} C, 1.00$ bar and $10 L$ , respectively is heated at constant volume until pressure is 10.0 bar; it then undergoes a reversible isothermal expansion until pressure is 1.00 bar. What is the total work $W$ , during this process?

1

- 23.02 \times 10^{3} J

2

- 14.0 \times 10^{3} J

3

14.0 \times 10^{3} J

4 Zero

Explanation:

For process 1-2: $V = C$
$W_{1} = - P Δ V$
$∴ Δ V = 0$
$∴ W_{1} = 0$
For process 2-3: $T = C$
$W_{2} = - 2.303 nRT \log \frac{P_{2}}{P_{3}}$
We know that, $P V = n R T$
$W_{2} = - 2.303 \times P \times V \times \log \frac{10}{1}$
$= - 2.303 \times 1 \times 10^{5} \times 10^{- 3} \log 10$
$= - 2.303 \times 10^{2} J$
$= - 23.03 \times 10^{3} J \approx 23.02 \times 10^{3} J$

AMU-2018

Thermodynamics

272426 For the given PV isotherms, which of the following is correct for $T_{1}, T_{2}, T_{3}$ ?

1

T_{1} < T_{2} < T_{3}

2

T_{3} < T_{2} < T_{1}

3

T_{2} < T_{3} < T_{1}

4

T_{1} > T_{3} > T_{2}

AMU-2018

Thermodynamics

272432 Calculate the entropy change for
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
${JK}^{- 1}$ mole $^{- 1}$ are $186.2, 188.7, 130.6, 197.6$ respectively. The entropy change is

1

- 46 {JK}^{- 1} {mole}^{- 1}

2

+ 46 {JK}^{- 1} {mole}^{- 1}

3

- 214.8 {JK}^{- 1} {mole}^{- 1}

4

+ 214.8 {JK}^{- 1} {mole}^{- 1}

Explanation:

Given,
${CH}_{4} (g) = 186.2 {JK}^{- 1} {mol}^{- 1}$
$H_{2} O (g) = 188.7 {JK}^{- 1} {mol}^{- 1}$
$H_{2} (g) = 130.6 {JK}^{- 1} {mol}^{- 1}$
$CO (g) = 197.6 {JK}^{- 1} {mol}^{- 1}$
The entropy change $(Δ S^{\circ})$ for the reaction,
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
$Entropy of a reaction = Entropy of products - Entropy$
$of reactants$
$Entropy change (Δ S^{\circ}) = (197.6 + 3 \times 130.6) - (186.2 + 188.7)$
$= 589.4 - 374.9$
$= 214.5 {JK}^{- 1} mole$

AMU-2005

Thermodynamics

272438 For the gaseous reaction $N_{2} O_{4} (g) \to 2 {NO}_{2} (g)$

1

Δ H = 0

2

Δ H = Δ U

3

Δ H < Δ U

4

Δ H > Δ U

Explanation:

$Δ H = Δ U + Δ n_{g} RT$
$Δ n_{g} = n_{p} - n_{R} = 2 - 1 = 1$
$∴ Δ H = Δ U + (1) RT$
$∴ Δ H > Δ U$

Assam CEE-2014

Thermodynamics

272423 The Gibbs free energy for the decomposition of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is $482.5 kJ / mol$ . The potential difference needed for electrolytic reduction of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is at least

1

2.5 V

2

5.0 V

3

1.25 V

4

4.5 V

Explanation:

Reaction formed,
$\frac{2}{3} {Al}_{2} O_{3} ⟶ \frac{4}{3} Al + O_{2}$
Where, $Δ_{r} G = + 482.5 kJ {mol}^{- 1}$
(i) $2 O^{2 -} + 4 e^{-} ⟶ \underset{(A + 2 mode)}{O_{2}] \times 3}$
(ii) ${Al}^{3 +} ⟶ Al + 3 e^{-}] \times 4$
Accurate reaction,
$\frac{4}{3} A 1 + 2 O^{2 -} ⟶ O_{2} + \frac{4}{3} Al$
So, number of electron transferred $= \frac{12}{3} = 4$
From formula,
$Δ G^{\circ} = - {nFE}^{\circ}$
$(482.5 \times 1000) J / mol = - 4 \times 96500 \times E^{\circ}$
$E^{\circ} = \frac{(482.5 \times 1000) J / mol}{4 \times 96500} = 1.25$

AMU-2019

Thermodynamics

272425 An ideal gas initially at temperature, pressure and volume, $27^{\circ} C, 1.00$ bar and $10 L$ , respectively is heated at constant volume until pressure is 10.0 bar; it then undergoes a reversible isothermal expansion until pressure is 1.00 bar. What is the total work $W$ , during this process?

1

- 23.02 \times 10^{3} J

2

- 14.0 \times 10^{3} J

3

14.0 \times 10^{3} J

4 Zero

Explanation:

For process 1-2: $V = C$
$W_{1} = - P Δ V$
$∴ Δ V = 0$
$∴ W_{1} = 0$
For process 2-3: $T = C$
$W_{2} = - 2.303 nRT \log \frac{P_{2}}{P_{3}}$
We know that, $P V = n R T$
$W_{2} = - 2.303 \times P \times V \times \log \frac{10}{1}$
$= - 2.303 \times 1 \times 10^{5} \times 10^{- 3} \log 10$
$= - 2.303 \times 10^{2} J$
$= - 23.03 \times 10^{3} J \approx 23.02 \times 10^{3} J$

AMU-2018

Thermodynamics

272426 For the given PV isotherms, which of the following is correct for $T_{1}, T_{2}, T_{3}$ ?

1

T_{1} < T_{2} < T_{3}

2

T_{3} < T_{2} < T_{1}

3

T_{2} < T_{3} < T_{1}

4

T_{1} > T_{3} > T_{2}

AMU-2018

Thermodynamics

272432 Calculate the entropy change for
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
${JK}^{- 1}$ mole $^{- 1}$ are $186.2, 188.7, 130.6, 197.6$ respectively. The entropy change is

1

- 46 {JK}^{- 1} {mole}^{- 1}

2

+ 46 {JK}^{- 1} {mole}^{- 1}

3

- 214.8 {JK}^{- 1} {mole}^{- 1}

4

+ 214.8 {JK}^{- 1} {mole}^{- 1}

Explanation:

Given,
${CH}_{4} (g) = 186.2 {JK}^{- 1} {mol}^{- 1}$
$H_{2} O (g) = 188.7 {JK}^{- 1} {mol}^{- 1}$
$H_{2} (g) = 130.6 {JK}^{- 1} {mol}^{- 1}$
$CO (g) = 197.6 {JK}^{- 1} {mol}^{- 1}$
The entropy change $(Δ S^{\circ})$ for the reaction,
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
$Entropy of a reaction = Entropy of products - Entropy$
$of reactants$
$Entropy change (Δ S^{\circ}) = (197.6 + 3 \times 130.6) - (186.2 + 188.7)$
$= 589.4 - 374.9$
$= 214.5 {JK}^{- 1} mole$

AMU-2005

Thermodynamics

272438 For the gaseous reaction $N_{2} O_{4} (g) \to 2 {NO}_{2} (g)$

1

Δ H = 0

2

Δ H = Δ U

3

Δ H < Δ U

4

Δ H > Δ U

Explanation:

$Δ H = Δ U + Δ n_{g} RT$
$Δ n_{g} = n_{p} - n_{R} = 2 - 1 = 1$
$∴ Δ H = Δ U + (1) RT$
$∴ Δ H > Δ U$

Assam CEE-2014

Thermodynamics

272423 The Gibbs free energy for the decomposition of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is $482.5 kJ / mol$ . The potential difference needed for electrolytic reduction of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is at least

1

2.5 V

2

5.0 V

3

1.25 V

4

4.5 V

Explanation:

Reaction formed,
$\frac{2}{3} {Al}_{2} O_{3} ⟶ \frac{4}{3} Al + O_{2}$
Where, $Δ_{r} G = + 482.5 kJ {mol}^{- 1}$
(i) $2 O^{2 -} + 4 e^{-} ⟶ \underset{(A + 2 mode)}{O_{2}] \times 3}$
(ii) ${Al}^{3 +} ⟶ Al + 3 e^{-}] \times 4$
Accurate reaction,
$\frac{4}{3} A 1 + 2 O^{2 -} ⟶ O_{2} + \frac{4}{3} Al$
So, number of electron transferred $= \frac{12}{3} = 4$
From formula,
$Δ G^{\circ} = - {nFE}^{\circ}$
$(482.5 \times 1000) J / mol = - 4 \times 96500 \times E^{\circ}$
$E^{\circ} = \frac{(482.5 \times 1000) J / mol}{4 \times 96500} = 1.25$

AMU-2019

Thermodynamics

272425 An ideal gas initially at temperature, pressure and volume, $27^{\circ} C, 1.00$ bar and $10 L$ , respectively is heated at constant volume until pressure is 10.0 bar; it then undergoes a reversible isothermal expansion until pressure is 1.00 bar. What is the total work $W$ , during this process?

1

- 23.02 \times 10^{3} J

2

- 14.0 \times 10^{3} J

3

14.0 \times 10^{3} J

4 Zero

Explanation:

For process 1-2: $V = C$
$W_{1} = - P Δ V$
$∴ Δ V = 0$
$∴ W_{1} = 0$
For process 2-3: $T = C$
$W_{2} = - 2.303 nRT \log \frac{P_{2}}{P_{3}}$
We know that, $P V = n R T$
$W_{2} = - 2.303 \times P \times V \times \log \frac{10}{1}$
$= - 2.303 \times 1 \times 10^{5} \times 10^{- 3} \log 10$
$= - 2.303 \times 10^{2} J$
$= - 23.03 \times 10^{3} J \approx 23.02 \times 10^{3} J$

AMU-2018

Thermodynamics

272426 For the given PV isotherms, which of the following is correct for $T_{1}, T_{2}, T_{3}$ ?

1

T_{1} < T_{2} < T_{3}

2

T_{3} < T_{2} < T_{1}

3

T_{2} < T_{3} < T_{1}

4

T_{1} > T_{3} > T_{2}

AMU-2018

Thermodynamics

272432 Calculate the entropy change for
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
${JK}^{- 1}$ mole $^{- 1}$ are $186.2, 188.7, 130.6, 197.6$ respectively. The entropy change is

1

- 46 {JK}^{- 1} {mole}^{- 1}

2

+ 46 {JK}^{- 1} {mole}^{- 1}

3

- 214.8 {JK}^{- 1} {mole}^{- 1}

4

+ 214.8 {JK}^{- 1} {mole}^{- 1}

Explanation:

Given,
${CH}_{4} (g) = 186.2 {JK}^{- 1} {mol}^{- 1}$
$H_{2} O (g) = 188.7 {JK}^{- 1} {mol}^{- 1}$
$H_{2} (g) = 130.6 {JK}^{- 1} {mol}^{- 1}$
$CO (g) = 197.6 {JK}^{- 1} {mol}^{- 1}$
The entropy change $(Δ S^{\circ})$ for the reaction,
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
$Entropy of a reaction = Entropy of products - Entropy$
$of reactants$
$Entropy change (Δ S^{\circ}) = (197.6 + 3 \times 130.6) - (186.2 + 188.7)$
$= 589.4 - 374.9$
$= 214.5 {JK}^{- 1} mole$

AMU-2005

Thermodynamics

272438 For the gaseous reaction $N_{2} O_{4} (g) \to 2 {NO}_{2} (g)$

1

Δ H = 0

2

Δ H = Δ U

3

Δ H < Δ U

4

Δ H > Δ U

Explanation:

$Δ H = Δ U + Δ n_{g} RT$
$Δ n_{g} = n_{p} - n_{R} = 2 - 1 = 1$
$∴ Δ H = Δ U + (1) RT$
$∴ Δ H > Δ U$

Assam CEE-2014

Thermodynamics

272423 The Gibbs free energy for the decomposition of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is $482.5 kJ / mol$ . The potential difference needed for electrolytic reduction of ${Al}_{2} O_{3}$ at $500^{\circ} C$ is at least

1

2.5 V

2

5.0 V

3

1.25 V

4

4.5 V

Explanation:

Reaction formed,
$\frac{2}{3} {Al}_{2} O_{3} ⟶ \frac{4}{3} Al + O_{2}$
Where, $Δ_{r} G = + 482.5 kJ {mol}^{- 1}$
(i) $2 O^{2 -} + 4 e^{-} ⟶ \underset{(A + 2 mode)}{O_{2}] \times 3}$
(ii) ${Al}^{3 +} ⟶ Al + 3 e^{-}] \times 4$
Accurate reaction,
$\frac{4}{3} A 1 + 2 O^{2 -} ⟶ O_{2} + \frac{4}{3} Al$
So, number of electron transferred $= \frac{12}{3} = 4$
From formula,
$Δ G^{\circ} = - {nFE}^{\circ}$
$(482.5 \times 1000) J / mol = - 4 \times 96500 \times E^{\circ}$
$E^{\circ} = \frac{(482.5 \times 1000) J / mol}{4 \times 96500} = 1.25$

AMU-2019

Thermodynamics

272425 An ideal gas initially at temperature, pressure and volume, $27^{\circ} C, 1.00$ bar and $10 L$ , respectively is heated at constant volume until pressure is 10.0 bar; it then undergoes a reversible isothermal expansion until pressure is 1.00 bar. What is the total work $W$ , during this process?

1

- 23.02 \times 10^{3} J

2

- 14.0 \times 10^{3} J

3

14.0 \times 10^{3} J

4 Zero

Explanation:

For process 1-2: $V = C$
$W_{1} = - P Δ V$
$∴ Δ V = 0$
$∴ W_{1} = 0$
For process 2-3: $T = C$
$W_{2} = - 2.303 nRT \log \frac{P_{2}}{P_{3}}$
We know that, $P V = n R T$
$W_{2} = - 2.303 \times P \times V \times \log \frac{10}{1}$
$= - 2.303 \times 1 \times 10^{5} \times 10^{- 3} \log 10$
$= - 2.303 \times 10^{2} J$
$= - 23.03 \times 10^{3} J \approx 23.02 \times 10^{3} J$

AMU-2018

Thermodynamics

272426 For the given PV isotherms, which of the following is correct for $T_{1}, T_{2}, T_{3}$ ?

1

T_{1} < T_{2} < T_{3}

2

T_{3} < T_{2} < T_{1}

3

T_{2} < T_{3} < T_{1}

4

T_{1} > T_{3} > T_{2}

AMU-2018

Thermodynamics

272432 Calculate the entropy change for
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
${JK}^{- 1}$ mole $^{- 1}$ are $186.2, 188.7, 130.6, 197.6$ respectively. The entropy change is

1

- 46 {JK}^{- 1} {mole}^{- 1}

2

+ 46 {JK}^{- 1} {mole}^{- 1}

3

- 214.8 {JK}^{- 1} {mole}^{- 1}

4

+ 214.8 {JK}^{- 1} {mole}^{- 1}

Explanation:

Given,
${CH}_{4} (g) = 186.2 {JK}^{- 1} {mol}^{- 1}$
$H_{2} O (g) = 188.7 {JK}^{- 1} {mol}^{- 1}$
$H_{2} (g) = 130.6 {JK}^{- 1} {mol}^{- 1}$
$CO (g) = 197.6 {JK}^{- 1} {mol}^{- 1}$
The entropy change $(Δ S^{\circ})$ for the reaction,
${CH}_{4} (g) + H_{2} O (g) \to 3 H_{2} (g) + CO (g)$
$Entropy of a reaction = Entropy of products - Entropy$
$of reactants$
$Entropy change (Δ S^{\circ}) = (197.6 + 3 \times 130.6) - (186.2 + 188.7)$
$= 589.4 - 374.9$
$= 214.5 {JK}^{- 1} mole$

AMU-2005

Thermodynamics

272438 For the gaseous reaction $N_{2} O_{4} (g) \to 2 {NO}_{2} (g)$

1

Δ H = 0

2

Δ H = Δ U

3

Δ H < Δ U

4

Δ H > Δ U

Explanation:

$Δ H = Δ U + Δ n_{g} RT$
$Δ n_{g} = n_{p} - n_{R} = 2 - 1 = 1$
$∴ Δ H = Δ U + (1) RT$
$∴ Δ H > Δ U$

Assam CEE-2014