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Thermodynamics

272689 Identify the correct statement regarding a spontaneous process.

1 For a spontaneous process in an isolated system, the change in entropy is positive

2 Endothermic processes are never spontaneous

3 Exothermic processes are always spontaneous

4 Lowering of energy in the reaction process is the only criterion for spontaneity

Explanation:

For a spontaneous process, $Δ G < 0$ $Δ G = Δ H - T Δ S < 0$
For isolated system, change in enthalpy is zero.
Hence, $Δ S$ must be positive for spontaneous process. Hence, option (a) is correct.

AIEEE-2007

Thermodynamics

272690 During which of the following processes, does entropy decrease?
A. Freezing of water to ice at $0^{\circ} C$ .
B. Freezing of water to ice at $- 10^{\circ} C$ .
C. $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3} (g)$
D. Adsorption of $CO (g)$ and lead surface.
E. Dissolution of $NaCl$ in water.

1 A, B, C and D

2

B

and

C

3

A

and

E

4

A, C

and

E

Explanation:

$A, B \to$ freezing of $H_{2} O$ will decreases entropy as particles more closer and forces of attraction will increases. Thus leads to decreases in randomness so entropy decreases.
$C \to$ No of molecule decreasing, so entropy decreases.
$D \to$ Adsorption will lead to a decreases in the randomness of gaseous particles.
At the adsorption $Δ H < 0, Δ g < 0, Δ S < 0$

Shift-II

Thermodynamics

272696 Five moles of an ideal gas at 1 bar and $298 K$ is expanded into vacuum to double the volume. The work done is

1

- RT In (V_{2} / V_{1})

2

- RT (V_{2} - V_{1})

3

C_{v} (T_{2} - T_{1})

4 zero

Explanation:

Expansion of a gas in vacuum $(P_{ext} = 0)$ is called free expansion. No work is done during free expansion, since as the expansion is done in vacuum that is in absence of external pressure hence work will be,
$W = - P_{ext} Δ V \Rightarrow W = 0$
Here, $(P_{e x t} = 0)$ because expansion takes place in vacuum.

[JEE Main 2020

Thermodynamics

272703 According to the first law of thermodynamics which of the following quantities represents the change in a state function?

1

qrev

2

q_{rev} / W_{rev}

3

W_{rev}

4

W_{rev}

5

q_{rev} + W_{Iev}

Explanation:

: State Function- Those quantities which is depend upon the initial and final states not of path followed.
According to the first law of thermodynamics (q $qrev +$ $W_{rev}$ ), represents the change in a state function.
$Δ E = q_{rev} + W_{rev} .$

**[JIPMER-2011]

Thermodynamics

272689 Identify the correct statement regarding a spontaneous process.

1 For a spontaneous process in an isolated system, the change in entropy is positive

2 Endothermic processes are never spontaneous

3 Exothermic processes are always spontaneous

4 Lowering of energy in the reaction process is the only criterion for spontaneity

Explanation:

For a spontaneous process, $Δ G < 0$ $Δ G = Δ H - T Δ S < 0$
For isolated system, change in enthalpy is zero.
Hence, $Δ S$ must be positive for spontaneous process. Hence, option (a) is correct.

AIEEE-2007

Thermodynamics

272690 During which of the following processes, does entropy decrease?
A. Freezing of water to ice at $0^{\circ} C$ .
B. Freezing of water to ice at $- 10^{\circ} C$ .
C. $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3} (g)$
D. Adsorption of $CO (g)$ and lead surface.
E. Dissolution of $NaCl$ in water.

1 A, B, C and D

2

B

and

C

3

A

and

E

4

A, C

and

E

Explanation:

$A, B \to$ freezing of $H_{2} O$ will decreases entropy as particles more closer and forces of attraction will increases. Thus leads to decreases in randomness so entropy decreases.
$C \to$ No of molecule decreasing, so entropy decreases.
$D \to$ Adsorption will lead to a decreases in the randomness of gaseous particles.
At the adsorption $Δ H < 0, Δ g < 0, Δ S < 0$

Shift-II

Thermodynamics

272696 Five moles of an ideal gas at 1 bar and $298 K$ is expanded into vacuum to double the volume. The work done is

1

- RT In (V_{2} / V_{1})

2

- RT (V_{2} - V_{1})

3

C_{v} (T_{2} - T_{1})

4 zero

Explanation:

Expansion of a gas in vacuum $(P_{ext} = 0)$ is called free expansion. No work is done during free expansion, since as the expansion is done in vacuum that is in absence of external pressure hence work will be,
$W = - P_{ext} Δ V \Rightarrow W = 0$
Here, $(P_{e x t} = 0)$ because expansion takes place in vacuum.

[JEE Main 2020

Thermodynamics

272703 According to the first law of thermodynamics which of the following quantities represents the change in a state function?

1

qrev

2

q_{rev} / W_{rev}

3

W_{rev}

4

W_{rev}

5

q_{rev} + W_{Iev}

Explanation:

: State Function- Those quantities which is depend upon the initial and final states not of path followed.
According to the first law of thermodynamics (q $qrev +$ $W_{rev}$ ), represents the change in a state function.
$Δ E = q_{rev} + W_{rev} .$

**[JIPMER-2011]

Thermodynamics

272689 Identify the correct statement regarding a spontaneous process.

1 For a spontaneous process in an isolated system, the change in entropy is positive

2 Endothermic processes are never spontaneous

3 Exothermic processes are always spontaneous

4 Lowering of energy in the reaction process is the only criterion for spontaneity

Explanation:

For a spontaneous process, $Δ G < 0$ $Δ G = Δ H - T Δ S < 0$
For isolated system, change in enthalpy is zero.
Hence, $Δ S$ must be positive for spontaneous process. Hence, option (a) is correct.

AIEEE-2007

Thermodynamics

272690 During which of the following processes, does entropy decrease?
A. Freezing of water to ice at $0^{\circ} C$ .
B. Freezing of water to ice at $- 10^{\circ} C$ .
C. $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3} (g)$
D. Adsorption of $CO (g)$ and lead surface.
E. Dissolution of $NaCl$ in water.

1 A, B, C and D

2

B

and

C

3

A

and

E

4

A, C

and

E

Explanation:

$A, B \to$ freezing of $H_{2} O$ will decreases entropy as particles more closer and forces of attraction will increases. Thus leads to decreases in randomness so entropy decreases.
$C \to$ No of molecule decreasing, so entropy decreases.
$D \to$ Adsorption will lead to a decreases in the randomness of gaseous particles.
At the adsorption $Δ H < 0, Δ g < 0, Δ S < 0$

Shift-II

Thermodynamics

272696 Five moles of an ideal gas at 1 bar and $298 K$ is expanded into vacuum to double the volume. The work done is

1

- RT In (V_{2} / V_{1})

2

- RT (V_{2} - V_{1})

3

C_{v} (T_{2} - T_{1})

4 zero

Explanation:

Expansion of a gas in vacuum $(P_{ext} = 0)$ is called free expansion. No work is done during free expansion, since as the expansion is done in vacuum that is in absence of external pressure hence work will be,
$W = - P_{ext} Δ V \Rightarrow W = 0$
Here, $(P_{e x t} = 0)$ because expansion takes place in vacuum.

[JEE Main 2020

Thermodynamics

272703 According to the first law of thermodynamics which of the following quantities represents the change in a state function?

1

qrev

2

q_{rev} / W_{rev}

3

W_{rev}

4

W_{rev}

5

q_{rev} + W_{Iev}

Explanation:

: State Function- Those quantities which is depend upon the initial and final states not of path followed.
According to the first law of thermodynamics (q $qrev +$ $W_{rev}$ ), represents the change in a state function.
$Δ E = q_{rev} + W_{rev} .$

**[JIPMER-2011]

Thermodynamics

272689 Identify the correct statement regarding a spontaneous process.

1 For a spontaneous process in an isolated system, the change in entropy is positive

2 Endothermic processes are never spontaneous

3 Exothermic processes are always spontaneous

4 Lowering of energy in the reaction process is the only criterion for spontaneity

Explanation:

For a spontaneous process, $Δ G < 0$ $Δ G = Δ H - T Δ S < 0$
For isolated system, change in enthalpy is zero.
Hence, $Δ S$ must be positive for spontaneous process. Hence, option (a) is correct.

AIEEE-2007

Thermodynamics

272690 During which of the following processes, does entropy decrease?
A. Freezing of water to ice at $0^{\circ} C$ .
B. Freezing of water to ice at $- 10^{\circ} C$ .
C. $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3} (g)$
D. Adsorption of $CO (g)$ and lead surface.
E. Dissolution of $NaCl$ in water.

1 A, B, C and D

2

B

and

C

3

A

and

E

4

A, C

and

E

Explanation:

$A, B \to$ freezing of $H_{2} O$ will decreases entropy as particles more closer and forces of attraction will increases. Thus leads to decreases in randomness so entropy decreases.
$C \to$ No of molecule decreasing, so entropy decreases.
$D \to$ Adsorption will lead to a decreases in the randomness of gaseous particles.
At the adsorption $Δ H < 0, Δ g < 0, Δ S < 0$

Shift-II

Thermodynamics

272696 Five moles of an ideal gas at 1 bar and $298 K$ is expanded into vacuum to double the volume. The work done is

1

- RT In (V_{2} / V_{1})

2

- RT (V_{2} - V_{1})

3

C_{v} (T_{2} - T_{1})

4 zero

Explanation:

Expansion of a gas in vacuum $(P_{ext} = 0)$ is called free expansion. No work is done during free expansion, since as the expansion is done in vacuum that is in absence of external pressure hence work will be,
$W = - P_{ext} Δ V \Rightarrow W = 0$
Here, $(P_{e x t} = 0)$ because expansion takes place in vacuum.

[JEE Main 2020

Thermodynamics

272703 According to the first law of thermodynamics which of the following quantities represents the change in a state function?

1

qrev

2

q_{rev} / W_{rev}

3

W_{rev}

4

W_{rev}

5

q_{rev} + W_{Iev}

Explanation:

: State Function- Those quantities which is depend upon the initial and final states not of path followed.
According to the first law of thermodynamics (q $qrev +$ $W_{rev}$ ), represents the change in a state function.
$Δ E = q_{rev} + W_{rev} .$

**[JIPMER-2011]

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