Thermodynamics — Page 8 | JEE Chemistry

Q71

For one mole of an ideal gas, which of these statements must be true? (a) U and H each depends only on temperature (b) Compressibility factor z is not equal to 1 (c) CP, m – CV, m = R (d) dU = CVdT for any process

A (a), (c) and (d)

B (a) and (c)

C (c) and (d)

D (b), (c) and (d)

Correct Answer

Option A

Solution

For 1 mole of ideal gas : 1.

Both internal energy (U) and Enthalpy (H) depends on temperature 2.

Compressibility factor Z = 1 3.

CP, m – CV, m = R 4. dU = CVdT for all process

Q72

The true statement amongst the following is :

A S is a function of temperature but

\Delta

S is not a function of temperature.

B Both S and

\Delta

S are not functions of temperature.

C Both

\Delta

S and S are functions of temperature.

D S is not a function of temperature but

\Delta

S is a function of temperature.

Correct Answer

Option C

Solution

\Delta S = \int {{{d{q_{rev}}} \over T}}

S = Kln(w) Both entropy and change in entropy are function of temperature.

Q73

If enthalpy of atomisation for Br2(1) is x kJ/mol and bond enthalpy for Br2 is y kJ/mol, the relation between them :

A does not exist

B is x < y

C is x > y

D is x = y

Correct Answer

Option C

Solution

$\therefore$

\Delta

Hatomisation =

\Delta

HVap + Bond Energy $\Rightarrow$ x =

\Delta

HVap + y $\Rightarrow$ x

>

y

Q74

The incorrect expression among the following is :

A

{{\Delta {G_{System}}} \over {\Delta {S_{Total}}}} = - T

(at constant P)

B

K = {{\Delta H^\circ - T\Delta S^\circ } \over {RT}}

C

K = {e^{ - \Delta G^\circ /RT}}

D For isothermal process

{w_{reversible}} = - nRT\ln {{{V_f}} \over {{V_i}}}

Correct Answer

Option B

Solution

Option (b) is incorrect.

\Delta

G

^\circ

= $-$ RT ln K

\Delta

H

^\circ

$-$ T

\Delta

S

^\circ

= $-$ RT ln K ln K =

\left[ {{{\Delta H^\circ - \Delta S^\circ } \over {RT}}} \right]

Q75

Which of the following relation is not correct?

A

\Delta \mathrm{H}=\Delta \mathrm{U}-\mathrm{P} \Delta \mathrm{V}

B

\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}

C

\Delta \mathrm{S}_{\text{sys }}+\Delta \mathrm{S}_{\text{surr }} \geqslant 0

D

\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}

Correct Answer

Option A

Solution

If $\mathrm{U}+\mathrm{Pv}$ (By definition) $\Delta 14=\Delta \mathrm{U}+\Delta(\mathrm{Pr})$ at constant pressure

\Delta \mathrm{H}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{V}

Q76

Given below are two statements: One is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A : The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state. Reason

\mathbf{R}

: The value of

\Delta G ^\Theta

becomes more on negative side as entropy is higher in liquid state than solid state. In the light of the above statements, choose the most appropriate answer from the options given below

A Both A and R are correct and R is the correct explanation of A

B Both A and R are correct but R is NOT the correct explanation of A

C A is correct but R is not correct

D A is not correct but R is correct

Correct Answer

Option A

Solution

Reduction of a metal oxide is easier if the metal is formed in a liquid state at the temperature of reduction because the entropy is higher if the metal is in a liquid state.

Q77

Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following :

A

\mathrm{q}=0, \Delta \mathrm{T}=0, \mathrm{w}=0

B

\mathrm{q}=0, \Delta \mathrm{T} \neq 0, \mathrm{w}=0

C

\mathrm{q} \neq 0, \Delta \mathrm{T}=0, \mathrm{w}=0

D

\mathrm{q}=0, \Delta \mathrm{T}<0, \mathrm{w} \neq 0

Correct Answer

Option A

Solution

Let's analyze what happens during a free (or unrestrained) expansion of an ideal gas under adiabatic conditions: Adiabatic process : By definition, in an adiabatic process, no heat is transferred between the system and its surroundings.

Thus, $q = 0$ .

Free expansion : In a free expansion, the gas expands into a vacuum or against zero external pressure.

Since work ( $w$ ) is defined as the product of pressure ( $P$ ) and volume change ( $\Delta V$ ), and since the external pressure is zero, we conclude that $w = 0$ .

Temperature change ( $\Delta T$ ) : Since work is zero and no heat is transferred, the first law of thermodynamics tells us that the change in internal energy ( $\Delta U$ ) is zero as well.

For an ideal gas, internal energy is a function of temperature only; thus, if the internal energy does not change, the temperature does not change.

Therefore, $\Delta T = 0$ .

Based on the above points, the free expansion of an ideal gas under adiabatic conditions is characterized by no heat transfer ( $q = 0$ ), no work done ( $w = 0$ ), and no change in temperature ( $\Delta T = 0$ ).

Therefore, the correct option is: Option A : $q = 0, \Delta T = 0, w = 0$

Q78

The effect of temperature on spontaneity of reactions are represented as : .tg .tg

\Delta

H

\Delta

S Temperature Spontaneity (A)

+

-

any T Non spontaneous (B)

+

+

low T spontaneous (C)

-

-

low T Non spontaneous (D)

-

+

any T spontaneous The incorrect combinations are :

A (A) and (C) only

B (B) and (D) only

C (A) and (D) only

D (B) and (C) only

Correct Answer

Option D

Solution

Let's analyze each combination using the Gibbs free energy equation:

\Delta G = \Delta H - T\Delta S

A reaction is spontaneous when

\Delta G Now, consider each case: (A)

\Delta H > 0

and

\Delta S Here,

\Delta G = +\text{(positive)} - T(-\text{(positive)}) = \text{positive} + T\text{(positive)}

This is positive for any temperature. Thus, the reaction is non-spontaneous for any T, which is correct. (B)

\Delta H > 0

and

\Delta S > 0

Now,

\Delta G = +\text{(positive)} - T\text{(positive)}.

For

\Delta G

to be negative, the temperature has to be high enough such that

T\Delta S > \Delta H.

The table incorrectly states that the reaction is spontaneous at low T. (C)

\Delta H In this case,

\Delta G = -\text{(positive)} - T(-\text{(positive)}) = -\text{(positive)} + T\text{(positive)}.

At very low temperatures, the term

T\Delta S

is small, making

\Delta G

negative (spontaneous). The table mistakenly indicates that the reaction is non-spontaneous at low T. (D)

\Delta H 0

Then,

\Delta G = -\text{(positive)} - T\text{(positive)}.

$$ This expression is negative at all temperatures, so the reaction is correctly marked as spontaneous.

Thus, the incorrect combinations are: (B): It should be spontaneous at high T, not low T.

(C): It should be spontaneous at low T, not non-spontaneous.

Therefore, the incorrect combinations are (B) and (C) only.

The correct answer is Option D.

Q79

Which of the following is not correct?

A

\Delta \mathrm{G}

is positive for a spontaneous reaction

B

\Delta \mathrm{G}

is positive for a non-spontaneous reaction

C

\Delta \mathrm{G}

is zero for a reversible reaction

D

\Delta \mathrm{G}

is negative for a spontaneous reaction

Correct Answer

Option A

Solution

The statement that is not correct among the given options is Option A. Option A states that

\Delta \mathrm{G}

is positive for a spontaneous reaction, which is incorrect.

The criterion for spontaneity in a chemical reaction is based on the Gibbs free energy change (

\Delta \mathrm{G}

) for the process.

A spontaneous reaction is one that occurs without needing continuous input of energy from an external source.

The correct relation is that for a spontaneous reaction,

\Delta \mathrm{G}

is negative (

\Delta \mathrm{G} Option B accurately states that

\Delta \mathrm{G}

is positive for a non-spontaneous reaction. A positive value of

\Delta \mathrm{G}

(

\Delta \mathrm{G} > 0

) indicates that the reaction is not spontaneous under the given conditions and requires external energy to proceed. Option C is correct in stating that

\Delta \mathrm{G}

is zero for a reversible reaction at equilibrium. When a reaction is at equilibrium, it has reached a state where the forward and reverse reactions occur at equal rates, and there is no net change in the composition of the system. At this point, the Gibbs free energy is at its minimum for the given conditions, and

\Delta \mathrm{G} = 0

, indicating that the system is in a state of maximum stability and no further net change can occur without the input or removal of energy. Option D correctly states that

\Delta \mathrm{G}$$ is negative for a spontaneous reaction.

As explained previously, a negative Gibbs free energy change signifies that a process or reaction can proceed spontaneously in the direction written.

Q80

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always

-57 \mathrm{~kJ} \mathrm{~mol}^{-1}

Reason (R) : Enthalpy of neutralisation is the amount of heat liberated when one mole of

\mathrm{H}^{+}

ions furnished by acid combine with one mole of

{ }^{-} \mathrm{OH}

ions furnished by base to form one mole of water. In the light of the above statements, choose the correct answer from the options given below.

A (A) is true but (R) is false

B Both (A) and (R) are true and (R) is the correct explanation of (A)

C Both (A) and (R) are true but (R) is not the correct explanation of (A)

D (A) is false but (R) is true

Correct Answer

Option B

Solution

Enthalpy of neutralisation of strong acids and bases is

-57 \mathrm{~kJ} / \mathrm{mol}

. which is fixed for reaction of 1 mole of

\mathrm{H}^{+}

with 1 mole of

\mathrm{OH}^{-}

to form 1 mole of water.