Chemistry - Thermodynamics - II Question with Solution | TestHub

Question Explanation: Match the thermodynamic processes with their corresponding conditions.

Concept: $\Delta G$, $Q$, $W$, $\Delta U$, $\Delta S$ for phase changes, free expansion, and cyclic processes.

Solution:

(A) H₂O(l) ⇌ H₂O(s) at 1 atm and 273 K.

Since this is an exothermic reaction, $q<0$.

Also, volume increases, so $w<0$.

H₂O(s) has lesser entropy compared to H₂O(l).

Therefore, $\Delta S_{sys}<0$ and $\Delta U<0$.

Also, since the reaction is at equilibrium, $\Delta G=0$.

Hence, the conditions are (R) & (T).

(B) Since it is an isolated condition, $q=0$.

Also, expansion occurs against a vacuum.

Therefore, $w=0$, which implies $\Delta U=0$.

Thus, the temperature remains constant.

Also, entropy increases.

For a physical process, the Gibbs free energy change is given by:

$$dG=VdP-SdT$$

Therefore, it will be $>0$.

Thus, the answer is (P), (Q), (S).

(C) Since it is an isolated container, $q=0$ and $w=0$.

Because gases are getting mixed, the entropy of the system is increasing.

Also, $\Delta U=0$ (since $\Delta U=q+w$).

Since mixing is spontaneous, $\Delta G<0$.

Thus, the answer is (P), (Q), (S).

(D) Since there is reversible heating and cooling, the initial and final states will be the same.

Hence, all state functions will remain the same.

Also, $q$ and $w$ for heating and cooling will be opposite of each other.

Therefore, overall $q=0$.

Thus, overall $w=0$.

Since mixing is spontaneous, $\Delta G<0$.

Therefore, the answer is (P), (Q), (S), (T).

Final Answer:

A → R, T; B → P, Q, S; C → P, Q, S; D → P, Q, S, T.