Chemistry - Thermodynamics - I Question with Solution | TestHub

(A) Reversible isothermal expansion

For an ideal gas, isothermal expansion implies $\Delta U=0$ (S).

A reversible process implies $\Delta S_{universe}=0$ (R).

Heat exchange is reversible, which means the surroundings' entropy is balanced (T).

So, A → R, S, T

(B) Reversible adiabatic expansion

Adiabatic implies $Q=0$, which means $\Delta S_{surrounding}=0$ (Q).

Reversible implies $\Delta S_{universe}=0$ (R).

For an ideal gas undergoing reversible adiabatic expansion, the entropy of the system remains constant, which is a path constraint (P, Q, R).

So, B → P, Q, R

(C) Adiabatic free expansion

Free expansion implies an irreversible process.

Adiabatic implies $Q=0$, which means $\Delta S_{surrounding}=0$ (Q).

For an ideal gas, $\Delta U=0$ (S).

The entropy of the universe increases (not zero), which corresponds to the listed set (Q, S, T).

So, C → Q, S, T

(D) Isothermal free expansion

For an ideal gas undergoing isothermal expansion, $\Delta U=0$ (S).

Free expansion implies an irreversible process, and $\Delta S_{surrounding}=0$ (Q).

The entropy increases in the system, matching the given set (Q, S, T).

So, D → Q, S, T