Question:

The CORRECT expression that corresponds to reversible and adiabatic expansion of an ideal gas is

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In an adiabatic process for an ideal gas, the change in internal energy is zero because there is no heat exchange, and the temperature remains constant.
Updated On: Dec 12, 2025
  • \( \Delta U = 0 \)
  • \( \Delta H = 0 \)
  • \( \Delta S = 0 \)
  • \( \Delta G = 0 \)
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The Correct Option is C

Solution and Explanation

Reversible adiabatic expansion of an ideal gas:

Condition 1: Adiabatic process

  • No heat transfer: $q = 0$

Condition 2: Reversible process

  • Process occurs through equilibrium states

Analyzing each thermodynamic quantity:

(A) ΔU = 0:

From the first law: $\Delta U = q + w$

For adiabatic process: $q = 0$, so $\Delta U = w$

In expansion, work is done by the gas: $w < 0$

Therefore: $\Delta U < 0$ (internal energy decreases)

INCORRECT 

(B) ΔH = 0:

For an ideal gas: $H = U + PV = U + nRT$

Since temperature changes during adiabatic expansion (gas cools), both U and T change.

Therefore: $\Delta H \neq 0$

INCORRECT 

(C) ΔS = 0:

For a reversible process: $dS = \frac{dq_{rev}}{T}$

For adiabatic process: $dq_{rev} = 0$

Therefore: $dS = 0$ and $\Delta S = 0$

CORRECT 

(D) ΔG = 0:

Gibbs free energy: $G = H - TS$

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

Since both H and T change, and the relationship is complex, $\Delta G \neq 0$ in general.

INCORRECT 

Answer: (C) ΔS = 0

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