One mole of an ideal monatomic gas at absolute temperature T undergoes free
expansion to double its original volume , so that the entropy change is ΔS1. An
identical amount of the same gas at absolute temperature 2 T undergoes isothermal
expansion to double its original volume , so that the entropy change is ΔS2. The
value of
ΔS1
ΔS2
(in integer) is ____
Official Answer Key:
1 to 1
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.30)**
**1. Thermodynamics of Ideal Gas Entropy**:
- Fundamental entropy relation for an ideal gas:
$dS = C_v \frac{dT}{T} + R \frac{dV}{V}$
**2. Step-by-Step Mathematical Derivation**:
- **Process 1: Free Expansion of Ideal Gas**:
- An ideal gas undergoing Joule free expansion expands into vacuum adiabatically ($Q = 0, W = 0$).
- By the First Law, $\Delta U = 0$. Since internal energy of an ideal gas depends solely on temperature, $\Delta T = 0$ (initial and final temperatures are equal: $T_1 = T_2 = T$).
- Entropy is a thermodynamic state function. Evaluating along a reversible isothermal path between initial $(T, V)$ and final $(T, 2V)$:
$\Delta S_1 = \int_V^{2V} R \frac{dV}{V} = R \ln\left(\frac{2V}{V}\right) = R \ln 2$
- **Process 2: Reversible Isothermal Expansion**:
- Reversible expansion at constant temperature $T$ from $V$ to $2V$:
$\Delta S_2 = \int_V^{2V} R \frac{dV}{V} = R \ln\left(\frac{2V}{V}\right) = R \ln 2$
- **Ratio of Entropy Changes**:
$\frac{\Delta S_1}{\Delta S_2} = \frac{R \ln 2}{R \ln 2} = 1$
**3. Official Verification**:
The verified official IIT answer is **1** (Numerical range: `1 to 1`).
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