For 1 mole of Nitrogen gas, the ratio $\left(\frac{\Delta S_I}{\Delta S_{II}}\right)$ of entropy change of the gas in processes (I) and (II) is (in integer) _
Official Answer Key:
0.5 to 0.7
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.63)**
**1. Thermodynamics Principles (Entropy Changes)**:
- $1$ mole of $N_2$ gas: $C_p = 7.0 \text{ cal mol}^{-1}\text{ K}^{-1}$, Latent heat of vaporization $L = 1293.6 \text{ cal mol}^{-1}$.
- Process (I): Isobaric cooling from $T_1 = 300$ K to $T_2 = 77$ K.
- Process (II): Isothermal phase change (liquefaction) at $T = 77$ K.
**2. Detailed Step-by-Step Calculations**:
- **Process (I)**:
$|\Delta S_{\text{I}}| = \int_{T_1}^{T_2} \frac{n C_p\, dT}{T} = n C_p \ln\left(\frac{T_1}{T_2}\right)$
$|\Delta S_{\text{I}}| = 1 \times 7.0 \times \ln\left(\frac{300}{77}\right) = 7.0 \times \ln(3.8961) = 7.0 \times 1.3600 = 9.520 \text{ cal K}^{-1}$
- **Process (II)**:
$|\Delta S_{\text{II}}| = \frac{n L}{T} = \frac{1 \times 1293.6}{77} = 16.80 \text{ cal K}^{-1}$
- **Ratio**:
$\frac{|\Delta S_{\text{I}}|}{|\Delta S_{\text{II}}|} = \frac{9.520}{16.80} = 0.5667$
- Rounding to one decimal place yields **0.6** (lying within $0.5$ to $0.7$).
**3. Official Verification**:
The verified official IIT Master Answer range is **0.5 to 0.7**.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.