GATE 2022 PH – Question 63
One mole of nitrogen is (I) cooled at 1 atm from 300 to 77 K, and (II) liquefied at 77 K. Find entropy-change ratio I/II to one decimal, using Cp=7 cal/(mol K), latent heat=1293.6 cal/mol.
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Correct answer: 0.59 to 0.61
Explanation
**Step 1: cooling the gas at constant pressure (300 K to 77 K).**
$$\Delta S_I=\int_{300}^{77}\frac{C_p}{T}dT=C_p\ln\frac{77}{300}=7\ln(0.2567)=7\times(-1.3601)=-9.52\text{ cal/(mol K)}.$$
**Step 2: liquefaction at 77 K** (heat $L=1293.6$ cal/mol is released, so the entropy of the nitrogen decreases):
$$\Delta S_{II}=-\frac{L}{T}=-\frac{1293.6}{77}=-16.8\text{ cal/(mol K)}.$$
**Ratio:**
$$\frac{\Delta S_I}{\Delta S_{II}}=\frac{-9.52}{-16.8}=0.567\approx\mathbf{0.6}.$$