ThermodynamicsHard

Question

The molar entropy of a constant volume sample of neon at 500 K if it is 46.2 cal /K-mol at 250 K is (ln 2 = 0.7)

Options

A.2.1 cal/K-mol
B.44.1 cal/K-mol
C.48.3 cal/K-mol
D.46.2 cal/K-mol

Solution

$\Delta S = n.C_{V,m}.\ln\frac{T_{2}}{T_{1}}$

$S_{500\text{ K}} - 46.2 = 1 \times \frac{3R}{2} \times \ln\frac{500}{250} $$$\therefore S_{\text{500 K}} = 48.3\text{ Cal/K-mol}$$

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