ThermodynamicsHard
Question
The entropy change of 2.0 moles of an ideal gas whose adiabatic exponent $\gamma$ = 1.50, if as a result of a certain process, the gas volume increased 2.0 times while the pressure dropped 4.0 times, is (ln 2 = 0.7)
Options
A.−11.64 J/K
B.+11.64 J/K
C.−34.92 J/K
D.+34.92 J/K
Solution
$\Delta S = n.C_{V,m}.\ln\frac{T_{2}}{T_{1}} + nR.\ln\frac{V_{2}}{V_{1}}$
$= n.C_{V,m}.\ln\frac{P_{2}}{P_{1}} + n.C_{P,m}.\ln\frac{V_{2}}{V_{1}} $$$= 2 \times \frac{R}{1.5 - 1} \times \ln\frac{1}{4} + 2 \times \frac{1.5R}{1.5 - 1} \times \ln 2 = - 11.64\text{ J/K}$$
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