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$Two moles of an ideal gas are heated at constant pressure from T_{1} = 300 K to T_{2} = 400 K . The molar heat capacity at constant volume is C_{V} = \frac{3}{2} R . The change in entropy of the gas is:$

A. $5 R ln (4/3)$
B. $2 R ln (4/3)$
C. $3 R ln (4/3)$
D. $\frac{5}{2} R ln (4/3)$

Show correct answer & step-by-step solution

Correct answer: A — $5 R ln (4/3)$

Solution

$For an ideal gas, C_{P} = C_{V} + R = \frac{3}{2} R + R = \frac{5}{2} R .$
$The change in entropy is Δ S = n C_{P} ln (\frac{T _{2}}{T _{1}}) .$
$Substituting the values, Δ S = 2 \times \frac{5}{2} R ln (\frac{400}{300}) .$
$This simplifies to Δ S = 5 R ln (\frac{4}{3}) .$
$Hence the answer is (A).$

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