A diatomic ideal gas has $C_v = \tfrac{5}{2}R$. Its $C_p$ and ratio $\gamma$ are:
A$\tfrac{5}{2}R$ and $\gamma = 1$, equal $C_v$ and $C_p$
B$\tfrac{3}{2}R$ and $\gamma = 5/3$, monatomic values here
C$\tfrac{9}{2}R$ and $\gamma = 2$, doubling all values
D$\tfrac{7}{2}R$ and $\gamma = 7/5$, since $C_p = C_v + R$
Answer & Solution
Correct answer: D. $\tfrac{7}{2}R$ and $\gamma = 7/5$, since $C_p = C_v + R$
$C_p = C_v + R = \tfrac{7}{2}R$; $\gamma = C_p/C_v = 7/5$.
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