A monatomic ideal gas has degrees of freedom $f = 3$. Its molar specific heat at constant volume is:
A$\tfrac{3}{2}R$, since $C_v = \tfrac{f}{2}R$ here
B$\tfrac{5}{2}R$, the diatomic value at constant volume
C$3R$, twice the correct equipartition result here
D$R$, the ideal gas constant without any factor
Answer & Solution
Correct answer: A. $\tfrac{3}{2}R$, since $C_v = \tfrac{f}{2}R$ here
$C_v = \tfrac{f}{2}R = \tfrac{3}{2}R$ for a monatomic gas.
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