Q168 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Physics
According to law of equi-partition of energy the molar specific heat of a rigid diatomic gas at constant pressure is
- A$\dfrac{9R}{2}$
- Correct answerB$\dfrac{7R}{2}$
- C$\dfrac{5R}{2}$
- D$\dfrac{3R}{2}$
Solution
For a rigid diatomic gas, the number of translational degrees of freedom is $3$ and the number of rotational degrees of freedom is $2$.
Total degrees of freedom, $f = 3 + 2 = 5$.
According to the law of equipartition of energy, the molar specific heat at constant volume is given by:
$C_v = \dfrac{f}{2} R = \dfrac{5}{2} R$
Using Mayer's relation, the molar specific heat at constant pressure is:
$C_p = C_v + R$
$C_p = \dfrac{5}{2} R + R = \dfrac{7}{2} R$
Answer: $\dfrac{7R}{2}$
Total degrees of freedom, $f = 3 + 2 = 5$.
According to the law of equipartition of energy, the molar specific heat at constant volume is given by:
$C_v = \dfrac{f}{2} R = \dfrac{5}{2} R$
Using Mayer's relation, the molar specific heat at constant pressure is:
$C_p = C_v + R$
$C_p = \dfrac{5}{2} R + R = \dfrac{7}{2} R$
Answer: $\dfrac{7R}{2}$
Hence, the correct answer is option: 'B'.