Q. At constant volume the specific heat of a gas is $\frac{3R}{2},$ then the value of $'\gamma'$ will be
A) $\frac{3}{2}$
B) $\frac{5}{2}$
C) $\frac{5}{3}$
D) None of the above
Explanation
(c) $C_{V} = \frac{R}{(\gamma - 1)} \Rightarrow \gamma = 1 + \frac{R}{C_{V}} = 1 + \frac{R}{\frac{3}{2}R} = \frac{5}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.