At ordinary temperatures, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy.
As a result of this, at higher temperature

Options

A. `C_(V) = (3 R)/(2)` for a monatomic gas
B. `C_(V) gt (3 R)/(2)` for a monatomic gas
C. `C_(V) lt (3 R)/(2)` for a diatomic gas
D. `C_(V) gt (3 R)/(2)` for a diatomic gas

Answer

A, D

Explanation

Vibrational kinetic energy of a monatomic gas `=0` at all temperature. So, `C_(v)=3 R//2` for a monoatomic gas at high temperatures also. In case of a diatomic gas `C_(v)=5 R//2` at low temperatures while , `C_(v)=5R//2` at high temperatures due to vibrational `KE`.