Mono atomic gas molecules has three translational degrees of freedom.

From law of equipartition of energy average energy of an molecule at temperature T is (3/2)K_{B}T

Total internal energy of one mole of such gas is
$U= \frac {3}{2}K_B T N$
$U = \frac {3}{2} RT$ ---(12)

If $C_V$ is molar specific heat at constant volume then
$C_V = \frac {dU}{dT}$
$ = \frac {3}{2}R$ ---(13)
Now for an ideal gas
$C_P - C_V = R$
$C_P$ - molar specific heat capacity at constant pressure
$C_P = \frac {5}{2} R$--- (14)
Thus for a mono atomic gas ratio of specific heats is
$ \gamma _{mono} = \frac {C_P}{C_V}= \frac {5}{3}$ ---(15)

(ii) Diatomic gases :

A diatomic gas molecule is treated as a rigid rotator like dumb-bell and has 5 degrees of freedom out of which three degrees of freedom are translational and two degrees of freedom are rotational.

Using law of equipartition of energy the total internal energy of one mole of diatomic gas is
$U= \frac {5}{2}K_BN T$
$= \frac {5}{2} RT$ ---(16)

Specific heats are thus
$C_V =\frac {dU}{dT}= \frac {5}{2}R$
Now for an ideal gas
$C_P - C_V = R$
$C_P$ - molar specific heat capacity at constant pressure
$C_P = \frac {7}{2} R$--- (14)
$ \gamma _{dia}= \frac {7}{2}$ (rigid rotator)

If diatomic molecule is not only rigid but also has an vibrational mode in addition, then
$U = \frac {7}{2} RT$
and $C_V=\frac {7}{2} R$
Now for an ideal gas
$C_P - C_V = R$
$C_P$ - molar specific heat capacity at constant pressure
$C_P=\frac {9}{2} R$
and $ \gamma _{dia}=\frac {C_P}{C_V}= \frac {9}{7}$