- 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)

- 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}$

**Notes****Assignments****Revision Notes**

Class 11 Maths Class 11 Physics Class 11 Chemistry Class 11 Biology

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