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Q.
If a gas has ‘n’ degrees of freedom, the ratio of the specific heats $\gamma$ of the gas is
Kinetic Theory
Solution:
Let us consider 1 mole of an ideal gas as a Kelvin temperature T. It has NA molecules (Avogadro’s number). The internal energy of an ideal gas is entirely kinetic. The average K.E. per molecule of an ideal gas $ = \frac{1}{2} n \, k \, T,$ where n is degrees of freedom.
Therefore the internal energy of one mole of an ideal gas would be
$U = N \left( \frac{1}{2} nkT\right) = \frac{1}{2} n RT \left(\because k = R/N_{A}\right) $
Now, $C_{v} = \frac{dU}{dT} = \frac{n}{2} R $
and $C_{P} = \frac{n}{2} R + R = \left(\frac{n}{2} + 1\right)R $
$\frac{C_{P}}{C_{v}} = \frac{\left(\frac{n}{2} + 1\right)R}{\frac{n}{2}R} = \left(1+ \frac{2}{n}\right)=\gamma$