Question

If a gas has ${}^{\prime }{n}^{\prime }$ degrees of freedom, the ratio of the specific heats $\gamma$ of the gas is

• A. $\frac{1+n}{2}$
• B. $1+\frac{n}{2}$
• C. $1+\frac{1}{n}$
• D. $1+\frac{2}{n}$

Answer 1

Answer: (D)

Let us consider 1 mole of an ideal gas at kelvin temperature $T.$ It has $N$ molecules (Avogadro's number). The internal energy of an ideal gas is entirely kinetic.
The average KE per molecule of a ideal gas is $\frac{1}{2}nkT$ (k is boltzman constant), where $n$ is degree of freedom. Therefore the internal energy of one mole of an gas would be
$E=N\left(\frac{1}{2}nkT\right)=\frac{1}{2}nRT$
$\left(\because k=\frac{R}{N}\right)$
Now, $C_V=\frac{dE}{dT}=\frac{n}{2}R$
and $C_p=\frac{n}{2}R+R=\left(\frac{n}{2}+1\right)R$
$\frac{C_p}{C_V}=\frac{\left(\frac{n}{2}+1\right)R}{\frac{n}{2}}$
$=\left(1+\frac{2}{n}\right)$