Question

If a gas has 'n' 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)

$1+\frac{2}{n}$

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
$E=N\left(\frac{1}{2}nKT\right)=\frac{1}{2}nRT$ $(\because k=\frac{R}{N})$
Now, $C_v=\frac{dE}{dT}=\frac{n}{2}R$
and $C_p=\frac{n}{2}R+R=(\frac{n}{2}+1)R$
$\frac{C_p}{C_v}=\frac{(\frac{n}{2}+1)R}{\frac{n}{2}}=(1+\frac{2}{n})$