Question

If a gas has n degrees of freedom ratio of specific heats of gas is?

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

Answer 1

The correct option is D $1+\frac{2}{n}$

Let us take 1 mole of the ideal gas at Kelvin temperature T. It has N molecules i.e. Avogadro's number. The internal energy of an ideal gas is only in the form of kinetic. The average KE per molecule of the ideal gas is

$\frac{1}{2}nkT$

Where k is Boltzmann constant,

n is the degree of freedom.

Therefore the internal energy of one mole of gas would be

$E=N\left(\frac{1}{2}nKT\right)$

Since $k=\frac{R}{N}$

So, $E=\frac{1}{2}nRT$

Now specific heat at constant volume,

$C_v=\frac{dE}{dT}=\frac{n}{2}R$

And specific heat at constant pressure,

$C_p=\frac{n}{2}R+R=\frac{n+2}{2}R$

Therefore, ratio of $C_p$ and $C_v$ will be

$\frac{C_p}{C_v}=\frac{\frac{n+2}{2}R}{\frac{n}{2}R}$

$\frac{C_p}{C_v}=\frac{\frac{n+2}{2}R}{\frac{n}{2}R}$

$\frac{C_p}{C_v}=1+\frac{2}{n}$

Thus option D is correct.