If a gas has ‘n’ degrees of freedom, the ratio of the specific heats γ of the gas is

Question

If a gas has ‘n’ degrees of freedom, the ratio of the specific heats &gamma; of the gas is (A) (1 + n/2)  (B)  1 + (n/2)  (C)  1 + (1/n)  (D)  1 + (2/n)

Tardigrade Admin · Accepted Answer

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 (\frac{1}{2} nk T) = \frac{1}{2} n RT (∵ k = R / N_{A})

Now,

C_{v} = \frac{d U}{d T} = \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} R} = (1 + \frac{2}{n}) = γ

Q. If a gas has ‘n’ degrees of freedom, the ratio of the specific heats $γ$ of the gas is

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

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

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

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

Q. If a gas has ‘n’ degrees of freedom, the ratio of the specific heats γ of the gas is

21+n​

1+2n​

1+n1​

1+n2​

Q. If a gas has ‘n’ degrees of freedom, the ratio of the specific heats $γ$ of the gas is

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

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

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

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