JEE PHYSICS MCQ: THERMODYNAMICS

1 Medium

1 of 1

Remembering +4 | -1

The ratio of the specific heats

\frac{C_p}{C_V}=\gamma

in terms of degrees of freedom

(n)

is given by

View answer and explanation

Correct option: C

\left(1+\frac{2}{n}\right)

Option C (

\left(1+\frac{2}{n}\right)

) is correct because the ratio of specific heats is related to the degrees of freedom by this formula.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →

0.25\,\,{m^3}

{10^3}N/{m^2},

T

P

V

\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)