JEE Main 2022 · 26 July · evening · Single correct · +4 / −1
A gas has n degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be :
- (a) n/(n+2)
- (b) (n+2)/n
- (c) n/(2n+2)
- (d) n/(n-2)
Answer and solution
Answer (a) n/(n+2)
Solution
Physics sir · 3 steps
Give each degree of freedom 1/2R per mole, then add R to go from C_V to C_P.
- Equipartition. Each degree of freedom carries 1/2kT of energy per molecule. So one mole of the gas has internal energy U = n/2RT, and C_V = dU/dT = n/2R.
- Mayer's relation. For an ideal gas C_P = C_V + R = n/2R + R = (n+2)/2R.
C_V/C_P = (n/2R)/((n+2)/2R) = n/(n+2)
- So. The ratio is n/(n+2), which is option (a). It is less than 1, as it must be, because C_P is always larger than C_V.
The idea to keep. With f degrees of freedom, C_V = f/2R, C_P = (f+2)/2R and γ = 1 + 2/f. Check which ratio is asked: C_V/C_P is just 1/γ.
Answer from the NTA final answer key. Solution by PrepWiser.