A gas has molar heat capacity `C = 37.55 J "mole"^(-1)K^(-1)`, in the process PT = constant, find the number of degree of freedom of the molecules of the gas.

Options

A. 2
B. 3
C. 5
D. 7

Answer

C

Explanation

Given, `C= 37.55 J mol^(-1) K^(-1)`
Also, pT= constant (K) …(i)
According to ideal gas equation, pV= RT
`rArr p= (RT)/(V)` …(ii)
Putting the value of p in Eq (i), we get
`(RT)/(V) xx T= K rArr V= (RT^(2))/(K)`
On differentiating above equation both sides, we get
`(dV)/(dT)= (2RT)/(K)` ...(iii)
But `(T)/(K)= (1)/(p)` [from Eq (i)]
Hence, Eq (iii) becomes
`(dV)/(dT)= (2R)/(p)`
So, `C= C_(V) + (pdV)/(dT)`
or `C =C_(V) + (p xx 2R)/(p)= C_(V) + 2R`
or `C_(V)= C-2R` ....(iv)
As `C_(V)= (nR)/(2)`, where n= number of degrees of freedom.
Putting the value of `C_(V)` in Eq (iv), we get
`(nR)/(2)= C-2R`
`n= (2(C-2R))/(R )= (2(37.55-2 xx 8.3))/(8.3)= 5.048= ~=5`