Question

The molar heat capacity in a process of diatomic gas if it does a work of $\frac{Q}{4}$ when a heat of Q is supplied to it is

• A. $\frac{2}{5}$R
• B. $\frac{5}{4}$R
• C. $\frac{10}{3}$R
• D. $\frac{6}{4}$R

Answer 1

Answer: (C)

$\frac{10}{3}$R

First law of Thermodynamics we have :-

$Q=w+\Delta U$ (1)

where,

Q = Heat supplied to the gas

w = Work done by the gas

ΔU = Change in the internal energy of the gas

Given :-

$w=Q/4$ (2)

From (1) & (2) we have :-

⟹$\Delta U=3Q/4$ (3)

Now,

$\Delta U=\frac{nR\Delta T}{\gamma -1}$ (4)

where,

n = No. of moles of gas

ΔT = Change in temperature

R = Universal Gas Constant

γ = Specific heat ratio

For a diatomic gas :-

$\gamma =7/5$ ( 5)

Using (3), (4) & (5) we get :-

⟹$n\Delta T=\frac{3Q}{10R}$ (6)

Also note that :-

$Q=nC\Delta T$ (7)

where,

C = Molar specific heat for the process

So now replace nΔT in (7) by (6) :-

⟹$\frac{3QC}{10R}=Q$

⟹$C=\frac{10R}{3}$