Question

A closed cylindrical vessel contains $N$ moles of an ideal diatomic gas at a temperature $T$. On supplying heat, the temperature remains same, but $n$ moles get dissociated into atoms. The heat supplied is:

• A. $\frac{5}{2}(N-n)RT$
• B. $\frac{5}{2}nRT$
• C. $\frac{1}{2}nRT$
• D. $\frac{3}{2}nRT$

Answer 1

The correct option is C $\frac{1}{2}nRT$

Solution:- (C) $\frac{1}{2}nRT$

Since the gas is enclosed in a vessel, during heating, volume of the gas remains constant.

$\therefore W=0$

$\therefore q=\Delta U.....\left(1\right)$

Now,

As we know that,

$\Delta U=n{C}_v\Delta T$

Initial internal energy of the gas is

$U_1=N\left(\frac{5}{2}R\right)T=\frac{5}{2}NRT$

Since $n$ moles dissociate into atoms, therefore after heating, the vessel contains $(N-n)$ moles of diatomic gas and $2n$ moles of mono atomic gas. Hence the internal energy of the gas after heating will be equal to

$U_2=(N-n)\left(\frac{5}{2}R\right)T+2n\left(\frac{3}{2}R\right)T$

$\Rightarrow U_2=\frac{1}{2}nRT+\frac{5}{2}NRT$

$\therefore \Delta U=U_2-U_1$

$\Rightarrow \Delta U=(\frac{1}{2}nRT+\frac{5}{2}NRT)-\frac{5}{2}NRT$

$\Rightarrow \Delta U=\frac{1}{2}nRT$

Now from ${eq}^n\left(1\right)$, we get

$q=\frac{1}{2}nRT$