Question

$N$ moles of an ideal diatomic gas are in a closed cylinder at temperature $T$. Suppose, on supplying heat to the gas, its temperature remains constant, but $n$ moles get dissociated into atoms. Then, heat supplied to the gas is -

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

Answer 1

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

Since the gas is enclosed in a vessel, during the heating process, volume of the gas remains constant. Hence, no work is done by the gas. It means heat supplied to the diatomic gas is used to increase its internal energy only.
Initial internal energy of the gas is
$U_1=N\left(\frac{5}{2}R\right)T=\frac{5}{2}NRT$
Since $n$ moles get dissociated into atoms, after heating, the vessel contains $(N-n)$ moles of diatomic gas and 2n moles of monoatomic gas. Hence, the internal energy of the gas after heating will be equal to
$U_2=(N-n)\left(\frac{5}{2}R\right)T+2n\left(\frac{3}{2}R\right)T=\frac{5}{2}NRT+\frac{1}{2}nRT$
Hence, heat supplied = Increase in internal energy
$=(U_2-U_1)=\frac{1}{2}nRT$
Thus, option (b) is the correct answer.