Initial internal energy of the gas is ${U_1} = N\,\left( {\frac{5}{2}R} \right)\,T$
Since $n$ moles get dissociated into atoms, therefore, after heating, vessel contains $(N - n)$ moles of diatomic gas and $2n$ moles of a mono-atomic gas. Hence the internal energy for 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, the heat supplied = increase in internal energy
$ = \,({U_2} - {U_1})\, = \,\frac{1}{2}\,nRT$

$(A)$ the process during the path $\mathrm{A} \rightarrow \mathrm{B}$ is isothermal
$(B)$ heat flows out of the gas during the path $\mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D}$
$(C)$ work done during the path $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}$ is zero
$(D)$ positive work is done by the gas in the cycle $ABCDA$
