$$N$$ moles of diatomic gas in a cylinder is at a temperature $$T$$. Heat is supplied to the cylinder such that the temperature remains constant but $$n$$ moles of the diatomic gas get converted into monoatomic gas. The change in the total kinetic energy of the gas is

We start with $$N$$ moles of an ideal diatomic gas at a fixed temperature $$T$$. For an ideal gas the total kinetic energy (translational + rotational) per mole equals $$\dfrac{f}{2}RT$$, where $$f$$ is the number of degrees of freedom.

For a diatomic molecule at ordinary temperatures the three translational and two rotational modes are active, so $$f=5$$. Hence the total kinetic energy of the initial sample is

$$ U_i \;=\; N\left(\dfrac{5}{2}RT\right)\;=\;\dfrac{5}{2}NRT . $$

Now heat is supplied while the temperature is kept constant, and $$n$$ moles (in the sense of diatomic molecules) dissociate completely into mono-atomic molecules. A single diatomic molecule $$XY$$ dissociates as

$$ XY \;\rightarrow\; X + Y , $$

so one mole of the diatomic gas produces two moles of mono-atomic gas. Therefore, after the dissociation:

• Remaining diatomic moles $$=\;N-n$$
• New mono-atomic moles $$=\;2n$$
• Total final moles $$=\;(N-n)+2n \;=\;N+n$$

The mono-atomic gas possesses only the three translational degrees of freedom, so for it $$f=3$$ and the kinetic energy per mole is $$\dfrac{3}{2}RT$$.

Hence the total kinetic energy after the change is

$$ \begin{aligned} U_f &= (N-n)\left(\dfrac{5}{2}RT\right) + (2n)\left(\dfrac{3}{2}RT\right) \\ &= \dfrac{5}{2}NRT - \dfrac{5}{2}nRT + 3nRT \\ &= \dfrac{5}{2}NRT + \left(3 - \dfrac{5}{2}\right)nRT \\ &= \dfrac{5}{2}NRT + \dfrac{1}{2}nRT . \end{aligned} $$

The change in total kinetic energy is therefore

$$ \Delta U = U_f - U_i = \left[\dfrac{5}{2}NRT + \dfrac{1}{2}nRT\right] - \dfrac{5}{2}NRT = \dfrac{1}{2}nRT . $$

Hence, the correct answer is Option D.