For an ideal gas the instantaneous change in pressure $$P$$ with volume $$V$$ is given by the equation $$\frac{dP}{dV} = -aP$$. If $$P = P_0$$ at $$V = 0$$ is the given boundary condition, then the maximum temperature one mole of gas can attain is: (Here $$\mathcal{R}$$ is the gas constant)

For one mole of an ideal gas the equation of state is stated first: $$PV=\mathcal{R}T$$.

We are also given the differential relation between pressure and volume: $$\frac{dP}{dV}=-aP$$, where $$a$$ is a positive constant.

Now we solve this first-order differential equation. We rewrite it as $$\frac{1}{P}\,dP=-a\,dV$$. Integrating both sides, we obtain $$\int\frac{1}{P}\,dP=-a\int dV$$ which gives $$\ln P=-aV+C$$, where $$C$$ is the constant of integration.

To determine $$C$$ we use the boundary condition $$P=P_0$$ when $$V=0$$. Substituting these values, $$\ln P_0=C$$. Hence $$C=\ln P_0$$, and the pressure as a function of volume is $$P(V)=P_0e^{-aV}$$.

Substituting this expression for pressure in the ideal-gas equation, we have $$P(V)\,V=\mathcal{R}T$$. Therefore $$\mathcal{R}T = P_0e^{-aV}\,V,$$ so the temperature is $$T(V)=\frac{P_0}{\mathcal{R}}\,V\,e^{-aV}.$$

To find the maximum attainable temperature we must maximize the function $$f(V)=V\,e^{-aV}$$ for $$V \ge 0$$.

We differentiate: $$\frac{df}{dV} = e^{-aV} - aV e^{-aV} = e^{-aV}\bigl(1-aV\bigr).$$ Setting $$\frac{df}{dV}=0$$ gives $$1-aV=0 \;\;\Longrightarrow\;\; V=\frac{1}{a}.$$ Because the exponential factor is always positive, this critical point corresponds to a maximum.

Substituting $$V=\frac{1}{a}$$ back into $$T(V)$$, we get $$T_{\max} = \frac{P_0}{\mathcal{R}}\left(\frac{1}{a}\right) e^{-a\left(\frac{1}{a}\right)} = \frac{P_0}{\mathcal{R}}\left(\frac{1}{a}\right) e^{-1} = \frac{P_0}{a e \mathcal{R}}.$$

Hence, the correct answer is Option D.