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Step-by-Step Solution
Step 1: Write down the given differential equation
The problem states that the instantaneous change in pressure p with respect to volume v is given by:
$$
\frac{dp}{dv} = -\,a\,p.
$$
Additionally, it is given that when $v=0$, $p = p_0$ (the boundary condition).
Step 2: Solve the differential equation for p
Separate the variables and integrate:
$$
\int_{p_0}^{p} \frac{dp}{p} \;=\; -\,a \int_{0}^{v} dv.
$$
Evaluating the integrals:
$$
\ln \left( \frac{p}{p_0} \right) \;=\; -\,a\,v.
$$
Exponentiating both sides gives:
$$
p \;=\; p_0 \, e^{-\,a\,v}.
$$
Step 3: Express the temperature in terms of p and v
For one mole of an ideal gas, the temperature T can be related to the product p·v by the ideal gas equation:
$$
T \;=\; \frac{p\,v}{n\,R} \quad\text{with}\quad n=1,
$$
which simplifies to
$$
T \;=\; \frac{p\,v}{R}.
$$
Substituting $p = p_0\,e^{-\,a\,v}$, we get:
$$
T \;=\; \frac{p_0\,v\,e^{-\,a\,v}}{R}.
$$
Step 4: Find the volume v that maximizes T
We need to set the derivative of T with respect to v equal to zero to locate the maximum temperature:
$$
\frac{dT}{dv} \;=\; 0.
$$
Compute the derivative:
$$
\frac{dT}{dv}
\;=\;
\frac{p_0}{R}
\left[
e^{-\,a\,v}
+
v \cdot \frac{d}{dv}\bigl(e^{-\,a\,v}\bigr)
\right].
$$
Since $\frac{d}{dv}(e^{-\,a\,v}) = -\,a\,e^{-\,a\,v}$, this becomes:
$$
\frac{dT}{dv}
\;=\;
\frac{p_0}{R}
\left[
e^{-\,a\,v}
-
a\,v\,e^{-\,a\,v}
\right]
\;=\;
\frac{p_0\,e^{-\,a\,v}}{R}
\bigl(1 - a\,v\bigr).
$$
Setting this equal to zero:
$$
\frac{p_0\,e^{-\,a\,v}}{R}
\bigl(1 - a\,v\bigr)
= 0.
$$
Since $p_0\,e^{-\,a\,v}/R \neq 0$, we require
$$
1 - a\,v = 0
\quad\Longrightarrow\quad
v = \frac{1}{a}.
$$
Step 5: Calculate the maximum temperature
Substituting $v = \tfrac{1}{a}$ into $T = \tfrac{p_0\,v\,e^{-\,a\,v}}{R}$:
$$
T_{\text{max}}
=
\frac{p_0 \,\bigl(\tfrac{1}{a}\bigr)\, e^{-\,a \cdot \tfrac{1}{a}}}{R}
\;=\;
\frac{p_0}{a\,R}
\,e^{-\,1}
\;=\;
\frac{p_0}{a\,e\,R}.
$$
Step 6: Conclude the correct answer
The maximum temperature one mole of gas can attain under the given conditions is
$$
\frac{p_0}{a\,e\,R}.
$$
This matches the correct answer,
Option (1):
$
\frac{p_0}{a\,e\,R}.
$
