For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation dp \over dv = - ap . If p = p0 at v =0 is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here R is the gas constant)

Heat and Thermodynamics PYQ — JEE Main 2021 · 31 Aug · Shift 1 · Q58 with Solution

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Correct answer: A

For the ideal gas,

\frac{dp}{dv}=-ap

where $a$ is a constant.

Also, boundary condition:

p=p_0 \quad \text{at} \quad v=0

Separate variables:

\frac{dp}{p}=-a\,dv

Integrating,

\int \frac{dp}{p} = -a \int dv

\ln p = -av + C

So,

p = Ce^{-av}

Using $p=p_0$ at $v=0$ ,

p_0 = Ce^{0} = C

Hence,

p = p_0 e^{-av}

For one mole,

pv = RT

Thus,

T = \frac{pv}{R} = \frac{p_0 v e^{-av}}{R}

So temperature as a function of volume is

T(v)=\frac{p_0}{R} v e^{-av}

We need to maximize

f(v)=v e^{-av}

Differentiate:

\frac{df}{dv}=e^{-av} + v(-a)e^{-av}

\frac{df}{dv}=e^{-av}(1-av)

For maximum,

\frac{df}{dv}=0

Since $e^{-av}\neq 0$ ,

1-av=0

v=\frac{1}{a}

Now,

T_{\max}=\frac{p_0}{R}\left(\frac{1}{a}\right)e^{-1}

T_{\max}=\frac{p_0}{aeR}

T_{\max}=\frac{p_0}{aeR}

This matches Option A.

Stored correct answer: A

My derived answer: A

They agree.