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

(B) $\frac{\mathrm{ap}_{0}}{\mathrm{eR}}$

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Heat & Thermodynamics

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}

v = 0

is the given boundary condition, then the maximum temperature one mole of gas can attain is :
(Here R is the gas constant)

(A) $0^{\circ} C$
(B) $\frac{ap _{0}}{eR}$
(C) $\frac{p _{0}}{a e R}$ ✓ Correct
(D)infinity

Answer: (C)

Solution

Separating variables in

\frac{d p}{d v} = - a p

with

p = p_{0}

v = 0

gives

p = p_{0} e^{- a v}

. For one mole,

p v = R T

, so

T = \frac{p _{0} v e ^{- a v}}{R}

. Setting

\frac{d T}{d v} = \frac{p _{0}}{R} e^{- a v} (1 - a v) = 0

gives

v = \frac{1}{a}

. The maximum temperature is

T_{m a x} = \frac{p _{0}}{R} \cdot \frac{1}{a} \cdot e^{- 1} = \frac{p _{0}}{a e R} .

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option

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