For an ideal gas the instantaneous change in pressure P with volume V is given by the equation (dP)/(dV) = - aP . If P = P _ 0 at V = 0 is the given boundary condition, then the maximum temperature one mole of gas can

Correct answer: D. P _ 0 aeR (dP)/(dV) = - aP ∫ _ P _ 0 ^ P (dP)/(P) = - ∫ _ 0 ^ v adV ⇒ ln P P _ 0 = - aV P = P _ 0 e ^ - aV Now, PV = RT (as n = 1 ) P + V(dP)/(dV) = R(dT)/(dV) [ for maximum T, (dT)/(dV) = 0 ] P - aPV = 0 ⇒ V = (1)/(a) Thus, T = (PV)/(R) = P _ 0 e ^ - 1 Ra = P _ 0 eRa

JEE Main 202131 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 31 August
Shift: Morning Shift
Subject: Physics
Chapter: Kinetic Theory of Gases

JEE Main 2021 · 31 August · Morning Shift · Physics · Kinetic Theory of Gases

For an ideal gas the instantaneous change in pressure P with volume...

For an ideal gas the instantaneous change in pressure

P

with volume

V

is given by the equation

\dfrac{dP}{dV} = - aP .

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)

Options

A $\dfrac{a{P}_{0}}{eR}$
Binfinity
C $0 ° C$
D $\dfrac{{P}_{0}}{aeR}$

Correct answer

D. $\dfrac{{P}_{0}}{aeR}$

Step-by-step solution

\dfrac{dP}{dV} = - aP

{∫}_{{P}_{0}}^{P}\dfrac{dP}{P} = - {∫}_{0}^{v}adV ⇒ ln\dfrac{P}{{P}_{0}} = - aV

P = {P}_{0}{e}^{- aV}

Now,

PV = RT

(as

n = 1

)

P + V\dfrac{dP}{dV} = R\dfrac{dT}{dV}

[ for maximum

T, \dfrac{dT}{dV} = 0

]

P - aPV = 0 ⇒ V = \dfrac{1}{a}

Thus,

T = \dfrac{PV}{R} = \dfrac{{P}_{0}{e}^{- 1}}{Ra} = \dfrac{{P}_{0}}{eRa}

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