One mole of an ideal gas passes through a process where pressure and volume obey the relation P = P _ o [1 - (1)/(2) ( V _ o V ) ^ 2 ] . Here P _ o and V _ o are constants. Calculate the change in the temperature of the

Correct answer: B. (5)/(4) P _ 0 V _ 0 R For ideal gas, PV = nRT Therefore, (nRT)/(V) = P _ 0 [1 - (1)/(2) ( V _ 0 V ) ^ 2 ] ⇒ T = P _ 0 V R [1 - (1)/(2) ( V _ 0 V ) ^ 2 ] Therefore, the change in temperature is: = T _ f - T _ i = P _ 0 ( 2V _ 0 ) R (1 - (1)/(2) ( V _ 0 2V _ 0 ) ^ 2 ) - P _ 0 V R (1 - V _ 0 ^ 2 2 V _ 0 ^ 2 ) = 5P _ 0 V _ 0 4R

JEE Main 201910 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 10 April
Shift: Evening Shift
Subject: Physics
Chapter: Thermodynamics

JEE Main 2019 · 10 April · Evening Shift · Physics · Thermodynamics

One mole of an ideal gas passes through a process where pressure...

One mole of an ideal gas passes through a process where pressure and volume obey the relation

P = {P}_{o}

\left[1 - \dfrac{1}{2}{

\left(\dfrac{{V}_{o}}{V}\right)}^{2}\right]

. Here

{P}_{o}

and

{V}_{o}

are constants. Calculate the change in the temperature of the gas if its volume changes from

{V}_{o}

2{V}_{o}

Options

A $\dfrac{1}{4}\dfrac{{P}_{0}{V}_{0}}{R}$
B $\dfrac{5}{4}\dfrac{{P}_{0}{V}_{0}}{R}$
C $\dfrac{1}{2}\dfrac{{P}_{0}{V}_{0}}{R}$
D $\dfrac{3}{4}\dfrac{{P}_{0}{V}_{0}}{R}$

Correct answer

B. $\dfrac{5}{4}\dfrac{{P}_{0}{V}_{0}}{R}$

Step-by-step solution

For ideal gas,

PV = nRT

Therefore,

\dfrac{nRT}{V} = {P}_{0}

\left[1 - \dfrac{1}{2}{

\left(\dfrac{{V}_{0}}{V}\right)}^{2}\right]

⇒ T = \dfrac{{P}_{0}V}{R}

\left[1 - \dfrac{1}{2}{

\left(\dfrac{{V}_{0}}{V}\right)}^{2}\right]

Therefore, the change in temperature is:

\DeltaT = {T}_{f} - {T}_{i}

= \dfrac{{P}_{0}\left({2V}_{0}\right)}

{R}

\left(1 - \dfrac{1}{2}{\left(\dfrac{{V}_{0}}{{2V}_{0}}\right)}^{2}\right) - \dfrac{{P}_{0}V}{R}\left(1 - \dfrac{{V}_{0}^{2}}{2{V}_{0}^{2}}\right)

= \dfrac{{5P}_{0}{V}_{0}}{4R}

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