Thermodynamics

JEE Main level, single correct, medium

158 of 190

Question 158 of 190

nn moles of an ideal gas undergoes a process A→BA\rightarrow B as shown in the figure. The maximum temperature of the gas during the process will be:

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Answer: (C) 9P0V04nR\dfrac{9{P}_{0}{V}_{0}}{4nR}

Temperature TT will be maximum where product of PVPV is max equation of line

P=−P0V0V+3P0P=-\dfrac{{P}_{0}}{{V}_{0}}V+3{P}_{0}

Multiplying both sides by VV,

PV=−P0V0V2+3P0VPV=-\dfrac{{P}_{0}}{{V}_{0}}{V}^{2}+3{P}_{0}V

For maxima, d(PV)dV=0\dfrac{d(PV)}{dV}=0

⇒−2P0V0V+3P0=0⇒V=3V02\Rightarrow \dfrac{-2{P}_{0}}{{V}_{0}}V+3{P}_{0}=0 \Rightarrow V=\dfrac{3{V}_{0}}{2}

andP=3P02\text{and}P=\dfrac{3{P}_{0}}{2}

⇒T=PVnR=9P0V04nR\Rightarrow T=\dfrac{PV}{nR}=\dfrac{9{P}_{0}{V}_{0}}{4 nR}

Alternate method :

Since initial and final temperature are equal.

Hence maximum temperature is at middle of line.

PV=nRTmax⁡PV=nR{T}_{\max }

⇒Tmax⁡=(3P02)(3V02)nR=9P0V04nR\Rightarrow {T}_{\max }=\dfrac{\left(\dfrac{3{P}_{0}}{2}\right)\left(\dfrac{3{V}_{0}}{2}\right)}{nR}=\dfrac{9{P}_{0}{V}_{0}}{4nR}

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