n moles of an ideal gas undergoes a process A → B as shown in the figure. The maximum temperature of…

n moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:
  1. 9 P0V02nR
  2. 9 P0V0nR
  3. 9 P0V04nR
  4. 3 P0V02nR

Solution

Temperature T will be maximum where product of PV is max equation of line

P=−P0V0V+3P0

Multiplying both sides by V,

PV=−P0V0V2+3P0V

For maxima,            d(PV) dV =0

⇒               −2 P 0 V 0 V+3 P 0 =0     ⇒  V= 3 V 0 2

and P=3P02

⇒   T= PV nR = 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=nR T max

⇒Tmax=3P023V02nR=9P0V04nR

Asked in: JEE Main 2016 (03 Apr)

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