Q37 / 160 · NTA Abhyas (JEE Main) 2020 · Physics · Thermodynamics

A cylindrical adiabatic container of total volume  2 V 0  divided into two equal parts by a conducting piston which is free to move as shown in figure. Each part  containing identical gas at pressure  P 0 . Initially temperature of left and right part is  4 T 0  and  T 0  respectively. An external force is applied on the piston of area ' A ' to keep the piston at rest. The value of external force required when thermal equilibrium is reached is.

Figure

 

  • A
    8 5 P 0 A
  • B
    2 5 P 0 A
  • C
    5 6 P 0 A
  • Correct answer
    D
    6 5 P 0 A

Solution

If final temp is T

P0V04RT0Cv4T0-T=P0V0RT0CvT-T0

4T0-T4=T-T0

4T0-T=4T-4T0

8T0=5T

T=8T05

Final pressure

In left Pf=TfT0Pi=85T04T0P0=25P0

In right Pf=85T0T0P0=85P0

Force=8P05-2P05A=6P0A5

Hence, the correct answer is option: 'D'.

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