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Following figure shows on adiabatic cyli...

Following figure shows on adiabatic cylindrical container of volume `V_(0)` divided by an adiabatic smooth piston (area of cross-section = A ) in two equal parts. An ideal gas `(C_(p)//C_(y)=lambda)` is at pressure `P_1` and temperature `T_1` in left part and gas at pressure `P_2` and temperature `T_2` in right part. The piston is slowly displaced and released at a position where it can stay in equilibrium. The final pressure of the two parts will be (Suppose x = displacement of the piston)

A

`P_2`

B

`P_1`

C

`(P_1((V_0)/(2))^gamma)/(((V_0)/(2)+Ax)^(gamma))`

D

`(P_2((V_0)/(2))^gamma)/(((V_0)/(2)+Ax)^(gamma))`

Text Solution

Verified by Experts

The correct Answer is:
C
Final volume of the left and right part
`V_L =(V_0)/(2) + Ax`
`V_R=(V_0)/(2)-Ax`
Since, container wall and piston are adiabatic in left side and gas undergoes adiabatic expansion on right side the gas undergoes adiabatic compression.
`P_1((V_0)/(2))^(gamma)=P_f((V_0)/(2)+Ax)^(gamma)` (for left side)
`P_2((V_0)/(2))^(gamma)=P_f((V_0)/(2)-Ax)^(gamma)` (for right side)
`:. (P_1)/(P_2) = (((V_0)/(2)+Ax)^(gamma))/(((V_0)/(2)-Ax)^(gamma)) implies Ax = (V_0)/(2)([P_1^(1//gamma)-P_2^(1//gamma)])/([P_1^(1//gamma)+P_2^(1//gamma)])`
Now, `P_f =(P_1((V_0)/(2))^(gamma))/([(V_0)/(2)+Ax]^(gamma)) `
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