1.

A piston divides a closed gas cylinder into two parts. Initially the piston is kept pressed such that one part has a pressure P and volume 5Vand the other part has pressure 8P and volume V, the piston is now left free. Find the new pressure and volume for the isothermal and adiabatic process.

Answer»

Solution :Final pressure will be same on both sides. Let it be `P^(.)` , with VOLUME `V^(.)` , in the left side and `(6V-V^(.))` in the right side.
(A) If the CHANGE is isothermal :
For the gas enclosed in the left chamber,
`8PxxV=P^(.)(6V-V)^(.) "…….." (2)`
Solving these for `V^(.) and P^(.)`, we get `V^(.)=30/13V and P^(.) = 13/6 P and (6V-V^(.))=48/13V`

(B) If the change is adiabatic : For the gas in the left chamber.
`P(5V)^(gamma)=P^(.)(V^(.))^(gamma) "............" (3)`
and for the gas in the right chamber, `8P(V)^(gamma)=P^(.)(6V-V^(.))^(gamma) "........" (4)`
Dividing (4) by (3),
`[(6V-V^(.))/V^(.)]^(3//2)=8/5^(3//2)or (6V)/V^(.)=1+4/5 [:. gamma=3/2]`
i.e., `V^(.)=10/3V`
Substituting it in EQN. (3),
`P^(.)=P[(5Vxx3)/(10V)]^(3//2)=(3sqrt3)/(2sqrt2)P=1.84P`
So `P^(.) = 1.84P, V^(.)=10/3 VAND (6V-V^(.))=8/3V` .


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