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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. |
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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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