Two cylinders are connected by a fixed diathermic partition `A` and a removable adiabatic partition `B` is placed adjacent to `A` as shown in the figure. Inititally n moles of an ideal monoatomic gas in present in both the cylinders an normal atmospheric pressure `p_(0)` . Both the gases occupy same volume `V_(0)` , initially. Now the piston of the left cylinder is compressed in adiabatic manner so that volume of the lift portion becomes `(V_(0))/(2)` and then the left piston clamped. Again the adiabatic slider `B` is removed so equilibrium. Assume all other surfaces except `A` to be adiabatic. For this situation, mark out the correct statement (s). A. Just after the removal of adiabatic separator `B` , the pressure in the lift and right chambers are `2^(gamma) p_(0)` and `p_(0)` , respectively.B. `A` fter the removal of adiabatic separtor `B` , the gas in right chamber expands under constant pressure process.C. Workdone by the gas of the right chamber on surroundings during its expansion is `0.22p_(0)V_(0)` .D. During the expansion of gas in right chamber, the energy transferred from the left chamber to right chamber is `0.55 p_(0)V_(0)` where `gamma=(5)/(3)` .

Two cylinders are connected by a fixed diathermic partition `A` and a

1.	Two cylinders are connected by a fixed diathermic partition `A` and a removable adiabatic partition `B` is placed adjacent to `A` as shown in the figure. Inititally n moles of an ideal monoatomic gas in present in both the cylinders an normal atmospheric pressure `p_(0)` . Both the gases occupy same volume `V_(0)` , initially. Now the piston of the left cylinder is compressed in adiabatic manner so that volume of the lift portion becomes `(V_(0))/(2)` and then the left piston clamped. Again the adiabatic slider `B` is removed so equilibrium. Assume all other surfaces except `A` to be adiabatic. For this situation, mark out the correct statement (s). A. Just after the removal of adiabatic separator `B` , the pressure in the lift and right chambers are `2^(gamma) p_(0)` and `p_(0)` , respectively.B. `A` fter the removal of adiabatic separtor `B` , the gas in right chamber expands under constant pressure process.C. Workdone by the gas of the right chamber on surroundings during its expansion is `0.22p_(0)V_(0)` .D. During the expansion of gas in right chamber, the energy transferred from the left chamber to right chamber is `0.55 p_(0)V_(0)` where `gamma=(5)/(3)` .
Answer» Correct Answer - A::B::C::D As the gas of the left chamber is compressed adiabatically to half to its initial volume. `p_(0)V_(0)=p_(1)(V_(0)//2)^(gamma) , impliesp_(1)=p_(0)xx2^(gamma)` i.e., pressure in the left chamber after the completion of adiabatic process is `p_(1)` while pressure in the right chamber remains same as it is isolated from surrondings or you say no energy transfer berween the right chamber and the surroundings occur. Due to this adiabatic compression, temperature will also change in the left chamber. `T_(0)V_(0)^(gamma-1)=T_(1)(V_(0)//2)^(gamma-1) , impliesT_(1)=2^(gamma-1)T_(0)` Where `T_(0)=(p_(0)V_(0))/(nR)` When adiabatic separator is removed, temperature on both sides are different, temperature on left is `T_(1)` while temperature on right is `Y_(0)` , so heat transfer takes place from left chamber to right chamber and the gas in cahmber starts expanding, as piston (right side) is free to move, the pressure remains constant as `p_(0)` , So , expansion in right chamber is taking place at constant temperature , During this expansion of gas in right chamber. the volume of gas in left chamber remains constant but temperature and pressure change, For the instant just a fter the removal of adiabatic separator we can take the pessure of gas in left chamber as the same just before the removal i.e., `p_(0)xx2^(gamma)`. Let the final temperature on both the side be `T`, then the energy transferred from left to right is used to incease the temperature of the tight part and to expand the gas. i.e., `nC_(v)(T_(1)-T)=nC_(p)(T-T_(0))` `=nC_(V)(T-T_(0))+W` `impliesT=(5T_(0)+3T_(1))/(8)` `=((5+3xx2^(gamma-1))/(8))T_(0)~=1.22T_(0)` So, the energy transferred is, `Q=nC_(V)(T_(1)-T)=(3nR)/(2)(2^(gamma-1)-1.22)T_(0)` `=0.55 p_(0)V_(0)` Work done, `W=p_(0)(V_(f-V_(i))=nR(T-T_(0))` `=0.22nRT_(0)~=0.22p_(0)V_(0)`

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