Question

Figure shows an adiabatic cylindrical tube of volume `(V_0)` divided in two parts by a frictionless adiabatic separator . Initially, the separator is kept in the middle, an ideal gas at pressure `(P_1)` and the temperatures `(T_1)` is injected into the left part and the another ideal gas at pressures `(P_2)` and temperature `(T_2)` is injected into the right part. `(C_p / C_v = gamma)` is the same for both the gases. The separator is slid slowly and is released at a position where it can stay in equilibrium. Find
(a) the volumes of the parts,
(b) the heat given to the gas in the left part and
(c) the final common pressure of the gases.

Answer 1

In equilibrium, pressure in two chambers are equal. Since no exchange of heat between two chambers or between vessel and surrounding, therefore adiabatic process in to chambers.
(a) chamber 1 `P_1(V_(0)//2)^(gamma)=PV_1^(gamma)` .(i)
chamber 2, `P_2(V_(0)//2)^(gamma)=PV_2^(gamma)`
`(i)//(ii)impliesP_1P_2=((V_1)/(V_2))^(gamma)`
`(V_1)/(V_2)=((P_1)/(P_2))^((1)/(gamma))impliesV_1=((P_1)/(P_2))^((1)/(gamma))V_2`
`V_1+V_2=V_0`
`((P_1)/(P_2))^((1)/(gamma))V_2+V_2=V_0`
`V_2=(V_0)/(((P_1)/(P_2))^(gamma)+1)=((P_2)^(1//gamma)V_0)/((P_1)^((1)/(gamma))+(P_2)^(1//gamma)`
`V_1=((P_1)/(P_2))^(1//gamma)`,`V_2=((P_1)^(1//gamma)V_0)/((P_1)^(1//gamma)+(P_2)^(1//gamma)`
(b) In adiabatic process `DeltaQ=0`
(c) `P_1((V_0)/(2))^(gamma)=PV_1^(gamma)`
`P=(P_1((V_0)/(2))^(gamma))/(V_1^(gamma))=(P_1((V_0)/(2))^(gamma))/((P_1V_0^(gamma))/([(P_1)^((1)/(gamma))+(P_2)^((1)/(gamma))]^(gamma)))`
`=[((P_1)^((1)/(gamma))+(P_2)^((1)/(gamma)))/(2)]^(gamma)`