Figure shows an adiabatic cylindrical tube of volume `(V_0)` divided in two parts by a frictionless adibatic 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.

Explanation

For an adiabatic process, `PV^gamma= Constant`
So `P_1V_1^gamma = P_2V_2^gamma` … (i)
According to the problem
`V_1 + V_2 = V_0`
Then the euation (i)
`P_1V_2^gamma = P_2(V_0-V_1)^gamma`
or `((P_1)/(P_2))^(1/gamma) = (V_0 -V_1)/(V_1)`
or `V_1 -P_1^(1/gamma) = V_0-P_2^(1/gamma)-V_1P_2^(1/gamma)`
or `V_1(P_1^(1/gamma) + P_2 ^(1/gamma)) = V_0 P_2^(1/gamma)`
or `V_1 = (P_2^(1/gamma)V_0)/(P_1^(1/gamma) + P_2^(1/gamma))`
`V_2 = (P_1^(1/gamma)V_0)/(P_1^(1/gamma) + P_2^(1/gamma))`
(b) Since the whole process takes place in adiabatic surroundings, the separator is adiabatic.
Hence heat given to the gas in the left part = 0
(c) There will be a common pressure 'p' when the equilibrium is reached.
`P_1V_1^gamma + P_2V_2^gamma = PV_0^gamma`
For equilibrium, `V_1=V_2= V_0/2`
Hence,
`P_1((V_0)/(2))^gamma + P_2((V_0)/(2))^gamma = P(V_0)^gamma`
`P = ((P_1^(1/gamma) + P_2^(1/gamma))/(2))^gamma`