The 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 temperature T_{1} is injected into the left part and another ideal gas at pressure p_{2} and temperature T_{2} is injected into the right part. C_{p}/C_{v} = γ 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 two parts (b) the heat given to the gas in the left part and (c) the final common pressure of the gases.

#### Solution

For an adiabatic process, PV^{γ} = Constant

So, P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ} ...(i)

According to the problem,

V_{1} + V_{2} = V_{0} ...(ii)

Using the relation in eq (ii) in eq (i), we get

P_{1}V_{1}^{γ} = P_{2}(V_{0} − V_{1})^{γ}

Or `("P"_1/"P"_2)^(1/gamma) = ("V"_0 -"V"_1)/"V"_1`

`"V"_1"P"_1^(1/gamma) = "V"_0"P"_2^(1/gamma) - "V"_1"P"_2^(1/gamma)`

`"V"_1( "P"_1^(1/gamma) + "P"_2^(1/gamma)) = "V"_0"P"_2^(1/gamma)`

`"V"_1 = ("P"_2^(1-gamma) "V"_0)/("P"_1^(1/gamma) +"P"_2^(1/gamma))`

Using equation (ii), we get

`"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 is 0.

(c) There will be a common pressure '*P*' when equilibrium is reached. The slid will move until the pressure on the two sides becomes equal.*P _{1}V_{1}^{γ} + P_{2}V_{2}^{γ} = PV_{0}^{γ}*

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`

Or `"P" = (("P" _1^(1/gamma) + "P"_2^(1/gamma))/2)^gamma`