Question

The figure shows an adiabatic cylindrical tube of volume V0 divided in two parts by a frictionless adiabatic separator. Initially, the separator is kept in the middle, an ideal gas at pressure p1 and temperature T1 is injected into the left part and another ideal gas at pressure p2 and temperature T2 is injected into the right part. Cp/Cv = 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.
Figure

Answer 1

Volume of the two parts is = Vo P1^1/γ / P1^1/γ + P2^1/γ, heat given to the gas in the left part is 0, final common pressure of the gases is P   = ( P1/γ1 + P1/γ2/2 )^γo

Explanation:

For Adiabatic process:

PV^γ = constant

(a) P1V1 = P2V2

V1 + V2 = Vo

V2 =  Vo - V1

P1V1^γ  =  P2( Vo - V1)^γ

(P1/P2)^1/γ  =  Vo - V1/ V1

V1P1^1/γ =  Vo P2^1/γ - V1P2^1/γ

V1( P1^1/γ + P2^1/γ) =  VoP2^1/γ

V1 =  VoP2^1/γ / P1^1/γ + P2^1/γ

V2 = Vo - V1 = Vo P1^1/γ / P1^1/γ + P2^1/γ

(b) Adiabatic Wall + Adiabatic separator

Heat given to the left part is 0

(c) P1V1^γ + P2V2^γ = PVo^γ

For equilibrium:

V1 = V2 = Vo/2

P1 (Vo/2)^γ +  P2 (Vo/2)^γ =  PVo^γ

P = P1/2^γ + P2/2^γ

  = ( P1/γ1 + P1/γ2/2 )^γo

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Answer 2

a) The volume of the two parts are $V_{1}=\frac{V_{0} P_{2} \bar{\gamma}}{P_{1}^{\frac{1}{\gamma}}+P_{2}^{\frac{1}{\gamma}}}$and  $V_{2}=\frac{V_{0} P_{1} \bar{\gamma}}{P_{1}^{\frac{1}{\gamma}}+P_{2}^{\frac{1}{\gamma}}}$  

b) The heat given to the gas in the left part is zero

c) The final common pressure of the gases is $\left(\frac{P_{1}^{\frac{1}{\gamma}}+P_{2}^{\frac{1}{\gamma}}}{2}\right)^{\gamma}$

Explanation:

Given Data

Volume of adiabatic cylindrical tube = $V_0$

$Pressure = P_1 , P_2$

$Temperature = T_1, T_2$

(a) For an adiabatic method,

$PV^{\gamma}= Constant$

So, $P1V1^{\gamma} = P2V2^{\gamma}$   ...(i)

According to the question

$V_1 + V_2 = V_0$      ...(ii)

Using the equation (ii) relation in equation (ii) we obtain

$P1V1^{\gamma} = P2(V0 - V1)^{\gamma}$

Or  

$\left(\frac{P_{1}}{P_{2}}\right)^{1 / \gamma}=\frac{V_{0}-V_{1}}{V_{1}}$

$V_{1} P_{1}^{\frac{1}{\gamma}}=V_{0} P_{2}^{\frac{1}{\gamma}}-V_{1} P_{2}^{\frac{1}{\gamma}}$

$V_{1}\left(P_{1}^{\frac{1}{\gamma}}+P_{2}^{\frac{1}{\gamma}}\right)=V_{0} P_{2}^{\frac{1}{\gamma}}$

$V_{1}=\frac{V_{0} P_{2} \bar{\gamma}}{P_{1}^{\frac{1}{\gamma}}+P_{2}^{\frac{1}{\gamma}}}$

Using equation (ii), we get  

$V_{2}=\frac{V_{0} P_{1} \bar{\gamma}}{P_{1}^{\frac{1}{\gamma}}+P_{2}^{\frac{1}{\gamma}}}$

(b) The separator is adiabatic, since the whole process takes place in adiabatic surroundings.  

Therefore, in the left part heat given to the gas is 0.

(c) Once equilibrium is reached there will be a specific pressure ' P'.  The slid moves until the pressure on

$\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}+\mathrm{P}_{2} \mathrm{V}_{1}^{\gamma}=\mathrm{PV}_{1}^{\gamma}$

$V_{1}=\frac{V_{0}}{2}$

$P_{1}\left(\frac{V_{0}}{2}\right)^{\gamma}+P_{2}\left(\frac{V_{0}}{2}\right)^{\gamma}=P_{1}\left(V_{0}\right)^{\gamma}$

$P=\left(\frac{P_{1}^{\frac{1}{\gamma}}+P_{2}^{\frac{1}{\gamma}}}{2}\right)^{\gamma}$

Thus the volume , final common pressure and amount of heat in the part has been found.