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Figure below shows a cylindrical tube of volume V 0 divided in two parts by a frictionless separator. The walls of the tube are adiabatic but the separator is conducting. Ideal gases are filled in the two parts. When the separator is kept in the middle, ten pressure are p 1 and p 2 in the left part and teh right part respectively. The separator os slowly slid and is released at aposition where it can stay in equilibrium. Find the volumes of the two parts.

Figure below shows a cylindrical tube of volume Vdivided in two parts by a frictionless separator. The walls of the tube are adiabatic but the separator is conducting. Ideal gases are filled in the two parts. When the separator is kept in the middle, ten pressure are pand pin the left part and teh right part respectively. The separator os slowly slid and is released at aposition where it can stay in equilibrium. Find the volumes of the two parts.

Question Image
Grade:11

1 Answers

Arun
25763 Points
one year ago
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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