CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Figure shows an adiabatic cylindrical tube of volume Vo divided in two parts by a frictionless adiabatic seperator. Initially, the seperator 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 injecteed into the right part. Cp/Cv=γ is the same for both the gases. The seperator is slid slowely and is released at a position where it can stay in equlibrium. 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.

Open in App
Solution

For an adiabatic process, PVγ = Constant

So P1Vγ1=P2Vγ2 ....(i)

According to the problem

V1+V2=Vo

Then the equation (i)

P1Vγ2=P2(VoV1)γ

(P1P2)1γ=VoV1V1

or V1P1γ1=VoP1γ2V1P1γ2

or V1(P1γ1+P1γ2)=VoP1γ2

or V1=P1γ2VoP1γ1+P1γ2

V2=P1γ1VoP1γ1+P1γ2

(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 ressure 'P' when the equlibrium is reached.

P1Vγ1+P2Vγ2=PVγo

For equilibrium, V1=V2=Vo2

Hence,

P1(Vo2)γ+P2(Vo2)γ=P(Vo)γ

or P = P1γ1+P1γ22γ


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Thermodynamic Processes
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon