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I have attempt the question.Howeve i had find out some contradiction but i could not understand where is the mistake i made.Below is my working and the answer given by the book

A cylinder is close at both ends and has insulating walls. It is divided into two compartments by a perfectly insulating partition that is perpendicular to the axis of the cylinder. Each compartment contains 1.00 mol of oxygen, which behaves as an ideal gas with γ=7/5. Initially the two compartments have equals volumes, and their temperatures are 550k and 250k. The partition is then allowed to move slowly until the pressure on its two sides are equal. Find the final temperatures in the two compartments.


Let P1 be the initial pressure in 1st compartment

      P2 be the final pressure in 1st compartment

      P3 be the initial pressure n 2nd compartment

      P4 be the final pressure in 2nd compartment

      V1 be the initial volume in 1st compartment

      V3 be the initial volume in 2nd compartment

      V2 be the final volume in 1st  compartment

      V4 be the final volume in 2nd compartment

      T1 be the initial temperature in 1st compartment

      T2 be the final temperature in 1st compartment

      T3 be the initial temperature in 2nd compartment

      T4 be the final temperature in 2nd compartment

       X be the length of the cylinder

       Y is the distance move by the partition and is an unknown which must be find and in    

       the range from 0 to 0.5


Solving the equation 1 and 2

P1V1γ=P2V2γ -1



Will get the ration V2=1.756V4 or T2=1.756T4

Than consider the length of the whole compartment be X since initially two compartment same volume, the partition divide the cylinder into 0.5X at both side and consider the distance move by the cylinder be YX. since the partition will move to the right, the volume of the 1st compartment when the partition stop is (0.5X+YX)A where A is the cross section of the partition and for the 2nd compartment will be (0.5X-YX)A solve (0.5X+YX)A=1.756(0.5X-YX)A will get Y=0.137 and use T1V1γ-1=T2V2γ-1 will get T2=499K and T4=284K



Let ?T=T2+T4-800

Differentiate this equation

T1 (0.5XA/(0.5+Y)XA)^0.4+T2(0.5XA/(0.5-Y)XA)^0.4-800=?T

When d(?T)/dT=0



But answer given

Since the work done by the adiabatically expanding gas is equal and opposite to the work done by the adiabatically compress gas.





And substitute T2=1.756T4 into the equation will get T2=510K and T4=290K.


The distance move by the partition until it stop, should be consider this way.


T1 (0.5X/(0.5+Y)X)^0.4+T2(0.5X/(0.5-Y)X)^0.4-800=?T

Partition stop when ?T=0

And solve the equation


To get the value of Y


At Y=0.137 where the pressure on both side is the same, T2+T4=784K but not 800K


   Y              T2+T4

0.000           800.00

0.050           790.18

0.100           784.65

0.130           783.43

0.137            783.38

0.150            783.54

0.200            787.41

0.250            797.57

0.258            800.00

 I had discuss the problem with several teacher but it seem that none of the answer i feel like making sense.

At 1st, i thought it may be due to the cylinder is not a uniform cylinder that why the formula cannot be apply. On 2nd thought, the adiabatic formula should be true for all form no wonder it is uniform or not uniform.

secondly, it can be due to approximation error but i have use scientific key in the formula calculate as accurate as possible still the value remain different. If it is cause by approximation formula can someone prrof it?

Since previously question answered that translational kinetic energy and kinetic energy is different, could the gas particle convert portion of its translational energy to kinetic energy so there is a net movement of gas particle toward the right side? causing the total internal energy in the system appear different(there is a "loss" of internal energy) because the translational kinetic energy is use to do work on the right partition and increase in kinetic energy?From the formula i find that the "loss" in internal energy is maximum when both side is same pressure.

6 years ago


Answers : (1)


Dear student,

There is an error in the differentiation part.Please check again...

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

6 years ago

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