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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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 0X to 0.5X

Solving the equation 1 and 2

P1V1γ=P2V2γ -1

P3V3γ=P4V4γ-2

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 Y. 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=(0.5X-YX)A will get Y=0.137 and use T1V1γ-1=T2V2γ-1 will get T2=499K and T4=284K

Or

Let ∆T=T2+T4-800

Differentiate this equation

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

When d∆T=0

Y=0.137X

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.

nR/γ-1(T1-T3)=-nR/γ-1(T4-T2)

T2+T4=T1+T3=800Kt

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

Is this method correct? I just think that this is wrong because from my calculation the partition will continue moving forward a bit when it reach the same pressure but this is contradict with thermodynamic law which state that partition will stop immediately when both the compartment reach the same pressure.

And I think that to find 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

T1(0.5X/(0.5+Y)X)^0.4+T2(0.5X/(0.5-Y)X)^0.4-800=0

To get the value of Y

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# Homework Help: Can someone tell me which is wright which is wrong answer?

Can you offer guidance or do you also need help?

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