Solving Adiabatic Process for Two Compartments

[chuakoktong]

TL;DR

Adiabatic process

I tried this question and this the answer given in 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 is 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
m be the percentage of the cylinder length that move in the process
A be the area of partition

Solving the equation 1 and 2
$P_1V_1^{\gamma}=P_2V_2^{\gamma} -1$
$P_3V_3^{\gamma}=P_4V_4^{\gamma}-2$

(By using PV=nRT)
Will get the ration $V_2=1.756V_4$ or $T_2=1.756T_4$

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

$(T_1-T_3)=- (T_4-T_2)$
$T_2+T_4=T_1+T_3=800K$

And substitute $T_2=1.756T_4$ into the equation will get $T_2=510K$ and $T_4=290K$

However, i think this answer is not correct?

The equation should be as below since it not possible to undergo adiabatic expansion on one side and adiabatic process on the other side?
The correct one should be adiabatic compression on the right partition?

$2.5p_1v_1-(2.5p_4v_4-2.5p_3v_3)=2.5p_2v_2$
$p_1v_1-(p_4v_4-p_3v_3)=p_2v_2$
$nR(550)-[nR(250)(\frac{0.5}{0.5-m})^{\gamma-1}-nR(250)]=p_2(0.5+m)XA$

and

$p_3v_3^{\gamma}=p_4v_4^{\gamma}$
$nR(250)(0.5XA)^{\gamma-1}=p_4(0.5XA-mXA)^{\gamma}$

$p_4=p_2$

$[550-250(\frac{0.5}{0.5-m})^{\gamma-1}+250](\frac{0.5}{0.5+m})(\frac{0.5}{0.5-m})^{-\gamma}=250$

solve the value of m to get the pressure on both side and hence temperature on both side?

 

[Chestermiller]

If the partition is "allowed to move slowly", what they meant to say is that it is "forced or constrained to move slowly." To accomplish this, you need to exert external force on the partition during its motion. This means that the amounts of work done by the two gases will not be equal and the change in internal energy of the overall system will not be zero.

This is similar to a question addressed in another recent thread: https://www.physicsforums.com/threa...diabatic-container.981529/page-2#post-6274658