# Find entropy change when two tanks equalize

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1. Mar 25, 2016

### MattHorbacz

1. The problem statement, all variables and given/known data
Two rigid, insulated tanks are connected with a pipe and valve. One tank has 0.5 kg air at 200 kPa, 300 K and the other has 0.75 kg air at 100 kPa, 400 K. The valve is opened, and the air comes to a single uniform state without any heat transfer. Find the final temperature and the change in entropy of the air.

2. Relevant equations
ds=Cv*ln(T2/T1)+R*ln(v2/v1)
PV=nRT
m1cvΔT1=-m2cvΔT2
????

3. The attempt at a solution
I am pretty lost. I solved for the total volume, mass, and specific volume, but have no idea how to solve for temperature. And I thought that there was no change in entropy without heat transfer. Even showing me what equations to use would be extremely helpful

Edit: just remembered that q_tank1=-q_tank2: m1c_vΔT1=-m2c_vΔT2. Solved for final temp to be 360 K

Last edited: Mar 25, 2016
2. Mar 25, 2016

### DrDu

$\Delta S= Q/T$ only holds for reversible processes. Your process is irreversible, so there may be a change in entropy although there is no heat exchange. Namely in your problem, the process is irreversible as the two containers have different temperature. But you may calculate the change of entropy of each of the containers as taken on their own they have a uniform temperature.

3. Mar 25, 2016

### MattHorbacz

I'm not sure if this was correct, but I calculated two ΔS, tank1 initial vs total and also tank2 initial vs total. I used the equation I listed above, then apparently you multiply each value for entropy by the mass then add the values...I think what I did was the right method, but I most definately messed up somewhere. Is this sort of the right method?

4. Mar 25, 2016

### DrDu

Sounds good to me.

5. Mar 26, 2016

### Staff: Mentor

The way to approach this problem is to recognize that, since the total volume of the two tanks does not change and no heat enters or leaves the combined system, there is no work done on the combined system, and there is no heat transfer to or from the system. This means that the overall change in internal energy is zero. Also, the final pressures in the two tanks must match. This is enough information to determine the final temperature and pressure. Once you know these, determining the change in entropy is straightforward.