# Just requiring understanding of thermodynamics solution

1. Apr 7, 2017

### matthewsyq1995

1. The problem statement, all variables and given/known data
A vessel divided into 2 parts by a partition, contains 4 mol of nitrogen gas at 348.15K and 30 bar on one side and 2.5mol of argon gas at 130C and 20bar on the other. If the partition is removedand the gases mixes adiabatically, and comletely, what is the change in entropy?

2. Relevant equations
Assuming Nitrogen is an ideal gas with Cv = (5R/2)
Assuming argon is an ideal gas with Cv = (3R/2)

3. The attempt at a solution
Ok basically i have the solution but i cant seem to understand the starting part where he got the final temperature. the professor told us to 1st bring the individual stream to mixture T and P. But i don't understand how he got the final temperature Tf.

How he did it was by equating
U(N2) + U(Ar) = 0
Then,
4 x (5R/2) x (Tf - 348.15) = 2.5 x (3R/2) x (403.15 - Tf)
Tf = 363.15K

It is my understanding that this method is basically equating the total change in internal energy of the system to be 0. Then using U=nCv(deltaT), he finds Tf.

However, I believe that the change in internal energy to be 0 should only apply for isothermal systems instead of adiabatic.

I understand the rest of his solution to find the entrpy except for this part. Do help me pls :)

2. Apr 7, 2017

### Staff: Mentor

If you take as your system the contents of the entire vessel (i.e., both gases in both compartments), then, since the vessel is rigid, no work is done on the contents. And, since the vessel is insulated, no heat is transferred to the contents. So, since $\Delta U=Q-W$, the change in internal energy of the total contents of the vessel is zero.