Thermodynamics: 2 gases in 2 chambers

1. Dec 2, 2014

Karol

1. The problem statement, all variables and given/known data
A stiff and isolated vessel is divided into 2 halves by an isolator divider. in each half there is a gas (i am not told if it's the same) with presures and temperatures T1, P1 and T2, P2. the divider is removed. Apply the first law of thermodynamics and draw conclusions.

2. Relevant equations
The first law: $Q=U_2-U_1+W$
Work: $W=c\cdot v$

3. The attempt at a solution
I don't know how to calculate internal energy U from temperature T and pressure P, but i am not sure it's needed.
If i choose the system as the 2 gases together, then there is no work done and no heat transfer so the initial and final internal energies equal: U2=U1, but i don't know to deduce from that the final temperature Tf and pressure Pf.
If i choose each chamber as a system, and i have 2 systems, i don't know if the gas expansion from each system into the other chamber is as if in the other chamber is vaccum or if there is work done by the expanshion against the pressure of the other gas, and i don't know if i am allowed to take the pressure for the work as the the mean of the pressure of the other gas and the final, unknown pressure Pf.
So i don't know if i am allowed to treat each gas as if the other one doesn't exist, so to use the concept of partial pressure, but i wasn't taught this topic of partial pressures so i guess it's not relevant
If i use 2 separate systems which expand, make work and transfer heat to one another, and also apply the single system approach, i know that the heat transferred and the work each system does is equal to that the other absorbes, but i don't advance into final conclusions for Tf and Pf.

2. Dec 2, 2014

Staff: Mentor

There are various approaches you can use to attack this problem, but your first approach is the simplest and, in my judgement, is the preferred approach (i.e., the change in internal energy of the combined system is zero). I'm assuming you are dealing with ideal gases at this stage. For a pure ideal gas, you may have learned that the change in internal energy is given by ΔU=nCvΔT, where n is the number of moles, Cv is the molar heat capacity at constant volume, and ΔT is the change in temperature. For a mixture of ideal gases, the internal energy is the sum of the internal energies of the pure gases at the same temperature.

Chet

3. Dec 2, 2014

Karol

The book teaches ΔU=nCvΔT after this chapter of the first law. is it possible to solve Tf and Pf without that?

4. Dec 2, 2014

No.

5. Dec 2, 2014

Karol

So i am left with the conclusion that the only conclusion i can make is that ΔU=0 and nothing more about Tf and Pf?

6. Dec 2, 2014

Staff: Mentor

I guess that's right. But still, that's a pretty powerful conclusion.

Chet