How Does Thermal Equilibrium Affect Temperature and Entropy in a Two-Gas System?

[pentaquark]

Homework Statement

A cylindrical container 80cm long is in two compartments separated by piston, clamped at 30cm from left end. The left compartment is filled with 1 mole of He at 5atm. Right compartment is filled with argon at 1atm. Both gases may be considered ideal. Cylinder is submerged in 1L of H20 @ 25°C. When piston is unclamped, a new equilibrium is reached with piston in new position. The entire system is thermally and mechanically isolated and you may ignore the heat capacities of the cylinder and piston.

What is the increase in Temp of the water at new equilibrium? How far from left of the cylinder will the piston come to rest? What is increase in total entropy of the system?

Homework Equations

What I think I can use, are entropic equations regarding the variables, E, T, P, V:

[tex]\frac{dS}{dE}[/tex]=[tex]\frac{1}{T}[/tex]

[tex]\frac{\partial{dS}}{\partial{dV}}[/tex]=[tex]\frac{P}{T}[/tex]

Since the equilibrium must be reached when entropy is max, then dS[tex]_{1}[/tex]+dS[tex]_{2}[/tex]=0

The Attempt at a Solution

Ok, so I'm guessing that we can relate the pressures and temperatures with

[tex]\frac{dS_{1}}{dV_{1}}[/tex]=[tex]\frac{P_{1}}{T}[/tex]

[tex]\frac{dS_{2}}{dV_{2}}[/tex]=[tex]\frac{P_{2}}{T}[/tex]

meaning, that the temps should be equal after equilibrium.

The displacement of the piston would be a ratio of the two volumes at the new equilibrium.

 

[BigJDubs]

For the increase in entropy, I'm guessing it would be the difference between the two entropies at the start and end of the process, i.e. dS=dS_{1}-dS_{2}.