# Ideal gas system with diathermic piston externally manipulated

An ideal monatomic gas is separated into two volumes $V_{1}$ and $V_{2}$ through a diathermic piston, such that each volume containing $N$ atoms and the two sides are at the same temperature $T_{0}$. The entire system is isolated from the outside by means of insulating walls.

The plunger is reversibly externally manipulated until the two gases are in thermodynamic equilibrium with each other.

After the problem has different questions, however, I have not been able to answer the following:

How is called the type of process? Ie is isothermal, adiabatic, etc. Show that $\Delta S_{1}$ = $-\Delta S_{2}$, where $\Delta S_{1}$ and $\Delta S_{2}$ are the changes in entropy of the two gases.

• "The entire system is isolated from the outside by means of diathermic walls." This seems like a contradiction to me. "Diathermic" means freely passing heat. I'm guessing that the author intended to say insulted walls instead of diathermic walls. – DavePhD Mar 24 '14 at 0:55
• Ohh, sorry, are insulating walls – Santi Carmesí Mar 24 '14 at 2:03

We know that for reversible processes in isolated systems the entropy remains constant. You have mentioned that the system is isolated and that the plunger is reversibly manipulated. So considering the 2 gases as one isolated system we get that the entropy of the system is a constant. So $\Delta$$S = 0 Now since there is a diathermic piston separating the gases, it means that one gas in itself is not an isolated system, because heat transfer can take place through diathermic piston. Thus considering each gas, there will be a non-zero change in entropy for each. So \Delta$$S(1)$ + $\Delta$$S(2) = \Delta$$S$ = $0$ Thus you get your result.