Question

An ideal monoatomic gas is separated into two volumes V1 and V2 by means of
of a diathermic piston, such that each volume contains N atoms and both parts are
they find at the same temperature T0. The complete system is isolated from the
exterior by means of insulating walls.
The piston is externally manipulated reversibly until the two gases are
they find in thermodynamic equilibrium one with the other.
The purpose is to find the final temperature and the work done, or by the system, or
in the surroundings.
To answer this problem, follow the steps described below and answer
the questions that are asked:
a) What is the name of the type of process described? Show that
ΔS1 = -ΔS2
where ΔS1 and ΔS2 are the changes in the entropies of the two gases.
b) Write the equilibrium conditions in the final state and find the final volumes
On each side.
c) Find the final temperature. (Hint: Use the First Law on each side and the
result of subsection a)).
d) Find the total work due to the manipulation of the piston. Does the system do work
Or does the external agent do work on the system? Explain

Answer 1

a) The process takes place in a reversible way while being thermally isolated from its surrounding (and not heat transfer into the system). This is by definition adiabatic process. We are assuming a massless piston here.

This means heat moves from one chamber to the other only via the diathermic wall. that is:

Having the differentials are defined appropriately, dived both sides by instantaneous T. They will be same on both sides since the wall conducts heat, We reach the definition of entropy, i.e.:

This, upon integration, gives your result.

b) You have the same gas of equal amount on both sides, the walls conduct heat and the piston makes sure the pressures stay same on both sides so as to reach the final thermal equilibrium. So, since P, n, R, T are same pn both sides, you have to have same V on both sides too such that V on each sides would be .

c) It is not clear from your question, but I am assuming you had temperature on both sides to start with. Upon reaching equilibrium, you have equal volumes.

Now using first law: .

Since the temperature did not change throughout and neither did any heat went into the system and the temperature did not change too (since the walls are diathermic) and you started with equal T.

So final temperature: .

d) The manipulation of the system extracted a positive work out of it. Which was entirely due to the equilibration of the pressures.

You can use the formula for work done for adiabatic processes as given here: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html