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# 10.2 Consider two systems, A and B. each composed of the same single particle type. The two systems arc contained in a chamber surrounded by rigid adiabatic walls and they are separated from each...

10.2 Consider two systems, A and B. each composed of the same single particle type. The two systems arc contained in a chamber surrounded by rigid adiabatic walls and they are separated from each other within the chamber by a rigid diathermal wall which is also permeable to the particles (see Fig. D.12). Show, using an argument similar to the one used in section 10.2, that the condition for equilibrium against particle exchange is the equality of the chemical potentials.
10.3 Consider the system of question 10.2. Suppose that the two systems, composed of the same single type of particle, are both in the same phase, e.g. a gas on each side of the separating wall. Show that the pressures arc equal. Would the pressures be equal if different
phases existed on either side of the wall? (Hint: For a system consisting of a single type of particle, 14 = G(T, P, = 0(T, P). If the phases are the same on either side of the wall, the function (/) must also be the same on either side.)

## Solution

Robert answered on Dec 31 2021
Problem 10.2
Consider two systems A and B each composed of the same single particle typr. The two systems are contained in a
chamber su
ounded by rigid adiabatic walls and they are searated from each other withing the chamber by a rigid
diathermal wall which is also permeable to the particles. Show using an argument similar to the one used in section
10.2, that the condition for equili
ium against partcile exchange is the euality of the chemical potential.
Solution
Rigid adiabatic wall: A wall which doesnot allow the heat exchange throug it.
Diathermic wall: A wall which allows the heat exchange through it.
Since the above question is based on single particle system, hence we can take an example of ice melting in water
then there will be only one type of particles that is of water only.
Let us suppose there are two type of phases A and B as shown in the diagram of the same substance that is water as
we have considered before.
Since the two phases are separated by the diathermic wall then they will be in thermal equili
ium with each other
hence we can take
TA = TB = T ( In thermal equili
ium )
Let the two phases have separate volume VA and VB and are seprated by a diathermic wall which is also permeable
then there can be heat and particle exchange across the walls. Let there be NA particles in pahse A having internal
energy UA and pressure PA and similarly for phase B let NB , UB and PB are the number of particles , internal energy
and pressure respectively. Hence
N = NA + NB ( total no of Particles)
V = VA + VB ( total Volume)
UA + UB = U ( total internal energy)
Since there will be no exchange of heat with the su
oundings then internal energy will remain constant.
Now the total entropy for the combined system will be
S = SA + SB
Where S is a function of U , N and V of the system.
Now from the second law of thermodynamics we know that enetropy should be maximum
So SA + SB = maximum.
As all the above mentioned quantities will be change by a very infinitesimal amount hence we...
SOLUTION.PDF