# Equilibrium equation if the barrier allows particle exchange

• I
• Kashmir
In summary: It doesn't matter if there's actually a physical wall there or not.In summary, two physical systems, A1 and A2, are separated by a wall. A1 has ##\Omega_{1}(N1,V1,E1)## possible microstates, and the macrostate of A2 is ##\Omega_{2}(N2,V2,E2)##. At any time ##t##, the subsystem ##A_{1}## is equally likely to be in anyone of the ##\Omega_{1}\left(E_{1}\right)## microstates while the subsystem ##A_{2}## is equally likely to be in anyone of the ##\Omega_{2}\left(E_{2}\right
Kashmir

"... two physical systems [seperated by wall], A1 and A2. A1 has ##\Omega_{1}(N1,V1,E1)## possible microstates, and the macrostate of A2 is ##\Omega_{2}(N2,V2,E2)## "

"... at any time ##t##, the subsystem ##A_{1}## is equally likely to be in anyone of the ##\Omega_{1}\left(E_{1}\right)## microstates while the subsystem ##A_{2}## is equally likely to be in anyone of the ##\Omega_{2}\left(E_{2}\right)## microstates; therefore, the composite system ##A^{(0)}## is equally likely to be in anyone of the
##
\Omega_{1}\left(E_{1}\right) \Omega_{2}\left(E_{2}\right)=\Omega_{1}\left(E_{1}\right) \Omega_{2}\left(E^{(0)}-E_{1}\right)=\Omega^{(0)}\left(E^{(0)}, E_{1}\right)
##"
"... if A1 and A2 came into contact through a wall that allowed an exchange of particles as well, the conditions for equilibrium would [include] the equality of the parameter ##\zeta_{1}## of subsystem ##A_{1}## and the parameter ##\zeta_{2}## of subsystem ##A_{2}## where, by definition,
##
\zeta \equiv\left(\frac{\partial \ln \Omega(N, V, E)}{\partial N}\right)_{V, E, N=\bar{N}}
##"

• So if we've a wall that allowed an exchange of particles we have from above equation:

##
\left(\frac{\partial \ln \Omega_1(N_1, V_1, E_1)}{\partial N_1}\right)_{V_1, E_1, N=\bar{N}}
=\left(\frac{\partial \ln \Omega_2(N_2, V_2, E_2)}{\partial N_2}\right)_{V_2, E_2, N=\bar{N}}
##

However having a wall that allows particles to be exchanged means no wall at all, then ##V_1,V_2## are not well defined, but the above equation uses them?

Kashmir said:
• So if we've a wall that allowed an exchange of particles we have from above equation:

##
\left(\frac{\partial \ln \Omega_1(N_1, V_1, E_1)}{\partial N_1}\right)_{V_1, E_1, N=\bar{N}}
=\left(\frac{\partial \ln \Omega_2(N_2, V_2, E_2)}{\partial N_2}\right)_{V_2, E_2, N=\bar{N}}
##
Does that equality come from the textbook?

Kashmir said:
However having a wall that allows particles to be exchanged means no wall at all, then ##V_1,V_2## are not well defined, but the above equation uses them?
You can still conceptually consider a fixed partitioning of the full volume, even if there is no physical partition.

In this case of a grand-canonical ensemble the equilibrium condition is that both ##T## and ##\mu## are equal, and that's expressed above using the microcanonical description.

DrClaude said:
Does that equality come from the textbook?
Yes.
DrClaude said:
You can still conceptually consider a fixed partitioning of the full volume, even if there is no physical partition.
You mean ##V_1= V_0## where ##V_0## is total volume ?

vanhees71 said:
In this case of a grand-canonical ensemble the equilibrium condition is that both ##T## and ##\mu## are equal, and that's expressed above using the microcanonical description.
Yes this an condition of equilibrium but the way the author derives it is confusing.

Kashmir said:
You mean ##V_1= V_0## where ##V_0## is total volume ?
No, I mean that you can split ##V_0## into ##V_1 + V_2##, even if there is no actual physical partition.

DrClaude said:
No, I mean that you can split ##V_0## into ##V_1 + V_2##, even if there is no actual physical partition.
##V_1## was defined as the volume separated by a wall in which ##N_1## particles were present.

Now if we lift the wall then how will
##V_1## be defined?

Kashmir said:
##V_1## was defined as the volume separated by a wall in which ##N_1## particles were present.

Now if we lift the wall then how will
##V_1## be defined?
##V_1## is still the same: it is the volume that used to be constrained by the wall. If you want, you can imagine that someone drew a line across the container, and call the space on one side of the line ##V_1##, and the space on the other side ##V_2##.

## 1. What is the equilibrium equation if the barrier allows particle exchange?

The equilibrium equation for a system with a barrier that allows particle exchange is known as the Gibbs-Duhem equation. It is given by ∑i nii = 0, where ni represents the number of particles of type i and μi represents the chemical potential of type i.

## 2. How is the equilibrium equation derived for a system with a barrier?

The Gibbs-Duhem equation can be derived from the fundamental thermodynamic relation, dU = TdS - PdV + ∑iμidni, where U is the internal energy, T is the temperature, S is the entropy, P is the pressure, and μi is the chemical potential of type i. By setting dU = 0 and rearranging the terms, we can obtain the equilibrium equation.

## 3. What does the equilibrium equation tell us about a system with a barrier?

The equilibrium equation for a system with a barrier tells us that at equilibrium, the chemical potentials of all types of particles must be equal. This means that the system has reached a state where there is no net transfer of particles between the two sides of the barrier, and the system is in a stable state.

## 4. How does the equilibrium equation change if the barrier is removed?

If the barrier is removed, the equilibrium equation simplifies to the Gibbs-Duhem equation for an open system, which is given by ∑i nii = VdP, where V is the volume and P is the pressure. This equation tells us that at equilibrium, the chemical potentials of all types of particles must be equal, and the pressure must also be constant.

## 5. Can the equilibrium equation be applied to any system with a barrier?

Yes, the equilibrium equation can be applied to any system with a barrier that allows particle exchange. This equation is a fundamental thermodynamic relation and is applicable to a wide range of systems, including gases, liquids, and solids. However, the specific form of the equation may vary depending on the properties of the system and the type of barrier present.

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