Entropy- change & the condition when a system reaches equilibrium

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

I didn't understand the last part.
At eqbm. $\Delta S = 0$. This means that the RHS of the eqnn. 14.25 is 0.
This doesn't mean that the following eqns. must hold true.
$(\frac 1 T_1 - \frac 1 T_2) =0,.............(1) \\ (\frac {p_1}{T_1} - \frac{p_2} {T_2}) = 0$............(2)
The above eqns. are only a possibile solution.

I need help here.

 

[Chestermiller]

This is really a crazy question. I'm curious: what book does this come from.

On another subject, Have you decided not to continue our discussion of the difference between reversible and irreversible processes in the other thread that we were working on?

 

[Pushoam]

This is really a crazy question. I'm curious: what book does this come from.

On another subject, Have you decided not to continue our discussion of the difference between reversible and irreversible processes in the other thread that we were working on?

I would like to continue.
Thank you for asking it.

 

[Lord Jestocost]

The example tries to illustrate the characteristics of the state of an unconstrained equilibrium^#: pressure and temperature throughout a system are location-independent. Assume that this is not the case. One could now select two elements of space in which the temperatures would be T₁ and T₂ and the corresponding pressures would be p₁ and p₂. By a virtual process the energy of the first element could be changed by ΔU₁ and its volume by ΔV₁. The corresponding variations for the second element of space would then be ΔU₂ = -ΔU₁, ΔV₂ = -ΔV₁. One would get for the total change in entropy:

ΔS = ΔS₁ + ΔS₂ = (1/T₁ - 1/T₂) * ΔU₁ + (p₁/T₁ - p₂/T₂) * ΔV₁

The virtual changes ΔU₁ and ΔV₁ are arbitrary and independent of each other. The inequality ΔS ≤ 0 can therefore only be satisfied for T₁ = T₂, p₁ = p₂ if ΔU₁, ΔV₁ are to have any values.

[EDIT] ^# Virtual changes of the state from unconstrained thermodynamic equilibrium of an isolated system, for example an ideal gas with internal energy U enclosed in an isolated volume V, satisfy the relations
ΔS ≤ 0 when ΔU = 0 and ΔV = 0, or
S = S_maximal when U = constant and V = constant.

 

[Chestermiller]

The example tries to illustrate the characteristics of the state of an unconstrained equilibrium: pressure and temperature throughout a system are location-independent. Assume that this is not the case. One could now select two elements of space in which the temperatures would be T₁ and T₂ and the corresponding pressures would be p₁ and p₂. By a virtual process the energy of the first element could be changed by ΔU₁ and its volume by ΔV₁. The corresponding variations for the second element of space would then be ΔU₂ = -ΔU₁, ΔV₂ = -ΔV₁. One would get for the total change in entropy:

ΔS = ΔS₁ + ΔS₂ = (1/T₁ - 1/T₂) * ΔU₁ + (p₁/T₁ - p₂/T₂) * ΔV₁

The virtual changes ΔU₁ and ΔV₁ are arbitrary and independent of each other. ΔS = 0 can therefore only be satisfied for T₁ = T₂, p₁ = p₂ if ΔU₁, ΔV₁ are to have any values.

Are you assuming that the changes in U and V are differential?

 

[Lord Jestocost]

Are you assuming that the changes in U and V are differential?

@Chestermiller
I have tried to translate the essential parts from the chapter "Thermodynamische Gleichgewichte" from the German book "Thermodynamik und Statistik" by Arnold Sommerfeld (I made some editing in comment #4 to clarify). He speaks about virtual changes using the symbol "δ" (I used "Δ" to be in line with the original post). So the answer to your question: The changes in U and V are not differential ones.

 

[Chestermiller]

I'm having trouble visualizing the initial and final states of this system (particularly the final state). Can you please describe this for an actual system? Is mass transferred between the two chambers? Is there diabetic barrier that can move and permit heat exchange between the chambers, and permit one chamber to expand while the other chamber contracts by an equal amount? Is the spontaneous transfer of heat and work not allowed to proceed to final equilibrium?

 

[Lord Jestocost]

Sommerfeld’s reasoning is an abstract one, it is - so to speak - a virtual experiment inside an isolated system. The presentation in “Concepts in Thermal Physics” by S. J. Blundell and K. M. Blundell is really confusing (to my mind even poor), but when I read this thread, it let me immediately think on "the state of an unconstrained equilibrium where pressure and temperature throughout the system are independent of space coordinates".

From "Lectures on Theoretical Physics: Thermodynamics and Statistical Mechanics" by Arnold Sommerfeld one reads in chapter 8 (Thermodynamic equilibria"):

"We have found in Sec. 6 C that the entropy of an isolated system cannot decrease. A system was called isolated when it absorbed no heat and performed no work. These conditions are equivalent to stating that the internal energy U and the volume V are kept constant (dU = 0, dV = 0). An isolated system will tend to a final state at which the entropy has a maximum if all constraints within the system are removed. We shall call this a state of unconstrained thermodynamic equilibrium.

A process which starts spontaneously from a state of unconstrained equilibrium is impossible; if this were not so the entropy would have to increase again in contradiction to our assumption that the entropy already has a maximum value. We can, however, consider virtual processes δ which are compatible with the restrictions dU = 0, dV = 0 and which cannot, evidently, occur spontaneously. (Example: let a vessel be filled with a gas whose pressure and temperature are constant; we now let half of the gas be heated to a temperature T + δT, the other half being cooled to T - δT.) Such virtual changes of state from unconstrained thermodynamic equilibrium satisfy the relations

(1) δS ≤ 0 when δU = 0; δV = 0,

or, in another form

(1a) S = S_max when U = const, V = const.

If it were possible to indicate a process with δU = 0, δV = 0 for which δS > 0, we would conclude that the initial state was not one of unconstrained equilibrium. We would conclude that there existed constraints whose removal caused the entropy to increase further. Equation (1), or eq. (1a), constitutes one of the two conditions of equilibrium established by Gibbs."

Based upon this, he then shows that it follows from the characteristics of the state of an unconstrained equilibrium that the pressure and temperature throughout a system are independent of space coordinates (considering virtual changes in two elements of space, excluding changes in concentration or masses).

 

[Chestermiller]

I submit that this analysis must only apply for differential changes in U and V. Otherwise, for the case of an ideal gas, the relationship for the entropy change is different.

 

[Lord Jestocost]

I think that Sommerfeld wants to show the following: Virtual changes (which cannot occur spontaneously) of a state from unconstrained thermodynamic equilibrium must satisfy the relations δS ≤ 0 when δU = 0 and δV = 0. In case there is a conceived imbalance inside a system (location-dependent temperature and pressure), this cannot be the state of an unconstrained thermodynamic equilibrium. When one abstractly images some virtual changes δU₁ and δV₁ which are arbitrary and independent of each other, the inequality δS ≤ 0 would only hold in case T₁ = T₂, p₁ = p₂ if δU₁, δV₁ are to have any values, viz. a system in the state of an unconstrained thermodynamic equilibrium must be an isothermal and isobaric system. This reasoning doesn't address spontaneous changes which have to be analyzed in differential form.

 

[Pushoam]

When one abstractly images some virtual changes δU1 and δV1 which are arbitrary and independent of each other, the inequality δS ≤ 0 would only hold in case T1 = T2, p1 = p2 if δU1, δV1 are to have any values, viz. a system in the state of an unconstrained thermodynamic equilibrium must be an isothermal and isobaric system.

What I understood is:
The example in OP assumes that $\Delta U ~and~\Delta V$ are arbitrary and this leads to the eq.2.

Thank you, all, for helping me.

 

[Chestermiller]

What I understood is:
The example in OP assumes that $\Delta U ~and~\Delta V$ are arbitrary and this leads to the eq.2.

Thank you, all, for helping me.

Not so fast. You understand incorrectly. I can prove that Eqn. 14.25 of the OP is generally valid only for vanishingly small values of $\Delta U$ and $\Delta V$. To prove this, I will show that it is not even valid for the case of an ideal gas, except in the limit of vanishingly small values of $\Delta U$ and $\Delta V$.

In the example provided, the internal energy of system 1 decreases by $\Delta U$, and, since the internal energy of an ideal gas is a function only of temperature, the temperature in the final state of system 1 is given by:
$$T_{1f}=T_1-\frac{\Delta U}{n_1C_{v1}}$$where $T_1$ is the initial temperature, $n_1$ is the number of moles of gas in system 1, and $C_{v_1}$ is the heat capacity at constant volume of the gas in system 1. Similarly, the volume in the final state of system 1 is given by:$$V_{1f}=V_1-\Delta V$$where $V_1$ is the initial volume of system 1. We know that the change in entropy of an ideal gas between an initial thermodynamic equilibrium state and a final thermodynamic equilibrium state is given by:
$$\Delta S=nC_{v}\ln{(T_f/T_i)}+nR\ln{(V_f/V_i)}$$ If we substitute the values for system 1 into this equation, we obtain:
$$\Delta S_1=n_1C_{v1}\ln{\left(1-\frac{\Delta U}{n_1C_{v_1}T_1}\right)}+n_1R\ln{\left(1-\frac{\Delta V}{V_1}\right)}$$But, for an ideal gas,$$V_1=\frac{n_1RT_1}{p_1}$$So,
$$\Delta S_1=n_1C_{v1}\ln{\left(1-\frac{\Delta U}{n_1C_{v1}T_1}\right)}+n_1R\ln{\left(1-\frac{p_1\Delta V}{n_1RT_1}\right)}$$
Similarly, for system 2,
$$\Delta S_2=n_2C_{v2}\ln{\left(1+\frac{\Delta U}{n_2C_{v2}T_2}\right)}+n_2R\ln{\left(1+\frac{p_2\Delta V}{n_2RT_2}\right)}$$If we combine these two equations, we obtain the total entropy change:$$\Delta S=n_1C_{v1}\ln{\left(1-\frac{\Delta U}{n_1C_{v1}T_1}\right)}+n_1R\ln{\left(1-\frac{p_1\Delta V}{n_1RT_1}\right)}+n_2C_{v2}\ln{\left(1+\frac{\Delta U}{n_2C_{v2}T_2}\right)}+n_2R\ln{\left(1+\frac{p_2\Delta V}{n_2RT_2}\right)}\tag{1}$$
Eqn. 14.25 of the OP clearly does not match this relationship, and no amount of mathematical manipulation will allow them to be matched. However, in the limit of $\Delta U$ and $\Delta V$ becoming vanishingly small, Eqn. 1 approaches:
$$\Delta S\rightarrow \left(\frac{1}{T_2}-\frac{1}{T_1}\right)\Delta U+\left(\frac{p_2}{T_2}-\frac{p_1}{T_1}\right)\Delta V$$
This is opposite in sign to Eqn. 14.25 of the OP. They couldn't even get the sign right.

 

[Lord Jestocost]

I think that my presentation of Sommerfeld’s reasoning needs some clarification. Consider an isolated system consisting of two „phases“ #1 (with T₁, p₁ and S₁(U₁, V₁)) and #2 (with T₂, p₂ and S₂(U₂, V₂)). The total entropy can thus be written as

S = S₁(U₁, V₁) + S₂(U₂, V₂)

and one would have dS = 1/T*dU + p/T*dV = 0 in unconstrained thermodynamic equilibrium.

Suppose now that there is a “virtual” change (i.e. anyone of an infinite number of possible changes) in the system such that δU = 0 and δV = 0. Any change in phase#1 (δU₁, δV₁) can be chosen as virtual changes as long as one also picks for phase#2, (δU₁ = -δU₂ and δV₁ = -δV₂). Considering differentials as virtual changes δ means that the system is “feeling out” the situation in the neighborhood of the equilibrium state. Thus, the symbol δ is used to signify a virtual infinitely small change, in contrast to d, which corresponds to an actual change. With this, the most general „virtual“ change of S can then be written as

δS = (1/T₁ - 1/T₂) * δU₁ + (p₁/T₁ - p₂/T₂) * δV₁.

 

[Chestermiller]

I think that my presentation of Sommerfeld’s reasoning needs some clarification. Consider an isolated system consisting of two „phases“ #1 (with T₁, p₁ and S₁(U₁, V₁)) and #2 (with T₂, p₂ and S₂(U₂, V₂)). The total entropy can thus be written as

S = S₁(U₁, V₁) + S₂(U₂, V₂)

and one would have dS = 1/T*dU + p/T*dV = 0 in unconstrained thermodynamic equilibrium.

Suppose now that there is a “virtual” change (i.e. anyone of an infinite number of possible changes) in the system such that δU = 0 and δV = 0. Any change in phase#1 (δU₁, δV₁) can be chosen as virtual changes as long as one also picks for phase#2, (δU₁ = -δU₂ and δV₁ = -δV₂). Considering differentials as virtual changes δ means that the system is “feeling out” the situation in the neighborhood of the equilibrium state. Thus, the symbol δ is used to signify a virtual infinitely small change, in contrast to d, which corresponds to an actual change. With this, the most general „virtual“ change of S can then be written as

δS = (1/T₁ - 1/T₂) * δU₁ + (p₁/T₁ - p₂/T₂) * δV₁.

This is exactly what I was getting at (with regard to the change is U and V having to be infinitely small). But, the sign of $\delta S$ in the book is incorrect; the statement in the book indicates that U and V of chamber 2 are increased and U and V of chamber 1 are decreased. So, your $\delta U_1$ is the same as the book's $-\Delta U$.

 

[Lord Jestocost]

This is exactly what I was getting at (with regard to the change is U and V having to be infinitely small). But, the sign of $\delta S$ in the book is incorrect; the statement in the book indicates that U and V of chamber 2 are increased and U and V of chamber 1 are decreased. So, your $\delta U_1$ is the same as the book's $-\Delta U$.

You are right, because I use always U₁ + δU₁. In case U₁ are decreased, that means δU₁ is neagtive.