Entropy and the Helmholtz Free Energy of a Mass-Piston System

Pendleton
Homework Statement:
Setup: Consider a container of volume V separated bisected by a piston of cross-sectional area A under a weight w. The piston is clamped midway, and the lower half of the container is filled with an ideal gas. The system (gas + weight) is in such thermal equilibrium with a reservoir at temperature T that P < w/A, where P is the gas pressure. The clamp is then released, and the piston descends a small distance δ, after which it is clamped again and the system equilibrates.

Question 1: During this process, what is the heat dQ transferred the reservoir to the ideal gas system? Correspondingly what is the entropy change of the system dS(s) and the entropy change for the universe dS(u)?

Question 2: What is the corresponding change dF of the Helmholtz Free Energy of the system?
Relevant Equations:
First Law: dU = dQ + dW
Combined First and Second Laws: dU = TdS - PdV
Fundamental Relation: TdS - PdV = dQ + dW

Helmholtz Free Energy: dF = -SdT - PdV + E(Xdx)
Attempt at a Solution:

Heat Absorbed By The System
By the first law of thermodynamics,
dU = dQ + dW

The system is of fixed volume and therefore mechanically isolated.
dW = 0

Therefore
dQ = dU

The change of energy of the system equals the change of energy of the gas plus the change of energy of the mass.
dU = dU(g) + dU(m)

The change of energy of the gas is zero because its temperature is constant.
dU(g) = 0

Therefore, the change of energy of the system equals the change of energy of the mass.
dU = dU(m)

The energy of the mass decreases by its weight w times its descent δ.
dU(m) = -wδ

Therefore the change of energy of the system is
dU = -wδ

Therefore, the heat absorbed by the system is
dQ = -wδ

Change of Entropy of The System
By the combined first and second law,
dU = TdS - PdV

The volume of our system is fixed.
dU = TdS

Recall that the change of energy of our system is
dU = -wδ

Therefore,
-wδ = TdS

Therefore, the change of entropy of our system is
dSs = -wδ/T

Change of Entropy of The Universe
By the fundamental thermodynamic relation,
TdS - PdV = dQ + dW

The reservoir is mechanically isolated.
dV = 0, dW = 0

Therefore,
TdS = dQ

The heat absorbed by the reservoir is the opposite of that absorbed by the system.
dQ = -(-wδ) = wδ

Therefore,
TdS = wδ

Therefore, the change of entropy of the universe equals
dSu = wδ/T

Change of Helmholtz Free Energy of The System
By the definition of change of Helmholtz free energy,
dF = -SdT -PdV + E(Xdx)

where E(Xdx) is the sum of the products of each non pressure-volume force X and its displacement dx. Here, that is just the weight of the mass times its descent δ.

E(Xdx) = -wδ

Therefore,
dF = -SdT -PdV -wδ

The temperature and volume of the system are constant.
dT = 0, dV = 0

Therefore, the change of the Helmholtz free energy of the system is
dF = -wδ

Mentor
What is the upper half of the container filled with? Is the system allowed to equilibrate again after the piston is reclamped (of before)?

Pendleton
What is the upper half of the container filled with? Is the system allowed to equilibrate again after the piston is reclamped (of before)?

The upper half of the cylinder is evacuated but for the mass. The system begins in equilibrium and, after the piston is clamped again, returns to it.

Mentor
This has not been done correctly. Taking as the system the piston plus gas, the work that this system does is zero. We should be using the full version of the 1st law of thermodynamic, which includes the kinetic- and potential energy changes of the system: $$\Delta E=\Delta U+\Delta (KE)+\Delta (PE)=Q-W$$Between the initial and final states of the system, there is no change in internal energy (the temperatures of the piston and gas do not change) and the change in kinetic energy is also zero. So the equation reduces to:
$$\Delta (PE)=-w\delta=Q$$

The change in entropy of the system is the same as that of the gas. For an ideal gas at constant temperature, $$\Delta S=nR\ln{(V_f/V_i)}=\frac{PV}{2T}\ln{\left(\frac{\frac{V}{2}-A\delta}{\frac{V}{2}}\right)}=\frac{PV}{2T}\ln{\left(1-\frac{2A\delta }{V}\right)}$$For small values of ##\delta##, this linearizes to: $$\Delta S = -\frac{PA\delta}{T}$$

The change in entropy of the surroundings is ##w\delta/T##. Therefore the change in entropy of the universe is: $$\Delta S_{universe}=\frac{w\delta}{T}-\frac{PA\delta}{T}=\frac{(w-PA)\delta}{T}$$

The change in Helmholtz free energy of the system for this isothermal change is just:
$$\Delta F=\Delta U-T\Delta S=PA\delta$$

Pendleton
Wow, thanks!

Change of Heat of The Gas
It seems we agree about the change of the heat of the gas.

Change of Entropy of The System
Why is the change of entropy of the system equal to the change of entropy of the gas? Is it because the change of entropy of any system equals the total changes of entropy of its components, and the change of entropy of the mass is zero? If so, is it zero because the mass has no temperature and cannot receive or emit heat?

That being said, your calculation of the entropy change is consistent with my calculation of the entropy change of an isothermally compressed gas in another problem. For the sake of my understanding of how to calculate it from multiple formulae, especially one as general as the combined first and second law, which I used, I have reproduced below.

The combined first and second laws state that
$$dU = TdS - PdV$$

The internal energy of the gas is constant
$$dU = 0$$

Therefore,
$$dS = PdV/T$$

The change of volume is
$$dV = -Aδ$$

Therefore,
$$dS_s = -PAδ/T$$

Is this correct?

Change of Entropy of The Reservoir
Is the change of entropy of the surroundings (i.e., the reservoir) ## dS = dQ/T ## because of the fundamental relation?
$$TdS - PdV = dQ + dW$$

The system is of fixed volume and mechanically isolated from its surroundings.
$$dV = 0, dW = 0$$

Therefore
$$TdS = dQ$$
$$dS = dQ/T$$

The heat into the reservoir is the opposite of the heat into the system.
$$dQ = -dQ_s = -(-wδ)$$

Therefore, the entropy of the reservoir changes by
$$dS_r = wδ/T$$

The change of entropy of the universe is the sum of the changes of the entropies of the system and the reservoir.
$$dS_u = dS_s + dS_r$$

Those quantities are
$$dS_s = -PAδ/T, ~ dS_r = wδ/T$$

Therefore,
$$dS_u = -PAδ/T + wδ/T$$
$$= (w - PA)δ/T$$

Change of Helmholtz Free Energy
Wikipedia says that the change of the Helmholtz Free Energy of a multi-component system is not ##dF = -SdT -PdV## but

$$dF = -SdT -PdV + \sum_{i=1}^N\ X_i dx_i$$

where ## X_i ## is the ##i^th## force along its respective ##x_i## axis in generalized coordinates. This equation is not obviously equivalent to what you wrote. Would you please tell me why I should differentiate directly from the definition of Helmholtz Free Energy rather than use the Wikipedian formula?

Gratefully,
Pendleton

Mentor
Wow, thanks!

Change of Heat of The Gas
It seems we agree about the change of the heat of the gas.

Change of Entropy of The System
Why is the change of entropy of the system equal to the change of entropy of the gas? Is it because the change of entropy of any system equals the total changes of entropy of its components, and the change of entropy of the mass is zero?
Yes.
If so, is it zero because the mass has no temperature and cannot receive or emit heat?
During the process, the mass gains and loses heat, but, in the end, its temperature is unchanged.
That being said, your calculation of the entropy change is consistent with my calculation of the entropy change of an isothermally compressed gas in another problem.
Since entropy is a physical property of the gas (rather than a fundamental characteristic of the process), to get the entropy change for an irreversible process like this, we must devise an alternate reversible path that takes the system between the same two end states as the irreversible process under consideration. This is why the entropy change is the same as for an isothermally compressed gas in your other problem. For more on how to determine the entropy change for an irreversible process, see the following Physics Forums Insights article: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
For the sake of my understanding of how to calculate it from multiple formulae, especially one as general as the combined first and second law, which I used, I have reproduced below.

The combined first and second laws state that
$$dU = TdS - PdV$$

The internal energy of the gas is constant
$$dU = 0$$

Therefore,
$$dS = PdV/T$$

The change of volume is
$$dV = -Aδ$$

Therefore,
$$dS_s = -PAδ/T$$

Is this correct?
This is correct provided ##\delta## is small.
Change of Entropy of The Reservoir
Is the change of entropy of the surroundings (i.e., the reservoir) ## dS = dQ/T ## because of the fundamental relation?
$$TdS - PdV = dQ + dW$$

The system is of fixed volume and mechanically isolated from its surroundings.
$$dV = 0, dW = 0$$

Therefore
$$TdS = dQ$$
$$dS = dQ/T$$

The heat into the reservoir is the opposite of the heat into the system.
$$dQ = -dQ_s = -(-wδ)$$

Therefore, the entropy of the reservoir changes by
$$dS_r = wδ/T$$
For an ideal isothermal reservoir (of infinite heat capacity], the change in entropy is always ##Q_{res}/T##
The change of entropy of the universe is the sum of the changes of the entropies of the system and the reservoir.
$$dS_u = dS_s + dS_r$$

Those quantities are
$$dS_s = -PAδ/T, ~ dS_r = wδ/T$$

Therefore,
$$dS_u = -PAδ/T + wδ/T$$
$$= (w - PA)δ/T$$
Correct
Change of Helmholtz Free Energy
Wikipedia says that the change of the Helmholtz Free Energy of a multi-component system is not ##dF = -SdT -PdV## but

$$dF = -SdT -PdV + \sum_{i=1}^N\ X_i dx_i$$

where ## X_i ## is the ##i^th## force along its respective ##x_i## axis in generalized coordinates. This equation is not obviously equivalent to what you wrote. Would you please tell me why I should differentiate directly from the definition of Helmholtz Free Energy rather than use the Wikipedian formula?

Gratefully,
Pendleton
The Wiki equation is for a differential change in F between two closely neighboring thermodynamic equilibrium states. So, to use it, one must integrate along a reversible path. Since, in this problem, T is constant, one would choose an isothermal reversible path between the two end states. So, for a reversible isothermal path, $$dF=-PdV=\frac{nRT}{V}dV$$and, so, integrating, $$\Delta F=-\frac{PV}{2}\ln(V_f/V_i)$$For small values of ##\delta##, this approaches the present result. Of course, it is much easier to just use $$\Delta F=\Delta U-T\Delta S$$

Last edited:
Pendleton
Yes.

During the process, the mass gains and loses heat, but, in the end, its temperature is unchanged.

Thanks very much. I understand now.

Since entropy is a physical property of the gas (rather than a fundamental characteristic of the process), to get the entropy change for an irreversible process like this, we must devise an alternate reversible path that takes the system between the same two end states as the irreversible process under consideration. This is why the entropy change is the same as for an isothermally compressed gas in your other problem. For more on how to determine the entropy change for an irreversible process, see the following Physics Forums Insights article: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Thanks! I checked it out already while looking at your profile earlier. Well-written!

The Wiki equation is for a differential change in F between two closely neighboring thermodynamic equilibrium states. So, to use it, one must integrate along a reversible path. Since, in this problem, T is constant, one would choose an isothermal reversible path between the two end states. So, for a reversible isothermal path, $$dF=-PdV=\frac{nRT}{V}dV$$and, so, integrating, $$\Delta F=-\frac{PV}{2}\ln(V_f/V_i)$$For small values of ##\delta##, this approaches the present result. Of course, it is much easier to just use $$\Delta F=\Delta U-T\Delta S$$

Let me develop my intuition, then. Entropy is a state function and therefore depends not on the path taken, unlike a path function, but rather the particular state of the system. Therefore, the change of entropy between two states is the same regardless of the process whereby that change is modeled. Thus, it resembles the potential energy due to a uniform conservative force such as gravity, in that the difference of potential energy between any two points in the force field is the same, regardless of the path taken between them. Indeed, every path integral of the force from one point to the other will yield the same value, that of the path integral of a straight line between them, and that integral equals the difference of the potential energies at its beginning and end points. I am reminded of a harsh lesson from high school physics: no matter how long your distance traveled, your displacement depends only on the difference between your initial and final positions along the axis.

In other words, the entropy of a system does not ‘remember’ its past values anymore than an axis ‘remembers’ orthogonal or back-and-forth motion, or gravitational potential energy ‘remembers’ horizontal travel or previous energy levels. Is this intuition satisfying and enlightening?

Mentor
Let me develop my intuition, then. Entropy is a state function and therefore depends not on the path taken, unlike a path function, but rather the particular state of the system. Therefore, the change of entropy between two states is the same regardless of the process whereby that change is modeled. Thus, it resembles the potential energy due to a uniform conservative force such as gravity, in that the difference of potential energy between any two points in the force field is the same, regardless of the path taken between them. Indeed, every path integral of the force from one point to the other will yield the same value, that of the path integral of a straight line between them, and that integral equals the difference of the potential energies at its beginning and end points. I am reminded of a harsh lesson from high school physics: no matter how long your distance traveled, your displacement depends only on the difference between your initial and final positions along the axis.

In other words, the entropy of a system does not ‘remember’ its past values anymore than an axis ‘remembers’ orthogonal or back-and-forth motion, or gravitational potential energy ‘remembers’ horizontal travel or previous energy levels. Is this intuition satisfying and enlightening?
Very well said.

Pendleton
Very well said.

Thanks so much! I now better understand this elegant yet mysterious concept and finally get my homework done.

Chestermiller