# Causal Link Between Entropy and Mechanical Work

Let us consider a system consisting of one mole of an ideal gas enclosed in a cylinder with a friction-less piston at one end. We can force the gas to perform mechanical work on its surroundings by adding heat to the gas isothermally. Now, in the act of adding heat, we necessarily increase the entropy of the system by $$\Delta S = \frac{q}{T}$$, where $$T$$ is the temperature of the system and $$q$$ is the amount of heat transferred to the system. Applying the First Law, we have $$q = w \tag{1}$$ where $$w$$ is the amount of mechanical work performed by the system on its surroundings during the process. Now, in terms of state functions, the above can be phrased as

$$\int_{S_i}^{S_f} T\,dS = \int_{V_i}^{V_f} p dV.$$ Evaluating the integrals on each side of the equation above gives

$$\Delta S = R\cdot \ln(V_f/V_i). \tag{2}$$

I find the above equation to be very interesting as it suggests a very unintuitive causal relationship between entropy and mechanical work - one interpretation is that the increase in entropy directly leads to the performance of mechanical work. However, this seems rather odd to me as we often think of entropy as 'unusable' energy. This being the case, how can increasing the entropy of a system cause said system to do useful work?

A related (and possibly ill-posed) question is the following - where does the increase in entropy come from? Does it come from the heat we added to the system or is it a direct result of the volume-expansion of the gas?

However, this seems rather odd to me as we often think of entropy as 'unusable' energy. This being the case, how can increasing the entropy of a system cause said system to do useful work?

Because when entropy increase is due to reversible heat transfer, the entropy increase is entropy transfer which does not result in "unusable energy".

What you are probably thinking about is the increase in entropy that occurs due to entropy generation, as opposed to entropy transfer, and is a result of irreversible work or irreversible heat transfer. That is what results in less "useable" energy for performing work.

That said, the problem is a completely reversible process is an idealization. All real processes are irreversible and generate entropy. Moreover, while it is possible in theory to completely convert heat to work in a reversible process it is not for a reversible cycle. That would violate the Kelvin-Planck statement of the second law. Some heat must always be rejected to a lower temperature in a cycle making it unavailable to perform useful work in the current environment.

A related (and possibly ill-posed) question is the following - where does the increase in entropy come from?

As indicated above, in the case of a reversible isothermal process it comes from the transfer of entropy between the system and surroundings.

Hope this helps.

You have assumed a reversible and isothermal process but a significant piece in the governing equation you used hides what really happens. The culprit here is that you assumed an "ideal" gas. As you will see, isothermal entropy transport is not the source of work, it does something else.

Recall the Gibbs-Duhem equation for a fluid in a closed system but note that you cannot get that from $$dU=TdS-pdV$$ because if you do so then the result is $$SdT-Vdp=0$$ which, in general, is obviously nonsense. Specifically, it implies that an isothermal process, $$dT=0$$, must also be an isobaric process. The error is that we have ignored the chemical potential and its contribution.

In fact, for any fluid, in general, for any simple single phase system the Gibbs energy conservation equation is $$dU=TdS-pdV+\mu dN$$, and combined with $$U=TS-pV+\mu N$$ follows the corresponding Gibbs-Duhem equation: $$SdT-Vdp+Nd\mu =0 \tag{1}.$$

Calculated properly, even in a closed system for which $$N=\text{const; }dN=0,$$ the chemical work terms stays! Since this holds for any two equilibrium states that are infinitesimally close to each other, it also must hold, per force, in an integrated form to finite differences between two arbitrary equilibrium states, say, $$\mathcal P_1$$ and $$\mathcal P_2$$ that are connected by a reversible path along which every state is an equilibrium one: $$\int_{\mathcal P_1}^{\mathcal P_2}SdT-\int_{\mathcal P_1}^{\mathcal P_2}Vdp+\int_{\mathcal P_1}^{\mathcal P_2}Nd\mu =0 \tag{2}$$ Every term in Eq (2) represents work, and that their total sum is zero means that in a reversible process the energetic interactions balance each other.

Now assume that the reversible path is also an isothermal one, $$dT=0$$, and taking into account that your system is assumed to be closed, $$N=\text{const}$$, then you have $$\int_{\mathcal P_1}^{\mathcal P_2}SdT-\int_{\mathcal P_1}^{\mathcal P_2}Vdp+\int_{\mathcal P_1}^{\mathcal P_2}Nd\mu \\= -\int_{\mathcal P_1}^{\mathcal P_2}Vdp+N\int_{\mathcal P_1}^{\mathcal P_2}d\mu \tag{3}=0$$ or in the case of a closed system, $$N=\text{const}$$, $$\int_{\mathcal P_1}^{\mathcal P_2}Vdp = N(\mu_2-\mu_1) \tag{4}$$

Integrate the differential identity $$-pdV=pdV-d(pV)$$ along the process $$-\int_{\mathcal P_1}^{\mathcal P_2}dV=\int_{\mathcal P_1}^{\mathcal P_2}Vdp+\int_{\mathcal P_1}^{\mathcal P_2}d(pV)$$ and you get $$-\int_{\mathcal P_1}^{\mathcal P_2}pdV=\int_{\mathcal P_1}^{\mathcal P_2}Vdp+(p_1V_1-p_2V_2) \tag{5}$$

Eq.(5) shows the work done on the environment. In the special case of an ideal gas described by $$pV=NRT$$ and going through an isothermal process, $$dT=0$$, $$pV=\text{const}$$ and $$p_1V_1-p_2V_2=0$$, so $$-\int_{\mathcal P_1}^{\mathcal P_2}pdV=\int_{\mathcal P_1}^{\mathcal P_2}Vdp \tag{6}.$$

What this tells you is that during its isothermal expansion/compression of the gas the chemical work, $$N(\mu_1-\mu_2)$$, is converted into mechanical work by changing the chemical energy of the fluid while changing its volume and pressure from $$V_1, p_1$$ to $$V_2, p_2$$.

Integrating (1) for an ideal gas, $$pV=NRT$$ while $$dT=0$$ gets you its chemical potential as $$\mu(p,T)=\mu_0(T)+RT\ln|\tfrac{p}{p_0}|$$ showing its logarithmic dependence on pressure at constant temperature. Because of the assumption that the process is isothermal, $$dT=0$$, the entropy transport between the reservoir and the system does not directly participate in the work delivery, instead its role is to keep the temperature constant, here, specifically, for the chemical potential, as the system works on its environment.

To arrive here we did not need to know that the internal energy of an ideal gas depends only on its temperature (Joule's law). This is important because the caloric equation of state, $$U=f_u(T,V)$$ is independent of its thermal equation of state $$T=f_T(p,V)$$, one does not follow from the other. And if $$U\ne f_u(T)$$ but also depends on $$V$$, say, then one cannot conclude that in an isothermal process the delivered work $$w$$ equals $$q$$.

An additional comment is in order. The differential form of the Gibbs-Duhem equation (1) is true to any portion of the fluid body, so it could be interpreted as $$\Delta SdT-\Delta Vdp+\Delta Nd\mu =0 \tag{1a}.$$ where $$\Delta S\div S =\Delta V\div V=\Delta N \div N,$$ and you could conceive (1a) as referring to a certain piece of the thermodynamic body going through a reversible process. When interpreted that way it could also mean that a certain piece of the size of $$\Delta S, \Delta V, \Delta N$$ having intensive parameters $$T, p, \mu$$ are being added and then the same amount is removed with $$T+dT, p+dp, \mu+d\mu$$ from the system. Then the Gibbs-Duhem equation gains an important new interpretation that is really illuminating: the equation (1a) or its integrated form $$\int_{\mathcal P_1}^{\mathcal P_2}\Delta S dT-\int_{\mathcal P_1}^{\mathcal P_2}\Delta Vdp+\int_{\mathcal P_1}^{\mathcal P_2}\Delta Nd\mu =0 \tag{2a}$$ expresses a universal work conservation among elementary work processes in which the system is sandwiched between two sets of sources and sinks, one source/sink pair for each of the extensive quantities. In a reversible process, work converts only to work and only work converts to work.

As regards your second question, since it is assumed that the process is reversible the total entropy does not change at any step, whatever entropy is absorbed from the constant temperature reservoir and is lost by it will increase the entropy of the system by exactly the same amount. If the process were irreversible then the system entropy would increase its entropy more than the amount it absorbed from the reservoir.

• I think you completely miss the level of the question. You ignore simple explanation based on the first law and once again write extensively about your idiosyncratic view of thermodynamics, using formal convention-dependent properties of entropy and chemical potential instead of simple worded physical laws. This is not pedagogic and is likely to confuse the person asking the question. Commented Sep 7, 2023 at 15:57
• > "In a reversible process, work converts only to work and only work converts to work." This sounds like a sophism, not explanation. In standard physics, part of the heat accepted in isothermal expansion converts to work. Part may convert to internal energy of the medium, but OP asks about ideal gas isothermal process, where all heat transforms to work. Commented Sep 7, 2023 at 15:58
• Please do not use simple questions about high school physics here to promote your views with needlessly complicated explanations, where there is obviously a simpler standard explanation. Commented Sep 7, 2023 at 16:00
• The explanation was rather cumbersome but I found it enlightening. Commented Sep 7, 2023 at 18:55
• I’ve never cared for the phrase “convert” heat into work. For one thing, I would say “heat” is a mechanism for transferring energy, not the energy itself. For the same reason I would say “work” is a mechanism for transferring energy, not the energy itself. So $Q$ and $W$ in the first law represent the amount of energy transfer by means of two different mechanisms. Commented Sep 8, 2023 at 20:29