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Given an free expansion of a gas to twice the volume. How exactly is the work 'lost' related to the entropy 'gained'?

Proposed solution:

The change in entropy would be, however I don't really understand this formula:

$$ \Delta S = nR \ln \frac{V}{V_0}= nR \ln 2$$

Work that could've been done is:

$$ W= \int_{V_0}^V P dV= \int_{V_0}^V \frac{nRT}{V} dV = nRT \ln \frac{V}{V_0 }= nRT \ln 2$$

So the energy (work) 'lost' to the universe is for example 300 Joules. Then the entropy gained is this energy lost per kelvin (e.g. at 30 kelvin, 10 joules per kelvin.). Why is entropy not just defined by energy 'lost' but specifically energy lost per kelvin?

Why is the entropy gained not related to temperature looking at to the work 'lost' in this process?

Is there a corresponding way to look at heat and entropy?

$$ U_{\text{lost}} = nRT \ln 2 $$

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As Denbigh has shown, for a closed system in contact with a single constant temperature reservoir held at the same temperature as the initial temperature of the system T (i.e., all heat transfer takes place at T), we have: $$\Delta U=Q-W$$and$$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is equal to the irreversible entropy generation. If we combine these two equations, we obtain: $$\Delta U-T\Delta S=\Delta A=-W-T\sigma$$where A is the Helmholtz free energy. Equivalently, $$W=-\Delta A-T\sigma$$From this, it follows that the maximum possible work between the same initial and final states is $$W_{max}=-\Delta A=-\Delta U+T\Delta S$$

In the free expansion case where $\Delta U=0$ and W = 0, this tells us that the lost work between the same two end states is $$W_{lost}=W_{max}=T\Delta S$$

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The change in entropy would be, however I don't really understand this formula:

$$ \Delta S = nR \ln \frac{V}{V_0}= nR \ln 2$$

The formula assumes (1) an insulated container ($Q=0$), (2) an ideal gas, and (3) the free expansion is against a vacuum so no work is done ($W=0$). Then, from the first law, $\Delta U=0$. For an ideal gas internal energy change depends only on temperature change, thus the final and initial temperature of the gas are the same ($\Delta T=0$)

Since entropy change doesn't depend on the process, to calculate the entropy change you can replace the free expansion with any convenient reversible process connecting the initial and final state that need not bear any resemblance to the actual process. In this case, a reversible isothermal expansion of an ideal gas is assumed where $Q=W$ and

$$Q=W=nRT\ln\frac{V_{final}}{V_{inital}}$$

and

$$\Delta S=\frac{Q}{T}=nR\ln\frac{V_{final}}{V_{inital}}$$

In your example

$$\Delta S=nR \ln 2$$.

Why is entropy not just defined by energy 'lost' but specifically energy lost per kelvin?

In the first place, entropy is not "defined" in terms of energy "lost". In the second place, no energy is "lost". There is just the lost opportunity to due work in the amount of $nRT\ln2$.

That said, it is OK in this example to think of the lost opportunity to do work in terms of lost work per degree Kelvin.

Hope this helps.

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  • $\begingroup$ I understand that it is not lost in general, but I am ofcourse speaking in opportunity. Why is this defined per kelvin? $\endgroup$
    – bananenheld
    Commented May 31, 2022 at 18:01
  • $\begingroup$ @bananenheld In this example it only happens to be per kelvin because $\Delta T=0$ and $\Delta PV=0$ so the lost work is that which could have been done for a reversible isothermal process between the initial and final state. But an irreversible process does not always result in lost work. An example is an ideal gas irreversible isochoric (constant volume) heat addition. In that case $$\Delta S=nC_{v}\ln\frac{T_{final}}{T_{initial}}$$ and there is no lost work. $\endgroup$
    – Bob D
    Commented May 31, 2022 at 19:21
  • $\begingroup$ thank you for your clear answer. I was wondering if this can be viewed as 'lost work' as well, because an 'isochoric' process cannot perform work. We could've used the added heat to perform work. Thus converting the heat into mechanical energy; which means no entropy would be gained. $\endgroup$
    – bananenheld
    Commented Jun 1, 2022 at 12:15
  • $\begingroup$ @bananenheld Remember you need to look at the opportunity to perform work between the SAME two equilibrium states. If the final and initial volumes are the same, the net work over by ANY path between the states, reversible or irreversible, will necessarily be zero. The path may involve a combination of positive work (adding heat) and negative work (rejecting heat), but the sum of the two has to be zero. $\endgroup$
    – Bob D
    Commented Jun 1, 2022 at 12:26
  • $\begingroup$ I don't really understand what you mean. There is a lost opportunity of work right? Because of isochoric process. $\endgroup$
    – bananenheld
    Commented Jun 1, 2022 at 12:44

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