Does the W in the 1st law include the external and also the internal work done by the gas.

because the author has given this below proof and reasoning in the Joule coefficient expansion and calculates it using the internal work done by a gas

His Reasoning

His proof

So his proof should mean and cause W to include the internal work done by gas also. Because it does so.

Does the W in the 1st law include the internal work done by the gas ?

  • $\begingroup$ Conclusion: The work done involved in the derivation is different from the one involved in the 1st law. Only the kinetic energy is equated equal to it and it has nothing to do with the q and w term in the law. It is a different version of expansion of internal energy. I suggest you to look for some other derivations and drop this one until you find a satisfactory answer. Internal work has been equated to change in kinetic energy, im not sure if this is correct in every case. $\endgroup$ – Mitchell Mar 29 '17 at 14:03

As per @Bhavya Sharma's assessment, I too disagree strongly with what this author presents.

Consider a rigid adiabatic container of volume $V_2$, with a barrier within the container separating one mole of a real gas in a sub-compartment of volume $V_1<V_2$ from a second sub-compartment of volume $V_2-V_1$ containing vacuum. The initial gas temperature is $T_1$. We remove the barrier and allow the gas to re-equilibrate to a new (unknown) temperature $T_2$. We wish to determine the new temperature.

We begin by applying the first law of thermodynamics $\Delta U=Q+W$, where Q is the amount of heat that enters the container, W is the amount of work that the surroundings (in this case, the container) does on our system (in this case, the gas), and $\Delta U$ is the change in internal energy of our gas. The amount of heat that enters the container is zero (since the container is adiabatic) and the amount of work that the surroundings do on our system is zero (since the displacements at the interface with the rigid surroundings are zero). Therefore, the change in internal energy is zero. $$\Delta U=0\tag{1a}$$ or, equivalently, $$U(T_2,V_2)=U(T_1,V_1)\tag{1b}$$where U is determined relative to some reference state, say $(T_{ref},\infty)$, where U is taken to be zero.

This is all the information we can gleen from applying the first law of thermodynamics to this problem. However, we know that U is a function of state, and we also know that, from the 2nd law of thermodynamics, it follows that: $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV\tag{2}$$If we know the heat capacity at constant volume as a function of temperature and pressure, and we know the equation of state for the real gas, we can integrate this equation over any arbitrary path to determine the change in internal energy from an initial state to a final state, and can thus determine the final temperature required to produce zero change in internal energy in our problem.

For some reason that doesn't make sense to me, the term in brackets in Eqn. 2 is sometimes referred to in the literature as the "internal pressure" $\pi(T,V)$ of the gas:$$\pi(T,V)=\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]\tag{3}$$ As far as I'm concerned, they might as well have called this the "hippopotamus." I think there is some molecular motivation for calling this the "internal pressure," but I have never been able to see through the arguments. This author takes the conceptualization one step further by calling $$-\pi(T,V)dV=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV\tag{4}$$ the "internal work." This too has no physical significance to me because it is totally separate from the first law of thermodynamics, and does not even relate to any kind of physical work involving force and displacement. I would prefer to call it "hippopotamus work." None of this terminology makes any sense to me, and to make matters worse, it is not even needed to solve problems of this kind. To make matters still worse, it has caused the OP to become confused into thinking that "internal work" is something that can actually be interpreted as real work in applying the laws of thermodynamics.

Out of the infinite number of possible (T,V) trajectories that can be used to integrate Eqn. 2 from the initial state to the final state of the system (to obtain $\Delta U$), the simplest trajectory involves an excursion through the ideal gas region ($V \rightarrow \infty$) . This is because, in the ideal gas region, $C_v$ is a function only of temperature $C_v^{IG}(T)$, and not of specific volume. Furthermore, the ideal gas heat capacity of the gas as a function of temperature is quite often known in advance. Any other path would require knowledge of both the temperature dependence and the volume dependence of $C_v$ (i.e., $C_v(T,V)$. So, to integrate Eqn. 2, we write: $$\Delta U=-\int_{V_1}^{\infty}{\pi (T_1,V')dV'}+\int_{T_1}^{T_2}{C_v^{IG}(T)dT}+\int_{V_2}^{\infty}{\pi (T_2,V')dV'}\tag{5}$$where use has been made here of Eqn. 3 for shorthand purposes only. The same result can also be expressed in the form of Eqn. 1b by writing: $$U(T,V)=\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}+\int_{V}^{\infty}{\pi (T,V')dV'}\tag{6}$$ For a van der Waals gas, $$\pi=-\frac{a}{V^2}$$and$$U(T,V)=\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}\tag{7a}$$or $$\Delta U=\int_{T_1}^{T_2}{C_v^{IG}(T)dT}-a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)\tag{7b}$$


Great question!! In the case where there is a small hole punched in the barrier (and the barrier is insulated), when the system equilibrates, the final temperatures in the two chambers will be different (assuming also that there is no heat transfer through the tiny opening in the barrier). But, there is still no work done on the combined system, and no change in internal energy of the combined system. But if you want to find out the final equilibrium state in each of the chambers, then you need to divide the gas into two subsystems: 1. The gas that has remained in the original chamber and 2. The gas that has gone into the vacuum chamber.

Imagine that there was a membrane that completely surrounded gas #1 during the process. The gas that has remained in the original chamber has experienced a very slow adiabatic (essentially reversible) expansion up to its new final volume (entirely filling the original containter). It has done adiabatic reversible work at its boundary, driving gas #2 out of the container. But this is work external to gas #1. If we knew the number of moles that got transferred through the opening, we could calculate the final state of gas #1. The conditions that we can use to determine the final states in both containers are that (1) the final pressures in the two chambers match and (2) the total change in internal energy for the combined system is zero.

  • $\begingroup$ +1: great explanation as usual! For a while even I was convinced that W contaiins internal work as well. Because if you look at it this way -Take 2 thermally insulated connected jars with a stop cork in between.One jar contains real gas at high pressure . The other is at 0 pressure.Open the stop cork the air goes into the other beaker and it's temperature falls ( real gas ).Q is 0 because they are thermally insulated. W(external) =0Nothing is displaced plus external pressure =0.So U ought to be 0. But it isn't.Will it not call for an internal work addition to W to allow Delta U to be non zero. $\endgroup$ – Shashaank Apr 1 '17 at 11:41
  • $\begingroup$ @Shashaank See my addendum on how this question is addressed. $\endgroup$ – Chet Miller Apr 1 '17 at 12:13
  • $\begingroup$ You mean to say that the total change in internal energy is 0. It decreases for the 1st jar (with the gas) and increases by the same amount for the other jar (in which the gas goes). But you say the final temperature will be different for the 2 parts . But the final temperature is constant equal for both the gasses. Isn't it ? Equilibrium means that....Only . Also on what physical thing does the 1st gas do work on. What is the physical force and the physical displacement . $\endgroup$ – Shashaank Apr 1 '17 at 15:28
  • $\begingroup$ Yes. The total change in internal energy is zero. Yes, it decreases for the 1st jar, and increase (not necessarily by the same amount) for the other jar. At equilibrium, the two jars will have different amounts of gas. In the case of a tiny hole in the barrier, the temperature will not be equal for both gases, unless heat can be transferred between the two compartments. In this case, equilibrium does not mean that, because the barrier constrains the heat transfer. The first gas does work on the 2nd gas, in exerting a pressure force on it and displacing it out of container 1. $\endgroup$ – Chet Miller Apr 1 '17 at 16:17
  • $\begingroup$ If the increase in the other is not equal how will the total change be 0 ? The above experiment that I gave is like the ones that Joule did. For a stop cork , I think there will be a heat transfer. But then this looks like that only. A part of the same gas doing work on its other (molecules pushing other molecules of the same gas) to push it out of the hole. This calls for an internal pressure. Isn't it like that only Essentially the 2 parts are the one one and the same thing. One part pushes the other calling for an internal work. It's quite confusing !! $\endgroup$ – Shashaank Apr 1 '17 at 16:48

According to the first law of thermodynamic,


In this equation, the work is done by the external agent. The work done by the gas is equal and opposite to the work done by the external agent.

$w=-P_{ext}\Delta{V}=-work_{gas}=work_ {ext. agent}$

Considering adiabatic process, $q=0$,

If the gas expands, then the gas does the effective +ve work in expanding the system, which means that the gas uses some of the internal energy to do this work. Thus for sych process the internal energy of the system decreases.

If the gas contracts, then the external agent does the effective +ve work and this energy gets stored in the system as the internal energy.

Use energy conservation, if you get confused.

All the signs used in the equations are with respect to the system. If internal energy of the system increases, $\Delta{U}$ gets a +ve sign and vice versa.

But in your case of free expansion, gas is allowed to expand into a vacuum. This happens quickly, so there is no heat transferred. No work is done, because the gas does not displace anything. According to the First Law, this means that:

$\Delta{E}_{int} = 0$

There is no change in internal energy, so the temperature stays the same.

But again this is only true for ideal gas.

For real gas the given details in your textbook are correct about the temperature. But the work done will be zero as the external pressure is 0.

  • 1
    $\begingroup$ But I think the author wants to say that the internal energy stays constant. The internal potential energy ( non zero for a real gas) increases & k.e lessesns causing the gas to cool. $\endgroup$ – Shashaank Mar 28 '17 at 16:14
  • $\begingroup$ thank you. But I think you have gone slightly wrong. Temperature is due to only the kinetic energy not the potential. According to the author ( which I seriously don't know is right or not because I have not heard of this internal work ) the P.E increases and the K.E decreases. The decrease in K.E results in the fall of the temperature ( as the author says in the last few lines , have a look). That's why , also otherwise your last line is wrong . Temperature doesn't stay constant for a real gas because as above P.E increases , K.E lessesns . For an ideal gas it's true. $\endgroup$ – Shashaank Mar 28 '17 at 16:35
  • $\begingroup$ How about now..? $\endgroup$ – Mitchell Mar 28 '17 at 16:42
  • $\begingroup$ Change in energy is 0 . That's is because energy is conserved. But the point is only K.E of the total internal energy gives the temperature not the P.E part. If P.E part & K.E part change such that total change in internal energy is 0 then off course the temperature can change. Which happens here. According to you then the temperature of the universe will never change $\endgroup$ – Shashaank Mar 28 '17 at 16:43
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    $\begingroup$ Your wish. Wait to see if anyone else has another opinion $\endgroup$ – Shashaank Mar 28 '17 at 17:03

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