# Work on the Entire System in Thermodynamics

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1. Jul 26, 2015

### 0pt618

A box is initially at rest on a frictionless surface. David then pushes it and the box moves across the frictionless surface.

Let's discuss this process in the context of thermodynamics, especially the First Law. One way to state it is:
ΔU=Q−W (where positive W is work done by the system, negative W is work done on the system.)

For the scenario described, let the system be the box. So ΔU refers to the change in internal energy of the box.

Question 1: Does ΔU = 0? This should be correct, because U refers to internal energy, which only includes microscopic kinetic and potential energy, and specifically excludes kinetic energy of motion of the system as a whole and the potential energy of the system as a whole.

Question 2: What is W in this case? Does the W in the First Law exclude work that affects the system as a whole? This definitely seems to be the case, but I have not been able to find any references on that provide a definition to this effect. Please provide references if available.

Last edited: Jul 26, 2015
2. Jul 26, 2015

### Staff: Mentor

The usual form of the first law, ΔU=Q−W is based on the assumption that the changes in KE and PE are negligible. The more general form of the first law is ΔU+Δ(KE)+Δ(PE)=Q−W. See Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics. So in your case, W is just what you would think it is, -Fd=-Δ(KE).

Chet

Last edited: Jul 26, 2015
3. Jul 27, 2015

### 0pt618

Hi Chet, this makes sense. But would you mind letting me know in which section or page the generalized First Law is stated in the Smith & Van Ness book you mentioned? I found the 7th edition in the library, but am having trouble finding it explicitly stated. I did find, on p. 23, Section 2.4, it is stated that "Closed systems often undergo processes during which only the internal energy of the system changes. For such processes, [ΔU=Q+W holds]", which is consistent with the assumption you mentioned.

Thanks,
David

4. Jul 27, 2015

### Staff: Mentor

It's right near the bottom of page 23, section 2.3. Do you not see it there?

Chet

5. Jul 27, 2015

### 0pt618

Yes, I see it now. Thanks.

In light of this, I think care needs to be taken to select either the generalized First Law or the more restrictive (but usual) First Law. For the example I gave, we can need to apply the generalized First Law: ΔU+Δ(KE)+Δ(PE)=Q−W and include the "pushing" work on the box in W as you said. Alternatively, if we wish to study only processes inside the box, we can use the usual form ΔU=Q-W if and only if we exclude the "pushing" work on the box, and set Δ(KE)+Δ(PE) to zero.

But sometimes it may not be so simple...if I kick a soccer ball, it may be hard to difficult to apply ΔU=Q-W to the ball, because the "kick" increase the KE of the ball as a whole, and also increases U due to the deformation of the ball.

6. Jul 27, 2015

### 256bits

Here's something for you to think about:

How would you apply the energy equation for an electrical situation such as,
The box has a resistive element inside connected to a battery.
The resistive element heats the interior of the box.

Do you think you should apply the heat as a Q?, since the interior is getting warmer.
But then no heat is crossing the boundary of the box, so that can't be it.
And no shaft work is crossing the boundary either, so it should not be a W in the normally considered sense as W=Fd.

The energy equation in the most general sense takes into account all forms of energy transfer.
The simplified versions, such as ΔU=Q−W for thermo, assume that certain "energies" are negligible.
In some particular cases the negligible may be important.

7. Jul 28, 2015

### Staff: Mentor

It depends on how you define your system. Just because the heater is inside the boundary of the box doesn't necessarily mean that it is part of your system. If you choose your system is the gas only, then the heater is part of the surroundings, and Q is positive. If the system is defined as the contents of the box, then electrical work is being done to cause the current to flow through the resistor (the inlet voltage is higher than the outlet voltage and current is flowing). So, in this case W is not zero. Either way, you get the same answer for the change in internal energy.

Chet

8. Jul 28, 2015

### 256bits

Defining the system. By choosing a particular system boundary, we end up seeing that a Q or W really do end up doing the same thing to the system.