Conceptual question about internal energy

[jdstokes]

Homework Statement

From the beginning of chapter 21 Haliday, Resnick and Walker:

`Suppose that you return to your chilly dwelling after snowshoeing through the woods on a cold winter day. Your first thought is to light a stove. But why, exactly, would you do that? Is it because the stove will increase the store of internal (thermal) energy of the air in the cabin, until eventually the air will have enough of that internal energy to keep you comfortable? As logical as this reasoning sounds, it is flawed, because the air's store of internal enrgy will not be changed by the stove. How can that be? And if it is so, why would you bother to light the stove?'

The Attempt at a Solution

I'm confused by what HRW are saying here. My understanding is that the stove releases heat into the constant-volume environment which raises the internal energy according to the first law of thermodynamics dU = Q - PdV = Q.

Am I missing something here?

 

[Andrew Mason]

Homework Statement

From the beginning of chapter 21 Haliday, Resnick and Walker:

`Suppose that you return to your chilly dwelling after snowshoeing through the woods on a cold winter day. Your first thought is to light a stove. But why, exactly, would you do that? Is it because the stove will increase the store of internal (thermal) energy of the air in the cabin, until eventually the air will have enough of that internal energy to keep you comfortable? As logical as this reasoning sounds, it is flawed, because the air's store of internal enrgy will not be changed by the stove. How can that be? And if it is so, why would you bother to light the stove?'

The Attempt at a Solution

I'm confused by what HRW are saying here. My understanding is that the stove releases heat into the constant-volume environment which raises the internal energy according to the first law of thermodynamics dU = Q - PdV = Q.

Am I missing something here?

This is really an ideal gas law problem, not a first law problem.

State the expression for the total internal energy of the air inside the cabin.

Now add heat, dQ. This increases the temperature by dT. P is constant because the atmospheric pressure does not change. The cabin's volume does not change. So PV does not change. If dQ heat is added and temperature increases by dT, what else has to change? Express that quantity as a function of temperature and substitute into your expression for total internal energy of the air in the cabin. You will see why the total internal energy does not change with temperature.

AM

 

[jdstokes]

Thanks for your reply Andrew,

U = \frac{3}{2}NkT.
PV = NkT \implies N = PV/kT \implies U = \frac{3}{2}PV.

That is a very interesting result. So there is less air in a hot room than a cold one? Does this effect of reduced number density contribute to the deaths of small pets in heated cars?

You mention that this is not a first law problem. I'm wondering what is the flaw in the logic which leads to the conclusion \Delta U = Q? Presumably the idenity W = -PdV fails because of the varying number density. What is the correct expression? W = - PdV + \mu dN?

 

[Andrew Mason]

You mention that this is not a first law problem. I'm wondering what is the flaw in the logic which leads to the conclusion \Delta U = Q? Presumably the idenity W = -PdV fails because of the varying number density. What is the correct expression? W = - PdV + \mu dN?

There is no flaw. The first law assumes that the mass does not change. If the mass changes, the first law does not apply. You could analyse the problem by keeping track of what happens to the air that leaves the room, and then the first law would apply. But that would be rather difficult.

AM

 

[jdstokes]

That's a fair point, but I think it is possible to save the first law for open systems by introducing an additional term, namely \mu dN. This makes it more `universal' in some sense.