First law of thermodynamics -- two different expressions for dQ

[laser1]

TL;DR Summary

dQ gives two different expressions, not sure why

Okay, so the first law $dU=dQ+dW$. We all know that $dU=C_VdT$. So that means that we have: $$dQ=C_VdT+PdV$$ Now I have a problem. We also have that $$dU=\left(\frac{\partial U}{\partial V}\right)_TdV+\left(\frac{\partial U}{\partial T}\right)_VdT$$ and substituting that in to $dQ=dU+dW$ gives $$dQ=C_VdT+\left[P+\left(\frac{\partial U}{\partial V}\right)_T\right]dV$$ which is almost but not quite the same as the earlier expression. The first one sure does look like a special case of the second one, when internal energy does not depend on volume at constant temperature, but which equation is more "correct"? Thanks!

 

[Philip Koeck]

TL;DR Summary: dQ gives two different expressions, not sure why

Okay, so the first law $dU=dQ+dW$. We all know that $dU=C_VdT$. So that means that we have: $$dQ=C_VdT+PdV$$ Now I have a problem. We also have that $$dU=\left(\frac{\partial U}{\partial V}\right)_TdV+\left(\frac{\partial U}{\partial T}\right)_VdT$$ and substituting that in to $dQ=dU+dW$ gives $$dQ=C_VdT+\left[P+\left(\frac{\partial U}{\partial V}\right)_T\right]dV$$ which is almost but not quite the same as the earlier expression. The first one sure does look like a special case of the second one, when internal energy does not depend on volume at constant temperature, but which equation is more "correct"? Thanks!

dU = C_V dT is only valid for an ideal gas.
For an ideal gas U doesn't depend on volume.

 

[laser1]

dU = C_V dT is only valid for an ideal gas.
For an ideal gas U doesn't depend on volume.

Can you tell me at which step does this derivation break down? $$dU=dQ+dW$$ Seems valid for a real gas. $$dU=dQ-PdV$$ This also seems valid for a real gas. Heat at constant volume is given by $dQ=C_VdT$ so just subbing in definitions. $$dV=0 \rightarrow dU=C_VdT$$ And I'm not sure where this breaks down, thanks

 

[Philip Koeck]

Can you tell me at which step does this derivation break down? $$dU=dQ+dW$$ Seems valid for a real gas. $$dU=dQ-PdV$$ This also seems valid for a real gas. Heat at constant volume is given by $dQ=C_VdT$ so just subbing in definitions. $$dV=0 \rightarrow dU=C_VdT$$ And I'm not sure where this breaks down, thanks

dW = P dV is only valid for an ideal gas. You do work against an external force, which is balanced by the internal pressure in case of a quasistatic (reversible) process.

To expand a real gas you also have to do work against the attractive forces between molecules.

Note also that the inner energy of an ideal gas can only depend on temperature since energy can only be stored as kinetic energy if there are no forces between molecules (other than when they collide).

PS: Actually I'm not completely sure about everything above.
Let's see what others say.

 

[anuttarasammyak]

TL;DR Summary: dQ gives two different expressions, not sure why

We all know that dU=CVdT.

Yes, in case
$$\frac{\partial U}{\partial V}|_T=0.$$
Ideal gas satisfies it.

 

[Philip Koeck]

dW = P dV is only valid for an ideal gas. You do work against an external force, which is balanced by the internal pressure in case of a quasistatic (reversible) process.

To expand a real gas you also have to do work against the attractive forces between molecules.

Note also that the inner energy of an ideal gas can only depend on temperature since energy can only be stored as kinetic energy if there are no forces between molecules (other than when they collide).

PS: Actually I'm not completely sure about everything above.
Let's see what others say.

Here's what I was wondering about:
The textbook I use says nothing about the type of gas when it develops the concept of PV-work.
I'm pretty certain, however, that it can only be valid for an ideal gas (where U is independent of V, see post 5).
One way of seeing this is that a free expansion of an ideal gas does no work. If there are intermolecular forces holding the gas together even a free expansion requires work. The attraction between molecules has to be overcome. So in a free expansion of a real gas work is done against the intermolecular forces and the inner energy will decrease during a free expansion.
Therefore dW = P dV, where P is only the external pressure, is only valid for ideal gases.

The textbook I use might be seriously sloppy not mentioning this.
Do people agree?

 

[anuttarasammyak]

Therefore dW = P dV, where P is only the external pressure, is only valid for ideal gases.

dW=P dV holds for any gas. dU=C_v dT of no dV term holds only for ideal gas which has no intermolecular force. For real gases derivative in post #5 which is not zero should contribute to dU by dV term.

 

[laser1]

dW=P dV holds for any gas. dU=C_v dT of no dV term holds only for ideal gas which has no intermolecular force. For real gases derivative in post #5 which is not zero should contribute to dU by dV term.

huh so you mean that at constant volume, dV is not zero? I don't see why! See my derivation in post 3, I might be misunderstanding you

 

[Chestermiller]

dW = P dV is only valid for an ideal gas.

This is not correct. It is valid for any real gas too.

I only use differentials for reversible changes, so I am assuming that is what we are dealing with here. I'm not going to go through all the previous posts,, but the correct relationships for a reversible change are:
$$dU=TdS-PdV$$
$$dU=C_VdT-\left[P-T\frac{\partial P}{\partial T}\right]dV$$
$$dS=\frac{dQ}{T}=\frac{C_V}{T}dT+\frac{\partial P}{\partial T}dV$$

 

[laser1]

I'm not going to go through all the previous posts

Do you mind checking what is wrong with my derivation in post #3 please?

 

[Chestermiller]

Do you mind checking what is wrong with my derivation in post #3 please?

The equations in post #3 seem consistent with my post.

 

[anuttarasammyak]

dV=0→dU=CVdT

holds for dV=0.

dQ=CVdT+[P+(∂U∂V)T]dV

holds for any dT and dV. The two are consistent.

So that means that we have: dQ=CVdT+PdV

does not hold.

 

[laser1]

Hmm...

The equations in post #3 seem consistent with my post.

So you agree that $dU=C_VdT$. Hence by the first law, $dQ=dU-dW$, and subbing in $dU=C_VdT$ and $dW=-PdV$ I get $dQ=C_VdT+PdV$ which everyone here claims is wrong for an ideal gas. What is happening?

 

[anuttarasammyak]

I agree that dU=C_v dT in condition that dV=0. In dQ=dU+dW, where W is work done to the environment in my preference of definition of sign, dV $\neq$0 in general. You are not able to use dU=C_v dT for this dV $\neq$0 process.

 

[laser1]

I agree that dU=C_v dT in condition that dV=0. In dQ=dU+dW, where W is work done to the environment in my preference of definition of sign, dV $\neq$0 in general. You are not able to use dU=C_v dT for this dV $\neq$0 process.

lol why does that matter, I already proved $dU=C_VdT$, it doesn't necessarily mean the process has to be at constant volume to use it

 

[anuttarasammyak]

This also seems valid for a real gas. Heat at constant volume is given by dQ=CVdT so just subbing in definitions. dV=0→dU=CVdT And I'm not sure where this breaks down, thanks

As bolded you proved it in condition that dV=0, didn't you ?

 

[Philip Koeck]

Can you tell me at which step does this derivation break down? $$dU=dQ+dW$$ Seems valid for a real gas. $$dU=dQ-PdV$$ This also seems valid for a real gas. Heat at constant volume is given by $dQ=C_VdT$ so just subbing in definitions. $$dV=0 \rightarrow dU=C_VdT$$ And I'm not sure where this breaks down, thanks

It must be the last step that's wrong as pointed out by others in this thread.
You're showing that dU = C_VdT for dV = 0, but that doesn't mean that this is true for all processes.
The fact that U is a state function doesn't help either, because that doesn't allow you to switch to a process with different end states.

So, for an ideal gas dU = C_VdT for any process.
For a real gas this is only the case for constant volume processes.

 

[weirdoguy]

it doesn't necessarily mean the process has to be at constant volume to use it

It does in general context. I think you are trying to push derivation you saw in the context of ideal gas in the general case. $dU=C_vdT$ is not true for general case. It is for ideal gas only.

 

[weirdoguy]

which everyone here claims is wrong for an ideal gas.

No, no one claims it's wrong for an ideal gas. It's wrong for real gas.

 

[Chestermiller]

Hmm...

So you agree that $dU=C_VdT$. Hence by the first law, $dQ=dU-dW$, and subbing in $dU=C_VdT$ and $dW=-PdV$ I get $dQ=C_VdT+PdV$ which everyone here claims is wrong for an ideal gas. What is happening?

No, not in general; only when dV=0 or an ideal gas. Otherwise, see my 2nd equation.

 

[laser1]

No, no one claims it's wrong for an ideal gas. It's wrong for real gas.

sorry, typo

 

[laser1]

Okay cool, I think I get it now. $$dU=C_VdT$$ iff $dV=0$. However, in ideal gases, $U=U(T)$ only, so this equation for $dU$ is valid no matter what the value of $V$ is. In real gases, that formula for $dU$ is only valid for $dV=0$ as $U=U(T, V)$.

 

[weirdoguy]

Yes

 

[anuttarasammyak]

I would prefer to say that

and substituting that in to dQ=dU+dW gives dQ=CVdT+[P+(∂U/∂V)T]dV which is almost but not quite the same as the earlier expression.

is the universal relation for systems including real gas and ideal gas. And ideal gas has the property of

(∂U/∂V)T=0.