Extremely challenging thermodynamics equation

[kntsy]

hi,
i actually posted it in physics section but no one knows so i think engineering experts would be stronger to this kind of equations.
1.
$$\color{blue}dq=dU+PdV$$
so
$$C_V = \left(\frac {\partial q}{\partial T}\right)_V= \left(\frac {\partial U}{\partial T}\right)_V$$
leads to
$$dU=C_{V}dT$$
so
$$\color{red}\left(\frac {\partial U}{\partial V}\right)_T = 0$$
meaning that internal energy of NONideal gas is a sole function of T?
2.
As
$$\color{blue}dq=dU+PdV$$
so
$$\left(\frac {\partial U}{\partial V}\right)_T= \left(\frac {\partial V}{\partial T}\right)_P \left(C_P-C_V\right) - P$$
why? even for ideal gas the change in volume results in change in internal energy:
$$\color{red}\left(\frac {\partial U}{\partial V}\right)_T \not= 0??$$

thanks for answering.

 

[Topher925]

[PLAIN]https://www.physicsforums.com/latex_images/28/2894908-3.png[/QUOTE]

How did you get to this? And why do you state that its for a NONideal gas?

even for ideal gas the change in volume results in change in internal energy:

Are you sure about this? Try doing a first law analysis of a finite volume of an ideal gas.

 

[Studiot]

Perhaps the working in this thread will help, see in particular post#7

https://www.physicsforums.com/showthread.php?t=408879&highlight=internal+energy

 

[boneh3ad]

$$dq=dU+PdV$$

$$\frac{1}{\partial T} \left(dq = dU+PdV \right)_V$$

$$\left(\frac{\partial q}{\partial T}\right)_V = \left(\frac{\partial U}{\partial T}\right)_V + \left(\frac{\partial (P dV)}{\partial T}\right)_V$$

$$\left(\frac{\partial q}{\partial T}\right)_V = \left(\frac{\partial U}{\partial T}\right)_V + P \left(\frac{\partial V}{\partial T}\right)_V$$

$$\left(\frac{\partial V}{\partial T}\right)_V \equiv 0$$

$$C_V \equiv \left(\frac {\partial q}{\partial T}\right)_V$$

$$C_V =\left(\frac {\partial U}{\partial T}\right)_V$$

$$\partial U = C_V \partial T$$

Maybe I missed something in my haste, but I don't know how he got from here to:

$$\left(\frac {\partial U}{\partial V}\right)_T = 0$$

The only thing I can think of is if he then held T constant and differentiated with respect to V, but that doesn't really make sense. You would then be holding both T and V constant, so nothing is changing.

Also I edited this to make it correct but it didn't update the post for some reason. Weird.

 

[kntsy]

Sorry for late response.

How did you get to this? And why do you state that its for a NONideal gas?

Because it is deduced from "nonideal-gas-equation"$dU=C_{V}dT$ and obtaining that internal energy is only sole function of temperature and independent of volume.

 

[kntsy]

$$dq=dU+PdV$$

$$\frac{1}{\partial T} \left(dq = dU+PdV \right)_V$$

$$\left(\frac{\partial q}{\partial T}\right)_V = \left(\frac{\partial U}{\partial T}\right)_V + \left(\frac{\partial (P dV)}{\partial T}\right)_V$$

$$\left(\frac{\partial q}{\partial T}\right)_V = \left(\frac{\partial U}{\partial T}\right)_V + P \left(\frac{\partial V}{\partial T}\right)_V$$

$$\left(\frac{\partial V}{\partial T}\right)_V \equiv 0$$

$$C_V \equiv \left(\frac {\partial q}{\partial T}\right)_V$$

$$C_V =\left(\frac {\partial U}{\partial T}\right)_V$$

$$\partial U = C_V \partial T$$

Maybe I missed something in my haste, but I don't know how he got from here to:

$$\left(\frac {\partial U}{\partial V}\right)_T = 0$$

The only thing I can think of is if he then held T constant and differentiated with respect to V, but that doesn't really make sense. You would then be holding both T and V constant, so nothing is changing.

Also I edited this to make it correct but it didn't update the post for some reason. Weird.

you also say that $\partial U = C_V \partial T$so why not $\left(\frac {\partial U}{\partial V}\right)_T = 0$

 

[kntsy]

Are you sure about this? Try doing a first law analysis of a finite volume of an ideal gas.

Yes i know that 1st law+kinetic theory leads to $U=\frac{3}{2}RT$, but the last equation i derived in the 1st post omitted kinetic theory and a strange result comes.

 

[boneh3ad]

Because it is deduced from "nonideal-gas-equation"

No it isn't. That isn't the "non-ideal gas equation." It is merely a derived relationship between U and T if V is held constant.

you also say that $\partial U = C_V \partial T$so why not $\left(\frac {\partial U}{\partial V}\right)_T = 0$

Because that step doesn't make sense. Start with:

$$\partial U = C_V \partial T$$

This equation is all you need to determine that internal energy is a function of only temperature at constant volume. Differentiating with respect to V is trivial and give you no new information. You can just look at this result and see that there are no pressure terms because you held them constant so they fell out when differentiating earlier.

Differentiating by V while holding T constant tells you no new information.

 

[kntsy]

No it isn't. That isn't the "non-ideal gas equation." It is merely a derived relationship between U and T if V is held constant.
Because that step doesn't make sense. Start with:

$$\partial U = C_V \partial T$$

This equation is all you need to determine that internal energy is a function of only temperature at constant volume. Differentiating with respect to V is trivial and give you no new information. You can just look at this result and see that there are no pressure terms because you held them constant so they fell out when differentiating earlier.

Differentiating by V while holding T constant tells you no new information.

So why whenever we see dU, we always substiture it by CvdT with no regard to the nature of process? How can we be so sure that change of pressure or change in temperuture does not change internal energy?

 

[boneh3ad]

So why whenever we see dU, we always substiture it by CvdT with no regard to the nature of process?

I am not sure what you mean here. The equation $dU = C_V dT$ is derived straight from the physical laws, as you know. It gets substituted in because you can't directly measure $dU$, but you can directly measure $dT$, so it makes any equations more useful.

How can we be so sure that change of pressure or change in temperuture does not change internal energy?

Well clearly, changing temperature does change $dU$ since $dT$ shows up in the equation. For example, if you integrate that, you get:

$$\Delta U = C_V \left(T-T_0\right)$$

We know, then, that pressure can change the internal energy in this case because a change in pressure would change the temperature (remember, you held V constant).

 

[kntsy]

I am not sure what you mean here. The equation $dU = C_V dT$ is derived straight from the physical laws, as you know. It gets substituted in because you can't directly measure $dU$, but you can directly measure $dT$, so it makes any equations more useful.



Well clearly, changing temperature does change $dU$ since $dT$ shows up in the equation. For example, if you integrate that, you get:

$$\Delta U = C_V \left(T-T_0\right)$$

We know, then, that pressure can change the internal energy in this case because a change in pressure would change the temperature (remember, you held V constant).

why we have to hold V constant? Can't we change P,V,T at the same time and determine what happens in internal energy?
Can i just allowing small change in volume so that change in P and T is still in the same direction?

 

[boneh3ad]

You can't allow small changes in volume and still use $C_V$. The definition is that volume is constant.