Partial differential of U at constant temperature?

[Outrageous]

U is internal energy
T is temperature
v is volume
U(T,v)
My book say (∂u/∂v) at constant temperature can be calculated from the equation of state.
How to calculate it?

Thank you

 

[Studiot]

Outrageous, since you consistently don't acknowledge posts by those offering help in your threads, I see no reason to post further help.

How to calculate it?

Try reading your book.

 

[Philip Wood]

What you need is this. You'll find its derivation in any standard thermodynamics text.

$\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V} - P.$

I agree, though, with Studiot. It is impolite not to acknowledge attempts to help. It's perfectly in order to add that you didn't understand the post, but better if you can say exactly what you didn't understand.

 

[Outrageous]

What you need is this. You'll find its derivation in any standard thermodynamics text.

$\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V} - P.$

I agree, though with Studiot. It is impolite not to acknowledge attempts to help. It's perfectly in order to add that you didn't understand the post, but better if you can say exactly what you didn't understand.

Thanks for replying. I am sorry , how can I acknowledge?
So can be calculated from the equation of state mean the answer will only have P,v ,T?

 

[Philip Wood]

You've acknowledged my post with the thanks.

P, V and T will be the only variables (for a given number of moles). You should try out the equation I gave you on (a) an ideal gas (b) a V der W gas. Then you'll get a better understanding of how to use it.

 

[Outrageous]

You've acknowledged my post with the thanks.

P, V and T will be the only variables (for a given number of moles). You should try out the equation I gave you on (a) an ideal gas (b) a V der W gas. Then you'll get a better understanding of how to use it.

So every time when I get the answer I think is correct then I should acknowledge by saying thank? That is all?$\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V} - P.$
For ideal gas I get (∂u/∂v) =RT/(v-b) -P
Do you mean this?

 

[Philip Wood]

You're right so far with the 'ideal gas' but what does RT/(v-b) - P equal? Something VERY simple!

[Incidentally, take b as zero for an ideal gas, but it won't affect your result in this case.]

Then go for the V der W gas.

 

[Outrageous]

Understand already.
Thank you

 

[Philip Wood]

Then you deduced that $\left(\frac{\partial U}{\partial V}\right)_T = 0$ for the ideal gas?

 

[Outrageous]

Seem like I am having exam here
$\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V} - P.$

To answer my question ,
Substitute U(T,v) into Tds=du+Pdv,
Then compare the result with s(T,v),
We will get (∂s/∂T) at constant volume ,and (∂s/∂v) at constant temperature.
Since they are equation of state ,then we know their second derivative will be the same.
Then we will get the formula given by you.

To prove (∂u/∂v) at constant T =0, for ideal gas
(∂u/∂v) =RT/(v-b) -P , I get (actually is from van der Waals ),
put b=0, and since it is ideal gas,P=RT/v
Substitute these two in, I will get the answer.
Correct?

Thank you so much for guiding me

 

[Philip Wood]

I'm so sorry about the exam! Just trying to help and got a bit overenthusiastic. Good luck with your studies!