Partial differential of U at constant temperature?

In summary: For ideal gas I get (∂u/∂v) =RT/(v-b) -PDo you mean this?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.Understand already.
  • #1
Outrageous
374
0
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
 
Science news on Phys.org
  • #2
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.
 
  • #3
What you need is this. You'll find its derivation in any standard thermodynamics text.

[itex]\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V} - P.[/itex]

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.
 
Last edited:
  • #4
Philip Wood said:
What you need is this. You'll find its derivation in any standard thermodynamics text.

[itex]\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V} - P.[/itex]

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?
 
  • #5
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.
 
  • #6
Philip Wood said:
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?[itex]\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V} - P.[/itex]
For ideal gas I get (∂u/∂v) =RT/(v-b) -P
Do you mean this?
 
  • #7
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.
 
  • #8
Understand already.
Thank you
 
  • #9
Then you deduced that [itex]\left(\frac{\partial U}{\partial V}\right)_T = 0[/itex] for the ideal gas?
 
  • #10
Seem like I am having exam here
[itex]\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V} - P.[/itex]

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:smile:
 
  • #11
I'm so sorry about the exam! Just trying to help and got a bit overenthusiastic. Good luck with your studies!
 

1. What is a partial differential of U at constant temperature?

A partial differential of U at constant temperature refers to the change in the internal energy of a system when one specific variable is changed while keeping the temperature constant. It is represented by the symbol ∂U/∂x, where x is the variable that is being changed.

2. How is the partial differential of U at constant temperature calculated?

The partial differential of U at constant temperature is calculated by taking the derivative of the internal energy function with respect to the variable that is being changed, while keeping the temperature constant. This can be done using mathematical techniques such as the chain rule or the product rule.

3. What is the importance of the partial differential of U at constant temperature?

The partial differential of U at constant temperature is important in thermodynamics as it helps us understand how the internal energy of a system changes in relation to specific variables while keeping the temperature constant. This information is crucial in determining the behavior and properties of a system.

4. Can the partial differential of U at constant temperature be negative?

Yes, the partial differential of U at constant temperature can be negative. This indicates that the internal energy of a system decreases as the variable being changed increases, while keeping the temperature constant. This is known as a negative relationship between the internal energy and the variable.

5. How does the partial differential of U at constant temperature differ from the total differential of U?

The partial differential of U at constant temperature only considers the change in internal energy when one specific variable is changed while keeping the temperature constant. On the other hand, the total differential of U takes into account the change in internal energy when multiple variables are changed simultaneously. The total differential is represented by the symbol dU, while the partial differential is represented by ∂U.

Similar threads

Replies
1
Views
533
  • Thermodynamics
Replies
4
Views
1K
Replies
1
Views
826
Replies
15
Views
1K
Replies
23
Views
1K
  • Thermodynamics
Replies
19
Views
1K
Replies
3
Views
958
Replies
13
Views
741
  • Thermodynamics
Replies
4
Views
4K
  • Thermodynamics
Replies
20
Views
1K
Back
Top