Partial integration misunderstanding

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SUMMARY

The discussion focuses on the calculation of the partial derivative of internal energy (U) with respect to temperature (T) while holding pressure (p) constant, expressed as \(\left(\frac{\partial U}{\partial T}\right)_p\). The correct expression derived is \(\left(\frac{\partial U}{\partial T}\right)_p=\left(\frac{\partial U}{\partial T}\right)_V+\left(\frac{\partial U}{\partial V}\right)_T\left(\frac{\partial V}{\partial T}\right)_p\). The participants clarify that this involves partial differentiation rather than integration, emphasizing the importance of treating U as a function of two variables initially.

PREREQUISITES
  • Understanding of thermodynamic concepts, specifically internal energy (U).
  • Familiarity with partial differentiation and mixed partial derivatives.
  • Knowledge of the relationships between temperature (T), volume (V), and pressure (p) in thermodynamics.
  • Experience with mathematical tools such as Mathematica for symbolic computation.
NEXT STEPS
  • Study the principles of thermodynamic derivatives, focusing on \(\left(\frac{\partial U}{\partial T}\right)_p\).
  • Learn about the application of partial differentiation in thermodynamics.
  • Explore the use of Mathematica for solving thermodynamic equations and derivatives.
  • Investigate the implications of holding variables constant in thermodynamic functions.
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Students and professionals in thermodynamics, particularly those studying internal energy calculations, as well as anyone looking to deepen their understanding of partial differentiation in physical sciences.

dingo_d
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Homework Statement


I'm doing some thermodynamics and I didn't quite get one thing:

I have a function: U=U(T,\ V(T,p)) and I need to find the expression:

\left(\frac{\partial U}{\partial T}\right)_p

The Attempt at a Solution



Now what I found in the book and what Mathematica gave me is:

\left(\frac{\partial U}{\partial T}\right)_p=\left(\frac{\partial U}{\partial T}\right)_V+\left(\frac{\partial U}{\partial V}\right)_T\left(\frac{\partial V}{\partial T}\right)_p


But I can't see why that is :\ I tried looking with the formulas for mixed partial derivatives but I must be missing sth out :\

Can someone point me in the right direction?
 
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dingo_d said:

Homework Statement


I'm doing some thermodynamics and I didn't quite get one thing:

I have a function: U=U(T,\ V(T,p)) and I need to find the expression:

\left(\frac{\partial U}{\partial T}\right)_p

The Attempt at a Solution



Now what I found in the book and what Mathematica gave me is:

\left(\frac{\partial U}{\partial T}\right)_p=\left(\frac{\partial U}{\partial T}\right)_V+\left(\frac{\partial U}{\partial V}\right)_T\left(\frac{\partial V}{\partial T}\right)_p


But I can't see why that is :\ I tried looking with the formulas for mixed partial derivatives but I must be missing sth out :\

Can someone point me in the right direction?
Whilst it isn't important to the question, this isn't partial integration, it is partial differentiation.

It is useful in this case to simplify the problem a little first. Start by treating U as a function of two variables. If

U = U\left(T,V\right) ,

(i.e. ignore the dependence of V on T) can you work out

\left(\frac{\partial U}{\partial T}\right)_p

?
 
Oh doh! diferentiation! Woops, my bad.

Is it:

<br /> \left(\frac{\partial U}{\partial T}\right)_p=\left(\frac{\partial U}{\partial V}\right)_p\left(\frac{\partial V}{\partial T}\right)_p<br />?

Ok so I have:<br /> dU=\left(\frac{\partial U}{\partial V}\right)_TdV+\left(\frac{\partial U}{\partial T}\right)_VdT<br />

Now I can diferentiate that with respect to T at constant p:<br /> \left(\frac{\partial U}{\partial T}\right)_p=\left(\frac{\partial U}{\partial V}\right)_T\left(\frac{\partial V}{\partial T}\right)_p+\left(\frac{\partial U}{\partial T}\right)_V\underbrace{\left(\frac{\partial T}{\partial T}\right)_p}_1<br />

Oh! That's it! XD Thnx :D
 
Last edited:

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