Heat capacity and its relation with internal energy

Click For Summary

Discussion Overview

The discussion revolves around the concept of heat capacity in thermodynamics, specifically the relationship between heat capacity at constant volume (Cv) and internal energy (U). Participants explore the definitions and derivations related to Cv, the distinction between dQ and dU, and the implications of these concepts in different thermodynamic conditions.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the transition from using dQ to dU in the definition of Cv, seeking clarification on the underlying reasoning.
  • Another participant emphasizes the importance of distinguishing between dQ and δQ, noting that δQ is not a state variable and should be treated differently in thermodynamic equations.
  • A participant explains the fundamental equations of thermodynamics and how they relate to the definitions of Cv and Cp, providing a mathematical derivation of Cv as the partial derivative of U with respect to T at constant volume.
  • It is noted that the statement about U being a function of T is only true for ideal gases, which was experimentally validated by Joule, indicating that this assumption may not apply universally.
  • Another participant clarifies that the first law of thermodynamics can be expressed in terms of Cv, leading to the conclusion that Cv can be defined as the partial derivative of U with respect to T at constant volume.

Areas of Agreement / Disagreement

Participants generally agree on the definitions and mathematical relationships discussed, but there is some initial confusion regarding the use of dQ versus dU. The discussion reflects a process of clarification rather than a resolution of differing views.

Contextual Notes

The discussion highlights the complexities of thermodynamic definitions and the importance of context, particularly regarding ideal gases versus other systems. Participants acknowledge the potential for misunderstanding in the application of these concepts.

Who May Find This Useful

This discussion may be useful for students and practitioners in thermodynamics, particularly those grappling with the concepts of heat capacity, internal energy, and the distinctions between different thermodynamic variables.

cooper607
Messages
49
Reaction score
0
hi everyone,
in thermodynamics, when we calculate the heat capacity in constant volume, we assume Cv=dQ/dT..

well, but at isothermal condition suddenly they came up with Cv=dU/dT...
so i am getting stuck with this concept how they replace dQ with dU?

i know U= internal energy is only a function of T, but what is the explanation behind it? please help with the derivation and clarification...

advanced regards.
 
Science news on Phys.org
You have to be careful with that "dQ" - there is no such thing as "Q" so there is no such thing as "dQ". Usually it is written "\delta Q", which is an abbreviation for "T dS" to emphasize this fact. You only use the "d" in front of state variables, and there is no "Q" state variable. A state variable X obeys \oint dX=0. In other words the integral around a closed path of a state variable is zero, and the integral between two points is not dependent on the path. \oint \delta Q is really \oint T dS and that's not necessarily zero.

The correct way to define the C's are TdS=C_V dT at constant volume and TdS=C_P dT at constant pressure.

You can write the fundamental equation (at constant number of particles) as dU=TdS-PdV and since, by the chain rule, dU=\left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV you can see that at constant volume, TdS= \left(\frac{\partial U}{\partial T}\right)_V dT so that C_V=\left(\frac{\partial U}{\partial T}\right)_V

You can also write the fundamental equation (at constant number of particles) as dH=TdS+VdP where H is the enthalpy (H=U+PV) and since, by the chain rule, dH=\left(\frac{\partial H}{\partial T}\right)_P dT + \left(\frac{\partial H}{\partial P}\right)_T dP you can see that at constant pressure, TdS= \left(\frac{\partial H}{\partial T}\right)_P dT so that C_P=\left(\frac{\partial H}{\partial T}\right)_P
 
i know U= internal energy is only a function of T, but what is the explanation behind it? please help with the derivation and clarification...

I think you are mixing up several things you have been told.
Don't worry this is not uncommon there is a lot to get hold of in thermodynamics.

So firstly your statement about U is only true for ideal gasses. It is one of the possible definitions of an ideal gas and was proved experimentally by Joule. However it is not relevant here.


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

As regards your query about Cv,

It is not necessary to invoke entropy and considering your definition of Cv you may not have met entropy anyway.

Since your version of the first law probably reads something like

dU = δQ - pdV and your definition of Cv is


{C_v} = {\left( {\frac{{\delta Q}}{{\partial T}}} \right)_V}

(RAP is correct it is not a good idea to use dQ)

So δQ = CvdT

Substituting into the first law

dU = CvdT - pdV

But at constant volume dV = 0

dU = CvdT

rearranging


{C_v} = {\left( {\frac{{dU}}{{dT}}} \right)_V}
 
oh now i got it... i was mixing up the things really.. thanks for the great help and explanation rap and studiot...

regards
 
oh now i got it... i was mixing up the things really.. thanks for the great help and explanation rap and studiot...

regards
 
Glad you got it sorted

:wink:
 

Similar threads

Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
18K
  • · Replies 2 ·
Replies
2
Views
3K