Partial Derivatives of U: Solving for Unknown Variables

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Homework Help Overview

The problem involves the application of partial derivatives in the context of thermodynamics, specifically relating to the internal energy function U and its dependence on entropy S and volume V. The discussion centers around the transformation of differentials into partial derivatives.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between differentials and partial derivatives, questioning the validity of substituting 'd' with '∂' in the context of the equation dU = TdS - PdV. There is discussion about expressing dU in terms of new variables X and Y.

Discussion Status

Several participants provide insights into the invariance of the first differential and its implications for expressing U in terms of different variables. There is an ongoing exploration of how to correctly manipulate the equations without reaching a definitive conclusion.

Contextual Notes

Participants are working within the constraints of thermodynamic principles and the definitions of differentials and partial derivatives, with some uncertainty regarding the assumptions made in the problem setup.

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Homework Statement



The problem is attached in the picture.


The Attempt at a Solution



I'm aware that:

dU = T dS - P dV

∫ dU = ∫ (T) dS - ∫ P dV

Are they assuming that T, P are constant so

U = TS - PV

∂U/∂X = T (∂S/∂X) - P (∂V/∂X)


Or is there a way to directly change dU to ∂U?
 

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dU is a differential. The first differential is invariant, so if you know what it looks like in terms of dS and dV, you can express it in terms of dX and dY if you express the former via the latter. That's exactly what they do.
 
voko said:
dU is a differential. The first differential is invariant, so if you know what it looks like in terms of dS and dV, you can express it in terms of dX and dY if you express the former via the latter. That's exactly what they do.

I'm not sure what you mean...My question is how did they "convert"

dU = T dS - P dV

into

(∂U/∂X) = T (∂S/∂X) - P (∂V/∂X) ?

Are you allowed to simply replace each 'd' by '∂' ?
 
The first differential is invariant. That means that for a function U(S, V), dU = TdS - PdV, if S and T are functions of (X, Y), with their differentials dS = idX + jdY, dV = kdX + ldY, we can simply plug these dS into dV into the original equation and we will get dU = T(idX + jdY) - P(kdX + ldY) = (Ti - Pk)dX + (Tj - Pl)dY.

Now, it is known that if we have a differential dU = AdX + BdY, then A is the partial derivative with respect to X, and B is the partial derivative with respect to Y. That is also true for i, j, k, l above.

Now combine all of this.
 
voko said:
The first differential is invariant. That means that for a function U(S, V), dU = TdS - PdV, if S and T are functions of (X, Y), with their differentials dS = idX + jdY, dV = kdX + ldY, we can simply plug these dS into dV into the original equation and we will get dU = T(idX + jdY) - P(kdX + ldY) = (Ti - Pk)dX + (Tj - Pl)dY.

Now, it is known that if we have a differential dU = AdX + BdY, then A is the partial derivative with respect to X, and B is the partial derivative with respect to Y. That is also true for i, j, k, l above.

Now combine all of this.

Got it! Thanks!
 

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