Deriving the Relationship between Pressure and Energy at Constant Entropy

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SUMMARY

The discussion focuses on deriving the relationship between pressure and energy at constant entropy, specifically proving that \(\left(\frac{\partial U}{\partial V}\right)_{S} = \sum n_{j} \frac{\partial \epsilon_{j}}{\partial V}\). The user clarifies that while the total internal energy \(U\) is expressed as \(U = \sum n_{j} \epsilon_{j}\), the partial derivative must be taken at constant entropy \(S\) and variable volume \(V\). It is established that the number of particles \(n_{j}\) remains constant, while the energy levels \(\epsilon_{j}\) can change with volume, necessitating the differentiation of \(\epsilon_{j}\) with respect to \(V\).

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spaghetti3451
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Hi,

I need to prove that [tex]\left(\frac{\partial U}{\partial V}\right)_{S} = \sum n_{j} \frac{\partial \epsilon_{j}}{\partial V}[/tex].
 
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In the bracket on the LHS, you have partial dU by dV. In the denominator on the RHS, you have partial dV.

This is my reasoning.

[tex]U = \sum n_{j} \epsilon_{j}[/tex].

We need to take the partial derivative of U at constant S.

At constant S and variable V, the number of particles at each energy level does not change but their energy levels may change. (I don't understand why this should be the case.) This means that [tex]n_{j}[/tex] does not vary, but [tex]\epsilon_{j}[/tex] does. Therefore, the derivative of [tex]\epsilon_{j}[/tex] is taken with respect to V?
 

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