Question

I just need help to solve a problem:

Show that (∂E/∂V)β, N + β(∂p/∂β)N, V = - p

PS: There is a bar over E and over p (this in both sides) - meaning that is an average.

Answer 1

Answer on Question #60826 - Physics - Molecular Physics | Thermodynamics

Show that $\left(\frac{\partial E}{\partial V}\right)_{\beta ,N} + \beta \left(\frac{\partial p}{\partial \beta}\right)_{N,V} = -p$

Solution:

$$d \beta = - \frac {\beta}{T} d T \Rightarrow \beta \left(\frac {\partial p}{\partial \beta}\right) = - \frac {\beta T}{\beta} \left(\frac {\partial p}{\partial T}\right) = - T \left(\frac {\partial p}{\partial T}\right) = - T \left(\frac {\partial S}{\partial V}\right)$$ $$d E = T d S - p d V \Rightarrow \frac {\partial E}{\partial V} = T \left(\frac {\partial S}{\partial V}\right) - p \Rightarrow \frac {\partial E}{\partial V} - T \left(\frac {\partial S}{\partial V}\right) = \frac {\partial E}{\partial V} + \beta \left(\frac {\partial p}{\partial \beta}\right) = - p$$ $$\frac {\partial E}{\partial V} = 0 \text{ (Joule's law)}$$ $$\beta \left(\frac {\partial p}{\partial \beta}\right) = - T \left(\frac {\partial S}{\partial V}\right) = - T \left(\frac {\partial p}{\partial T}\right) = - T \frac {R d T}{V d T} = - R \frac {T}{V} = - p$$ $$0 - p = - p \Rightarrow \left(\frac {\partial E}{\partial V}\right) + \beta \left(\frac {\partial p}{\partial \beta}\right) = - p$$

https://www.AssignmentExpert.com