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I know that the entropy of isothermal-isobaric ensemble is given by:

$$S = -k \sum_{i=1}^M p_i \ln p_i \quad \textrm{where $p_i$ must be normalized} \quad \sum_{i=1}^M p_i = 1 \, .$$

The average energy is

$$\sum_{i=1}^M p_i \varepsilon_i = \langle E \rangle$$

and the average volume is

$$\sum_{i=1}^M p_i V_i = \langle V \rangle \, .$$

Some authors say that the probability of finding and state $i$ is given by

$$p_i = \frac{1}{Q} \, \exp (-\beta \varepsilon_i - \gamma V_i)$$

where $\beta$ and $\varepsilon$ are Lagrange multipliers.

I need to physically interpret these two terms. I compared

$$S = k \, \ln \, Y + k \beta \langle E \rangle + k \gamma \langle V \rangle$$

with

$$S = - \frac{G}{T} + \frac{\langle E \rangle}{T} + \frac{P \langle V \rangle}{T}$$

Where I can obtain that

$$G = -kT \, \ln Y, \quad \gamma = \frac{P}{kT} \quad \textrm{and} \quad \beta = \frac{1}{kT} \, .$$

How can I obtain this equation using Lagrange?

$$p_i = \frac{1}{Q} \, \exp (-\beta \varepsilon_i - \gamma V_i)$$

I need some idea to open this equation, given that the physical interpretation of this parameters were done.

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I am not sure I clearly get the question. You can derive the expression of $p_i$ by maximizing the entropy $S$ under the constraints of your system (here being fixed average energy and volume) by Lagrange multipliers. Solving the saddle points equations leads to your result. Is that what you need ?

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  • $\begingroup$ Yes... It is! I'll try to do that... any other advice will be very helpful. $\endgroup$ – Rodrigo Monteiro Aug 24 '16 at 14:57

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