Fluctuations
I did this in the case of the ideal gas and the real gas.
Note: I couldn't upload images in the post comments, that's why I did it in this post.
Work for an ideal gas in a 3D container (corrected)
Using the differential al work calculated on exercise 6.2: 1. How can we use the result to obtain DW for an ideal gas in a 3D container? What is the interpretation of the parameter "a" in that case. Solution. The differential of work calculated on exercise 6.2 is Here, d represents the dimension of the box. So The parameter a is the lenght, according to equation (67). Because we are working on 3 dimensions, a³=V . This might come from the definition of entropy*. 2. What do we get for DW for d=3 and n --> infinity? And how is this result related to the expected form DW = N k_B T dV/V? Solution. On the last expression we get DW for three dimensional case. Now, applying the limit on infinity, we expect that the differential of work would become on well potential one. So,
I can see that you compute the first and the second derivative of ln(Omega). But what does it mean? The result of your calculation should be what?
ResponderBorrarEven in the case of the first derivative I am confused. Acording to our formula, pressure should be equal to beta^{-1} partial-derivative-wiith-respect to Volume of ln(Omega). But it seems you forgot beta^{-1}