T08: 1(a,b)
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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,
Hi, could you please do bigger the images? It is a little bit difficult to read it.
ResponderBorrar@Victor, you can do that in the editor. You just click on the figure, and choose between different options for its size. Here, "tamaño original". In fact, I just did that by myself.
ResponderBorrarYou have two small mistakes in part (b) -- the last two equations. Note for instance that the argument of the logarithm in the last equation is not intensive! Can you tell what are these mistakes? Can anybody else tell?
ResponderBorrar