T04.2 (b) & (c)
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,
In 4.2(b) you consider the accumulated number of states N(E), but you should consider the number of states with energy close to E, that is Sigma(E,J), as clearly said in the assignment. In the expresion for Sigma(E,J), n is not the number of spins, it is the number of links between the spins.
ResponderBorrarIt also not correct that the accumulated number of states N(E) is 2^N, independent of energy. It seems you did not understand that part of the lecture notes.
Ok, I can see that I'm wrong
ResponderBorrarI can have another chance to make it right?
Yes, please
ResponderBorrar