T07.3(a,b)
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,
At the bottom of page 10, you write "Applying the Clairaut-theorem"? I never heard that name? Can you say where it comes from?
ResponderBorrarIn any case, part (a) is absolutely correct.
The first task of part (b) is also correct. However, then you do something which is conceptually wrong. The problem 7.3 is about a real gas. So you should not assume that the gas is ideal. After all, we want to use the formula for calculating the internal energy for a real gas. Hence, on the LHS you should keep U(T,V), on the RHS you can -- in principle -- calculate the integral from the knowledge of the equation of state. So the only unknown part is U(T,V_0). So now, we consider the limit V_0 to infty. Then the physical question arises, what happens to U(T,V_0) which is the internal energy of the real gas, if V_0 goes to infinity. You should have found an answer to that question.
I had read it here: https://es.wikipedia.org/wiki/Teorema_de_Clairaut
ResponderBorrarOh, I see, I used what we want to demonstrate instead to do it. Thank you, professor, I'll check it.