Check the blog REGULARLY

Some of you have been replying too late, when asked to post their results. Please check the blog regularly, at least Monday, Wednesday and Friday (during the regular hours of the course). I will not only ask you to present solutions to the homework, but also ask you to answer additional questions or do small exercises, as if we would have our regular class session.
At the end of the semester, your participation in the coures (via posts to this blog) will count.

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Work for an ideal gas in a 3D container (corrected)

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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,   
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homework T06.3, a) and b)

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a) the code is: import numpy as np import matplotlib.pyplot as plt R=8.314 T=300 N=1 Va=1 Vb=2*Va g=5/3 Pa=N*R*T/Va Pb=Pa*(Va**g)/(Vb**g) l=np.arange(0,1,0.001) P1=l*Pb+(1-l)*Pa V1=l*Vb+(1-l)*Va V2=np.zeros(1000) P2=np.zeros(1000) for i in range (0,len(V2)):     V2[i]=Vb-i/1000     P2[i]=Pa*(Va**g)/(V2[i]**g)   c=P1[500]*(V1[500])**g V3=np.zeros(1000) P3=np.zeros(1000) for i in range (0,len(V2)):     V3[i]=Vb-i/1000     P3[i]=c/(V3[i]**g) plt.plot(V1,P1) plt.plot(V2,P2) plt.plot(V3,P3) plt.plot(V1[500],P1[500],"o") plt.show() and the plot is: pd: i'm sorry i don't know how to pass some latex to html so i put it like images
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