Dudas sobre la tarea 7, aquí T07.3

Cuando traten de obtener la condición de las derivadas cruzadas (para que DQ/T sea un diferencial exacto) tienen que usar "T" y "V" como variables independientes. Es decir

dU = dU(T,V) = (partial U/partial T) dT + (partial U/partial V) dV

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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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