Física Estadística

Due to the problems I had with preparing the homework 08, and due to several requests, the new deadline for submitting that assignment is Tuesday, 5'th of may.

For a given real gas, with known equation of state, we obtained an expression for its internal energy, U(T,V). The only unknown part in this expression 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. Has anybody found an answer?

Hello I want to ask you if we could have more time to hand the homework. Maybe for the next tuesday, May 5th. Thank you beforehand. Cheers. 

my answer to b) i think is very poor and unfinished

   

T07.1 (a)    : Raúl (but please expand you final result and simplify further) T07.1 (b,c) : Guillermo T07.2 (a,b) : Jesús Alberto T07.2 (c)    : Sebastian T07.3 (a,b) : Bianca T07.3 (c)    : Raúl

With my previous post, I hope you can verify, correct and possibly simplify your results for problem T06(b). Then it will not be so difficult to solve part (c) also. For the whole cycle, we have where Q_abs is the amount of heat absorbed from the environment, and Q_del is the amount of heat delivered to the environment. Note the minus-sign and make sure you understand the reason for it to appear in this expression. So as an excercise for the next class, use these formulas to calculate the work performed during the cycle and the efficiency of the cycle.

The expression for lambda, which yields the point on the straight line where the system changes from absorbing to delivering heat to the environment is not as complicated as it seems. But one has to simplify the expression, and possibly introduce some short-form notation:     Here Delta_p = p_B - p_A, Delta_V = V_B - V_A and c_p = R - c_V. For the coordinates of the point C, we then obtain Please check, whether your expressions coincide with this result.

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

Citlalli and Alan: Can you please upload your solutions for the problems 1(a,b) and 3(a,b).  You may arrange by yourself who wants to post which solution. Also please, anybody who thinks she has the correct solution for problem 3(c). Please post it

I am still working on two additional problems for Tarea 08. But meanwhile I want to share with you the first problem:

For the work in the exercise 6.2 I found that the work differential for an ideal fas in a powe-potential tramp was: We can use this to get the work differential for an ideal gas in a 3D container by subtitude d=3 thatis the dimensions, obtaining: Seeing this we can say that the parameter a is going to represent the volume: know, if we take the limit when n goes to infinity, we can write:  That applying the limit is: or that is also equal to: That is the work differential for an ideal gas in a 3D container, which relates to the equation mentioned in the exercise as it was for a 1D container.

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? As I wrote on my homework: Here a= 0.7. For values of 𝑥 greater than 𝑎 the curves develop higher slopes towards infinity. This means 𝑉(𝑥) tends to infinity faster after this point, though greater values of 𝑛 also make 𝑉(𝑥) tend to zero when −𝑎 < 𝑥 < 𝑎. On the other hand, 𝑒𝑡𝑎 changes the magnitude (the scale) of the potential but not its rate of growth. Thus it is concluded that as 𝑛 → ∞, 𝑉(𝑥) → ∞ when |𝑥| ≥ 𝑎 and 𝑉(𝑥) → 0 when |𝑥| < 𝑎. The work differential found for this system is: Taking all of this into account it can be understood that a is an effective limit (or wall) of the system for big values of n. Thus if we were to form a 3-dimensional container of equal length in all dimensions we could write DW with a ³= V instead of a .  To write a valid form of the complete differentia...