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Akalash72

Akalash72

dickofan

Akalash72

dickofan

Problem 3.13



> P:=(v,T)->8*T/(3*v-1)-3/(T*v^2);



                                  8 T      3  





                 P := (v, T) -> ------- - ----





                                3 v - 1      2





                                          T v 



> p1:=(nu,epsilon)=P(1+nu,1+epsilon)-1;



                      8 (1 + epsilon)              3               





p1 := (nu, epsilon) = --------------- - ----------------------- - 1





                         2 + 3 nu                             2    





                                        (1 + epsilon) (1 + nu)     



> aux:=series(8*(1+epsilon)/(2+3*nu)-3/(1+epsilon)/(1+nu)^2-1,nu=0,4):aux;



/                     3     \   /                      6     \   





|3 + 4 epsilon - -----------| + |-6 - 6 epsilon + -----------| nu





\                1 + epsilon/   \                 1 + epsilon/   











     /      9 1                    \   2   /   12 1       27   27        \   3





   + |- ----------- + 9 + 9 epsilon| nu  + |----------- - -- - -- epsilon| nu 





     \  1 + epsilon                /       \1 + epsilon   2    2         /    











      /  4\





   + O\nu /



> aux1:=(epsilon)->3+4*epsilon-3/(1+epsilon)+(-6-6*epsilon+6/(1+epsilon*nu+(-9*1/(1+epsilon)+9+9*epsilon)*nu^2 +(12*1/(1+epsilon)-27/2-27/2*epsilon)*nu^3;



                                        3     





aux1 := epsilon -> 3 + 4 epsilon - -----------





                                   1 + epsilon











     /                      6     \      /      9 1                    \   2





   + |-6 - 6 epsilon + -----------| nu + |- ----------- + 9 + 9 epsilon| nu 





     \                 1 + epsilon/      \  1 + epsilon                /    











     /   12 1       27   27        \   3





   + |----------- - -- - -- epsilon| nu 





     \1 + epsilon   2    2         /    



> aux2:=series(aux1(epsilonepsilon=0,2);



          3   3   /    51   3        2        \            /       2\





aux2 := - - nu  + |7 - -- nu  + 18 nu  - 12 nu| epsilon + O\epsilon /





          2       \    2                      /                      



Simplified equation of state



> p:=(nu,epsilon)->-3/2*nu^3+(7-12*nu)*epsilon;



                              3   3                      





      p := (nu, epsilon) -> - - nu  + (7 - 12 nu) epsilon





                              2                          



Critical point - epsilon=0; 



> d1:=(nu,epsilon)->subs(y=nu,diff(p(y,epsilony: d1(nu,epsilon);



                        9   2             





                      - - nu  - 12 epsilon





                        2                 



The curve is symmetric. That leads to vg=-vl=v



> solve(p(v,epsilon)= p(-v,epsilonv);;



           0, 2 sqrt(-2 epsilon -2 sqrt(-2 epsilon)



Critical exponent beta=1/2



Along the coexitence curve, v^2=-8epsilon. exponent gamma=1



> 



Function of pressure



> P1:=(rho,T)->8*T*rho/(3-rho)-3*rho^2/T: 



Near the critical point







> P1(1+xi,1);



                    8 (1 + xi)             2





                    ---------- - 3 (1 + xi) 





                      2 - xi                



> series(8*(1+xi)/(2-xi)-3*(1+xi)^2,xi=0);



                   3   3   3   4   3   5    /  6\





               1 + - xi  + - xi  + - xi  + O\xi /





                   2       4       8             



the exponent delta=3



> 



  

Akalash72

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