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Mathematics Calculus Equation Solving Wolfram Language

Using Solve but cannot get an answer

George Ip

Posted 2 months ago

Good morning, I am trying to solve a homework problem but could not get Mathematica to get an answer. Any suggestion? Clear["Global`"] R = 8.314 ;(J/mol/K) N1Ai = 0.5; N1Bi = 1; N2A = 0.75; N2B = 0.5; V = 5;(Liters) Teq = 272.7; (K) Uf = 9353.25;(J) dSdN1A = D[(N1A + N2A)A + (N1A + N2A)R Log[UA^(3/2)V/((N1A + N2A)^(5/2))] - N1ARLog[N1A/(N1A + N2A)] - N2ARLog[N2A/(N1A + N2A)], N1A]; UA = ((3/2)(N1A + N2A)RTeq); dSdN1B = D[(N1B + N2B)A + (N1B + N2B)R* Log[UB^(3/2)V/((N1B + N2B)^(5/2))] - N1BRLog[N1B/(N1B + N2B)] - N2BRLog[N2B/(N1B + N2B)], N1B]; UB = ((3/2)(N1B + N2B)RTeq); dSdN1A dSdN1B Solve[dSdN1A == dSdN1B && N1A + N1B == N1Ai + N1Bi, {N1A, N1B}] Here are the outputs dSdN1A= -20.785 + A + 6.2355/(0.75 + N1A) - 8.314 (0.75 + N1A) (-(N1A/(0.75 + N1A)^2) + 1/(0.75 + N1A)) + 8.314 Log[991630./(0.75 + N1A)] - 8.314 Log[N1A/(0.75 + N1A)] dSdN1B= -20.785 + A + 4.157/(0.5 + N1B) - 8.314 (0.5 + N1B) (-(N1B/(0.5 + N1B)^2) + 1/(0.5 + N1B)) + 8.314 Log[991630./(0.5 + N1B)] - 8.314 Log[N1B/(0.5 + N1B)] Solve::svars: Equations may not give solutions for all "solve" variables. {{N1B -> 1.5 - 1. N1A}} Attachments: 2_8_1_a.nb

Good morning,
I am trying to solve a homework problem but could not get Mathematica to get an answer. Any suggestion?

Clear["Global`*"]

R = 8.314 ;(*J/mol/K*)
N1Ai = 0.5; 
N1Bi = 1;
N2A = 0.75;
N2B = 0.5;
V = 5;(*Liters*)
Teq = 272.7; (*K*)
Uf = 9353.25;(*J*)


dSdN1A = 
  D[(N1A + N2A)*A + (N1A + N2A)*R*
     Log[UA^(3/2)*V/((N1A + N2A)^(5/2))] - 
    N1A*R*Log[N1A/(N1A + N2A)] - N2A*R*Log[N2A/(N1A + N2A)], N1A];

UA = ((3/2)*(N1A + N2A)*R*Teq);

dSdN1B = 
  D[(N1B + N2B)*A + (N1B + N2B)*R*
     Log[UB^(3/2)*V/((N1B + N2B)^(5/2))] - 
    N1B*R*Log[N1B/(N1B + N2B)] - N2B*R*Log[N2B/(N1B + N2B)], N1B];

UB = ((3/2)*(N1B + N2B)*R*Teq);

dSdN1A
dSdN1B

Solve[dSdN1A == dSdN1B && N1A + N1B == N1Ai + N1Bi, {N1A, N1B}]

Here are the outputs

dSdN1A=

-20.785 + A + 6.2355/(0.75 + N1A) - 
 8.314 (0.75 + N1A) (-(N1A/(0.75 + N1A)^2) + 1/(0.75 + N1A)) + 
 8.314 Log[991630./(0.75 + N1A)] - 8.314 Log[N1A/(0.75 + N1A)]

dSdN1B=

-20.785 + A + 4.157/(0.5 + N1B) - 
 8.314 (0.5 + N1B) (-(N1B/(0.5 + N1B)^2) + 1/(0.5 + N1B)) + 
 8.314 Log[991630./(0.5 + N1B)] - 8.314 Log[N1B/(0.5 + N1B)]

Solve::svars: Equations may not give solutions for all "solve" variables.

{{N1B -> 1.5 - 1. N1A}}

POSTED BY: George Ip

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 2 months ago

You forgot the multiplication star in `N1BRLog` and `N1ARLog`. A part from that, I get different output.

POSTED BY: Gianluca Gorni

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 2 months ago

Mathematica ignores your first equation, I wonder why. However, a plot suggests that your first equation is simpler than it may appear from the formula: ContourPlot[{dSdN1A == dSdN1B, N1A + N1B == N1Ai + N1Bi}, {N1A, -1, 2}, {N1B, -1, 2}] I would simplify your equations first: dSdN1A == dSdN1B; Rationalize[%, 0] FullSimplify[%, N1A > 0 && N1B > 0] eq1New = ApplySides[Exp, %] and N1A + N1B == N1Ai + N1Bi; eq2New = Rationalize[%] and finally Solve[{eq1New, eq2New}]

Mathematica ignores your first equation, I wonder why. However, a plot suggests that your first equation is simpler than it may appear from the formula:

ContourPlot[{dSdN1A == dSdN1B, N1A + N1B == N1Ai + N1Bi},
 {N1A, -1, 2}, {N1B, -1, 2}]

I would simplify your equations first:

dSdN1A == dSdN1B;
Rationalize[%, 0]
FullSimplify[%, N1A > 0 && N1B > 0]
eq1New = ApplySides[Exp, %]

and

 N1A + N1B == N1Ai + N1Bi;
eq2New = Rationalize[%]

and finally

Solve[{eq1New, eq2New}]

POSTED BY: Gianluca Gorni

Bill Nelson

Posted 2 months ago

Changing Solve to Reduce and no other changes Reduce[dSdN1inA==dSdN1inB&&N1A+N1B==N1Ai+N1Bi, {N1A,N1B}] returns a warning that can be ignored Reduce: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result. and then N1A==0.75&&N1B==0.75

Changing Solve to Reduce and no other changes

Reduce[dSdN1inA==dSdN1inB&&N1A+N1B==N1Ai+N1Bi, {N1A,N1B}]

returns a warning that can be ignored

Reduce: Reduce was unable to solve the system with inexact coefficients.
The answer was obtained by solving a corresponding exact system and numericizing the result.

and then

N1A==0.75&&N1B==0.75

POSTED BY: Bill Nelson

George Ip

Posted 2 months ago

Can you share your output? I made sure the multiplication stars are there but it still gives me the same output. I think when I make this post it took out some of the stars. See my attached code. Clear["Global`"] R = 8.314 ;(J/mol/K) N1Ai = 0.5; N1Bi = 1; N2A = 0.75; N2B = 0.5; V = 5;(Liters) Teq = 272.7; (K) Uf = 9353.25;(J) dSdN1A = D[(N1A + N2A)A + (N1A + N2A)R Log[UA^(3/2)V/((N1A + N2A)^(5/2))] - N1ARLog[N1A/(N1A + N2A)] - N2ARLog[N2A/(N1A + N2A)], N1A]; UA = ((3/2)(N1A + N2A)RTeq); dSdN1B = D[(N1B + N2B)A + (N1B + N2B)R* Log[UB^(3/2)V/((N1B + N2B)^(5/2))] - N1BRLog[N1B/(N1B + N2B)] - N2BRLog[N2B/(N1B + N2B)], N1B]; UB = ((3/2)(N1B + N2B)RTeq); Solve[dSdN1A == dSdN1B && N1A + N1B == N1Ai + N1Bi, {N1A, N1B}] Attachments: 2_8_1_a.nb

Can you share your output?
I made sure the multiplication stars are there but it still gives me the same output.
I think when I make this post it took out some of the stars. See my attached code.

Clear["Global`*"]

R = 8.314 ;(*J/mol/K*)
N1Ai = 0.5; 
N1Bi = 1;
N2A = 0.75;
N2B = 0.5;
V = 5;(*Liters*)
Teq = 272.7; (*K*)
Uf = 9353.25;(*J*)


dSdN1A = 
  D[(N1A + N2A)*A + (N1A + N2A)*R*
     Log[UA^(3/2)*V/((N1A + N2A)^(5/2))] - 
    N1A*R*Log[N1A/(N1A + N2A)] - N2A*R*Log[N2A/(N1A + N2A)], N1A];

UA = ((3/2)*(N1A + N2A)*R*Teq);

dSdN1B = 
  D[(N1B + N2B)*A + (N1B + N2B)*R*
     Log[UB^(3/2)*V/((N1B + N2B)^(5/2))] - 
    N1B*R*Log[N1B/(N1B + N2B)] - N2B*R*Log[N2B/(N1B + N2B)], N1B];

UB = ((3/2)*(N1B + N2B)*R*Teq);

Solve[dSdN1A == dSdN1B && N1A + N1B == N1Ai + N1Bi, {N1A, N1B}]

POSTED BY: George Ip

George Ip

Posted 2 months ago

I used your method and plug the answers back into the equations for both dSdN1A and dSdN1B. They do equal to each other!

POSTED BY: George Ip

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