Question

Investigate the chemical potential μ upon a change in volume. We will use the fact that

Answer 1

Answer:

$\mu_{Vf}-\mu_{Vi}=-3*10^{-21} J/mol$  

Explanation:

Let's rewrite F in terms of N.

$F=32NkT+C-T(Nkln(V-bN)+C)$

$F=32NkT+C-TNk\ln{(V-bN)}-TC$

C is a constant value.

We know that the chemical potential μ is the partial derivative of F with respect to N.

$\frac{\partial F}{\partial N}=32kT-Tk(ln(V-bN)+\frac{N}{V−bN}(-b))$

$\frac{\partial F}{\partial N}=kT(32-(ln(V-bN)+\frac{N}{V−bN}(-b)))$

$\frac{\partial F}{\partial N}=kT(32-ln(V-bN)+\frac{Nb}{V−bN})$

We can find now the chemical potential μ(Vf).

$\mu_{Vf}=kT(32-ln(Vf-bN)+\frac{Nb}{Vf−bN})$    

• k is the Boltzmann constant  $1.38*10^{-23} m^{2}kgs^{-2}K^{-1}$  
• N=20 moles
• T is temperature 300 K
• Vi initial volume 0.01 m3
• Vf final volume 0.02 m3

$\mu_{Vf}=1.38*10^{-23}*300*(32-ln(0.02-(9*10^{-29}*20*6.022*10^{23}))+\frac{20*6.022*10^{23}*9*10^{-29}}{0.02-(9*10^{-29}*20*6.022*10^{23}}))$    

$\mu_{Vf}= 1.49*10^{-19} J/mol$    

We can do the same to find μ(Vi).

$\mu_{Vi}=1.38*10^{-23}*300*(32-ln(0.01-(9*10^{-29}*20*6.022*10^{23}))+\frac{20*6.022*10^{23}*9*10^{-29}}{0.01-(9*10^{-29}*20*6.022*10^{23}}))$    

$\mu_{Vi}=1.52*10^{-19} J/mol$  

   

Therefore µ(Vf) - µ(Vi) will be:

$\mu_{Vf}-\mu_{Vi}=-3*10^{-21} J/mol$  

I hope it helps you!