Chemical potential of water using the Van der Waals model

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Homework Statement



Obtain the chemical potential of water as a function of temperature and volume using the Van der Waals model.

Homework Equations



μ=∂U∂N

The Attempt at a Solution



I don't really understand how to do this at all. Any help would be greatly appreciated.
 
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For a pure substance, how is the chemical potential related to the gibbs free energy per mole?

Chet
 
By this relationship: $$\mu= \frac{G}{n}$$
 
It's me said:
By this relationship: $$\mu= \frac{G}{n}$$
So, if you could calculate the gibbs free energy per mole as a function of temperature and volume for a van der walls gas, you would have your answer. Suppose you took the starting state of g = 0 as water vapor at 25 C and the corresponding equilibrium vapor pressure (i.e., in the ideal gas region). Could you determine g at the same pressure and a higher temperature T (i.e., within the ideal gas region)?

Chet
 
I'm sorry I don't understand how I could determine that.
 
It's me said:
I'm sorry I don't understand how I could determine that.
Well, you need to go back to your textbook and find out how to determine that change in free energy with temperature at constant pressure.

Chet
 
It is this relation? $$dG=-SdT+\mu dn$$
 
It's me said:
It is this relation? $$dG=-SdT+\mu dn$$
No. The number of moles should also be held constant.

Chet
 
Can you express S as a function of G, H, and T? If so, substitute it into your equation for dG.

Chet
 
  • #10
$$G=H-ST$$ $$S=\frac{H-G}{T}$$ $$dG=-SdT$$ $$\rightarrow dG=-(\frac{H-G}{T})dT$$
 
  • #11
It's me said:
$$G=H-ST$$ $$S=\frac{H-G}{T}$$ $$dG=-SdT$$ $$\rightarrow dG=-(\frac{H-G}{T})dT$$
Good. So, if we rearrange this, we get:
$$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$
Do you know how to get H as a function of T for a gas in the ideal gas region? Once you know that, you can integrate this equation to get G as a function of T at constant (low) pressure in the ideal gas region. Can you figure out what to do next?

Chet
 
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