- #1
WWCY
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Homework Statement
Two systems in diffusive equilibrium have equal chemical potentials. We can use this fact to solve the following problem. We begin with a closed system consisting of a liquid such as water in diffusive equilibrium with its vapour. At the start, only the liquid and its vapour are present. Then we pump in an inert gas (that is, a gas that doesn’t react chemically with the liquid or vapour) to increase the pressure in the container (we’re also assuming that everything is at the same temperature, so no heat flow occurs). Assuming that the inert gas doesn’t dissolve in the liquid, and that the liquid and its vapour remain in diffusive equilibrium, the chemical potentials of the liquid and vapour must change by the same amount: dµ`= dµv. What happens to the partial pressure Pv of the vapour?
I had a look at solutions but there are a few pretty important concepts that I couldn't get, and need help with:
1) I have learned Dalton's law, that states that for ideal gases, the total pressure of a system is the sum of the partial pressures of its component gases.
So if I start with Gas ##A## in a volume ##V##, it has a pressure of ##P_A = \frac{N_A KT}{V}##.
If I stick in another Gas, Gas ##B##, then the partial pressures are now (I believe ) ##P_A = \frac{N_A KT}{V}## and ##P_B = \frac{N_B KT}{V}##,
that is to say the Partial Pressure of Gas ##A## "doesn't change".
Why then, would the partial pressure of water vapour ##P_v## in the problem above vary with the addition of additional gas / total pressure?
2) If I wanted to find the chemical potential of the liquid as a function of some pressure, what pressure would it be a function of?
i.e. What is the ##P_?## in ##\mu (T,P_?)## and why? I understand that the identity ##\mu = G/N## allows me to avoid this issue but it's something I would prefer to clarify.
Thanks in advance!