Consider a closed system consisting of liquid water in equilibrium with its vapo

Question

Consider a closed system consisting of liquid water in equilibrium with its vapor (with vapor pressure P) at 298K. What is the relationship between the chemical potentials of the liquid and the vapor in this system? From the relationship in part a and other information about chemical potentials, derive the following expression:mu^.1 - muDegree g = RTIn p/pdegree State any assumptions you make. Show how the left hand side of the above equation can be calculated using the DeltafGdegree values for liquid water and water vapor. Make sure you explain this clearly because it is definitely not true that mu1 = Deltaf Gdegree(liq), for example. Use your result from part c and the data in Appendix B of Chang to calculate the vapor pressure of water at 298K. Compare your result to the experimental value of 3.169 Times 10^-2 bar. Would you expect a similar calculation to be more or less accurate for hexane than for water? Explain.

Explanation / Answer

For a closed system, when the liquid is in equilibrium with its vapor.

So, the chemical potential of liquid is equal to the chemical potential of vapor phase

let u(l) be the chemical potential for liquid and u(g) is chemical potenatila for vapor phase then,

So, at equilibrium, u(l) = u(g)

The equation is as follows:

u(l) = u(l)o + RT lnP/Po

u(g) = u(g)o + RTlnP/Po

Here, u(l)o and u(g)o are standard state chemical potential, P is standard state pressure, T is the temperature, R is the gas constant.

Now, according to Raoult's law,

u(l) = u(l)o + RTlnX

Here, X is the mole fraction.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.