Suppose you have a liquid (say, water) in equilibrium with its gas phase, inside some closed container. You then pump in an inert gas (say, air), thus raising the pressure exerted on the liquid. What happens?
(a) For the liquid to remain in diffusive equilibrium with its gas phase, the chemical potentials of each must change by the same amount: Use this fact and equation 5.40 to derive a differential equation for the equilibrium vapour pressure, Pv as a function of the total pressure P. (Treat the gases as ideal, and assume that none of the inert gas dissolves in the liquid.)
(b) Solve the differential equation to obtain
where the ratio V/N in the exponent is that of the liquid. (The term Pv(Pv) is just the vapour pressure in the absence of the inert gas.) Thus, the presence of the inert gas leads to a slight increase in the vapour pressure: It causes more of the liquid to evaporate.
(c) Calculate the percent increase in vapour pressure when air at atmospheric pressure is added to a system of water and water vapour in equilibrium at 25°C. Argue more generally that the increase in vapour pressure due to the presence of an inert gas will be negligible except under extreme conditions.
The vapour pressure is 0.07 percent higher than it would be in the absence of air.
Using the equation 5.40, the chemical potential for the gas is as follows.
k is Boltzmann's constant
T is temperature
P is the partial pressure
Differentiating the equation with respect to P
The chemical potential for liquid is
Where G is Gibbs free energy
Differentiating the equation with respect to pressure P
The volume V is equal to the differentiation of the Gibbs free energy with respect to pressure.
Substitute V for
The change in chemical potentials for the two phases, liquid and gas, must be equal in equilibrium.
Hence, the differential equation is
As a result, the differential equation solution found in part (a) is
The vapour pressure of water at a temperature of 25°C is as follows.
As a result, when the total pressure is 1 bar, the vapour pressure is minimal.
The value of Nk is equal to the gas constant for one mole of N, and the volume is
Substituting the values
According to the above calculation, the vapour pressure is 0.07 percent higher than it would be in the absence of air.
Because the volume of the liquid is substantially smaller than that of the gas, the quantity is always less than 1.
At the critical moment, the ratio is comparable to one.
Derive the van't Hoff equation,
which gives the dependence of the equilibrium constant on temperature." Here is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of; solve the equation in this case to obtain
Assume that the air you exhale is at 35°C, with a relative humidity of 90%. This air immediately mixes with environmental air at 5°C and unknown relative humidity; during the mixing, a variety of intermediate temperatures and water vapour percentages temporarily occur. If you are able to "see your breath" due to the formation of cloud droplets during this mixing, what can you conclude about the relative humidity of your environment? (Refer to the vapour pressure graph drawn in Problem 5.42.)
Seawater has a salinity of , meaning that if you boil away a kilogram of seawater, when you're finished you'll have of solids (mostly localid="1647507373105" ) left in the pot. When dissolved, sodium chloride dissociates into separate and ions.
(a) Calculate the osmotic pressure difference between seawater and fresh water. Assume for simplicity that all the dissolved salts in seawater are .
(b) If you apply a pressure difference greater than the osmotic pressure to a solution separated from pure solvent by a semipermeable membrane, you get reverse osmosis: a flow of solvent out of the solution. This process can be used to desalinate seawater. Calculate the minimum work required to desalinate one liter of seawater. Discuss some reasons why the actual work required would be greater than the minimum.
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