In Problem 1 you calculated the atmospheric temperature gradient required for unsaturated air to spontaneously undergo convection. When a rising air mass becomes saturated, however, the condensing water droplets will give up energy, thus slowing the adiabatic cooling process

(a) Use the first, law of thermodynamics to show that, as condensation forms during adiabatic expansion, the temperature of an air mass changes by

where nw is the number of moles of water vapour present, L is the latent heat of vaporization per mole, and I’ve set γ = 7/5 for air. You may assume that the H2O makes up only a small fraction of the air mass.

(b) Assuming that the air is always saturated during this process, the ratio nw/n is a known function of temperature and pressure, Carefully express dnw/dz in terms of dT/dz, dP/dz, and the vapour pressure Pv(T). Use the Clausius-Clapeyron relation to eliminate dPv/dT.

(c) Combine the results of parts (a) and (b) to obtain a formula relating the temperature gradient, dT/dz, to the pressure gradient, dP/dz. Eliminate the latter using the “barometric equation” from Problem 1.16. You should finally obtain.

where M is the mass of a mole of air. The prefactor is just the dry adiabatic lapse rate calculated in Problem 1.40, while the rest of the expression gives the correction due to heating from the condensing water vapor. The whole result is called the wet adiabatic lapse rate; it is the critical temperature gradient above which saturated air will spontaneously convect.

(d) Calculate the wet adiabatic lapse rate at atmospheric pressure (1 bar) and 25°C, then at atmospheric pressure and 0°C. Explain why the results are different, and discuss their implications. What happens at higher altitudes, where the pressure is lower?

Problem 1:

In Problem you calculated the pressure of earth’s atmosphere as a function of altitude, assuming constant temperature. Ordinarily, however, the temperature of the bottommost 10–15 km of the atmosphere (called the troposphere) decreases with increasing altitude, due to heating from the ground (which is warmed by sunlight). If the temperature gradient |dT/dz| exceeds a certain critical value, convection will occur: Warm, low-density air will rise, while cool, high-density air sinks. The decrease of pressure with altitude causes a rising air mass to expand adiabatically and thus to cool. The condition for convection to occur is that the rising air mass must remain warmer than the surrounding air despite this adiabatic cooling.

(a) Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation

(b) Assume that dT/dz is just at the critical value for convection to begin, so that the vertical forces on a convecting air mass are always approximately in balance. Use the result of Problem 2 (b) to find a formula for dT/dz in this case. The result should be a constant, independent of temperature and pressure, which evaluates to approximately - 10°C/km. This fundamental meteorological quantity is known as the dry adiabatic lapse rate.

Problem 2:

The exponential atmosphere.

(a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest , the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air.

(b) Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass m of the air molecules. (The information needed to calculate m is given in Problem 3.) Show, then, that the pressure obeys the differential equation

called the barometric equation.

(c) Assuming that the temperature of the atmosphere is independent of height (not a great assumption but not terrible either), solve the barometric equation to obtain the pressure as a function of height: P(z) = P(O)e−mgz/kT. Show also that the density obeys a similar equation.

(d) Estimate the pressure, in atmospheres, at the following locations: Ogden, Utah (4700 ft or 1430 m above sea level); Leadville, Colorado (10,150 ft , 3090 m) ; Mt. Whitney, California (14,500 ft, 4420 m); Mt. Everest, Nepal/ Tibet (29,000 ft, 8850 m). (Assume that the pressure at sea level is 1 atm.)

Problem 3: Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), 02 (21%), and argon (1%).

Chemistry of Solutions Lecture no. 7 Thursday, September 15, 2016 Topics Covered: Le Châtelier’s principle, the nature of acids and bases, acid strength, the pH scale, and calculating the pH of strong acid solutions. Le Châtelier's Principle If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. A chemical reaction is not a sentient being, if a substance is added the rates of the forward and reverse reactions are disturbed. The net reaction occurs in one direction, until the system reaches equilibrium again, At the new equilibrium position, the reactants and products have different concentrations (compared to their initial concentrations). 1. Changing co