Consider a monatomic ideal gas that lives at a height z above sea level, so each molecule has potential energy mgz in addition to its kinetic energy.

(a) Show that the chemical potential is the same as if the gas were at sea level, plus an additional term mgz:

(You can derive this result from either the definition μ = –T(∂S/∂N)U,V or the formula μ = (∂U/∂N)S,V.)

(b) Suppose you have two chunks of helium gas, one at sea level and one at height z, each having the same temperature and volume. Assuming that they are in diffusive equilibrium, show that the number of molecules in the higher chunk is

N(z) = N(0)e‒mgz/kT,

in agreement with the result of below Problem 1.

Problem 1:

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.) 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(0)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 2:

Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), O2 (21%), and argon (1%).

October 18, 2016 Lecture 16: El Nino & Hurricanes - Things to Remember o Upwelling and Downwelling at Coasts Onshore Wind = Downwelling Offshore Wind = Upwelling Upwelling = nutrients = high productivity o Global patterns of upwelling in Marine Productivity 1. Coastal (100m) 2. Equatorial (200m) Right hand rule (water will have a right-hand tendency to turn in the northern hemisphere) Left hand rule (water will pull away from the equator to the left) The winds blowing apart from each other at the equator would create a