 p/p=  lnp=lnp 1 -lnp 2 =ln (p 1 /p 2 ) ln (p 1 /p 2 )=-g  z /RT o p 1 /p 2 =exp[-g  z /RT o ] p 1 = p 2 exp[-g  z /RT o ] T o = mean temperature for layer Atmospheric pressure decreases with height as the mass of overlying air decreases. At sea level it decreases by ~1% for each 80 m-- it decreases by 3-4% riding up in the elevator to the top of a tall office building; by ~20% living in Denver, and by 40% climbing to the top of Mr. Rainier. As a comparison – in an intense hurricane the surface pressure can decrease up to ~10% compared to the undisturbed surrounding air. where p 1 0 z 2, p 2 z 1, p 1 R d = dry gas constant =287 J K -1 kg -1, g = 9.8 ms -2 Fig

Is the force on area A 1 = force on area A 2 ? A1A1 A2A2 high compressibility gasses low compressibility solids, liquids h density remains constant for all pressures

Atm. scientists do NOT measure density What do they typically measure? pressure temperature How do we get rid of density in hydrostatic eqn? p =  RT atm ~ ideal (collisionless) gas  p/  z = -  g = -pg/RT  p/p=-g  z /RT What is  p/p anyway? lnx  p/p=  lnp=lnp 1 -lnp 2 Take small  x x  x/x= /0.75=.133 ln0.8-ln0.7=0.133 Note that lnx 1 -lnx 2 = ln(x 1 /x 2 )