{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 3 Solutions

# For the parameters given the mean molecular weight is

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: κλ(0.03 kg/m3)(0.3 m) = 2/3 => κλ = 74 m2/kg Not surprisingly, clay is a lot more opaque than water! (3) Looking through gaseous water [25 pts] (a) [10 pt] Water vapor in the air can also absorb starlight, although in this case its density is not constant with height. Show that atmospheric pressure scales vertically as P (z ) = P (0)e−z/H where H= kT gµmH is the scale height. Compute this scale height assuming standard temperature and pressure (STP): T = 0 ºC = 273 K and P = 100 kPa = 1 bar; and an atmosphere composed (by volume) of 77% N2, 21% O2, 1% H2O and 1% Ar. Consider a slab of gas with area A and vertical width dz, so that the mass of this slab m = ρAdz. The balance of forces  ­ air pressure acting on the top and bottom of the slab, and the slab’s own weight  ­ keep the slab in place only if Ftop − Fbottom + g ρAdz = 0 dP ⇒ = −g ρ dz where we’ve used P = F/A. Using the ideal gas law, this right side be written in terms of pressure: dP P µmH dP gµmH = −g ⇒ =− dz dz kT P kT With H as defined above, we get the pressure equation as shown. For the parameters given, the mean molecular weight is µ = (0.77)(28) + (0.21)(3...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern