Recent content by Jilly

Hmmmm 0.015 is pressure at sea level... so 0.015 should be P(0) Is it that I'm simply misunderstanding the question and I should assume that temperature of 15C is constant? But how would I find pressure at the height the particle is at? And if I'm not assuming the temperature is constant...

The equation I posted is equation 2 on that page.

I just don't understand what I'm not seeing to solve this... What am I missing? =\

Well, in order to calculate for 'h' I'd need P(h), and Temperature at P(h) ...

Apologies, misread on the calc. The Calc for pressure is still 0.015. Number of moles should equal 0.000621

A. Given that the average diameter of all particles in a sample is 16 nm, te density of all particles is 1.25 x 10^-4 g/cm^3, and the density of the sample of air is 100 microgram/cm^3, how many particles are in 1 L of this air sample? V(particle) =2.14x10^-18 cm^3 x density = 2.675x10^-22 g 1...

Log P(sub h) = (-28.97*981*50*10^5)/ (2.303*[8.314x10^7]*273.15) LogP(sub h) = -2.72 p(sub h) = 0.00192 atm n=PV/RT=(.00192atm*1cm^3)/ (273.15K*82.057cm^3atm•K*-1•mol^-1) n = 8.566 x 10^-8 mol = 5.16 x 10^16 molecules, Yes?

Thank you!

The site autorotated the photo, here's an adjustment.

Okay so, this second equation was derived from the first using an 8 km scale height and is true under the conditions of atm pressure units, according to my textbook, that's whereni acquired all of my information, I've attached a photo of the page, maybe you can see what I'm not understanding...

*5.23x10^-276

Homework Statement At an altitude of 50 km, the average atmospheric temperature is 0C. what is the average number of air molecules per cm^3 of air at this atmosphere? Given formula: P(sub h) = P(sub 0) x e^(-Mgh/RT) = -(Mgh x 10^5)/2.303RT Assuming P = 1 atm at sea level and h in km...