Physical Chemistry: Adiabatic Expansion of Clouds at Altitude

[whitebuffalo]

Homework Statement
Cloud moves from alt. 2000m (P = 0.802atm) to 3500m (P = 0.602) when it encounters a mountain. It expands adiabatically. The initial temp is 288K, C_P,m for the air is 28.86 J/Kmol (assume ideal). What is the final temp and will it drop it's moisture?The attempt at a solution
This is what I did. C_V,m = C_P,m - R
C_V,mdT = -\frac{RT}{V}dV
\intC_V,mdT = \int-\frac{RT}{V}dV divide by T
C_V,m\int\frac{1}{T}dT = -R\int\frac{1}{V}dV
C_V,mln(\frac{T2}{T1}) = R ln(\frac{V1}{V2}) rearrange
T2 = T1(V1/V2)^R/C_V,m this to solve for T2, but 2 variables, so...

P₁V₁^\gamma = P₂V₂^\gamma solve for the ratio V1/V2
\frac{V1}{V2} = (\frac{P2}{P1})^1/\gamma

T2 = T1 ((\frac{P2}{P1})^1/\gamma)^R/C_V,m

C_V,m = C_P,m - R = 20.546 \gamma = C_P,m/C_V,m = 1.405

T2 = 288 ((\frac{0.602}{0.802})^1/1.405)^8.314/20.546 = 265KSomeone please tell me if I did this right, if not, what did I do wrong...and how do I know if it will drop moisture?

This is my first HW question in Pchem I and only the 2nd week of the semester, so please explain anything so I can understand it.

 

[vela]

You should reverse the order of your first and second steps. Divide by T before integrating. Once the integral signs are there, you can't just divide to move variables from one side of the equation to the other. Your solution otherwise looks okay, but you could have done it more directly by simply using just the ideal gas law and the relationship PV^\gamma=c, where c is a constant.

I don't know about the moisture. Perhaps someone else can chime in on that part of the problem.