Calculating reversible adiabatic expansion

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Azivegu
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I have some homework that I am just not able to figure out.

There is a 2kg parcel of air at T1=32°C and p1= 0.8 atm
This parcel of air expands until it is at T2=12°C

Also, during the expansion 3.2 g of water will condense due to the dT (ΔvapH (water) = 40.65kJ mol-1

Also: Cv=0.718 kJ kg-1K-1I understand that dU+dW=dQ=0
and that W=-pdV and w=CvΔT

The steps should be calculating the expansion work, then subtracting the heat from condensation to figure out the work, but I just keep drawing blanks. Can anybody help me? I am not looking for an answer, but mainly the steps to calculate it.
 
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Sorry, I believe I misunderstood the template being that this is my first time here. I hope that the following is more clear.

Homework Statement


Calculate the expansion work done by a reversibly adiabatic balloon.
Mass air parcel: 2 kg
T1=32°C = 305K
p1= 0.8 atm = 81060 Pa
T2= 12°C = 285K
Cv= 0.718 kJ kg-1K-1

Also, during expansion 3.2g water vapor will condense due to the falling temperature.
The vaporization energy of water is ΔvapH(water) = 40.65kJ mole-1

Homework Equations


dU+dW=dQ=0
W=-pdV
W=CvΔT

The Attempt at a Solution


I have made multiple attempts, but so far none have been successful. I think this is mostly because I am not sure what steps to use.
 
Your relevant equations suggest that you kind-of have the right idea. Irrespective of whether the process is reversible, if Q = 0, you must have that ΔU=-W. (Your equation for dW = -pdV is incorrect. It should be dW = +pdV).

The question is, "do you have enough information in the problem statement to calculate ΔU?" You certainly have enough information to get the change in internal energy of the air. What is that equal to? In the case of the water, it isn't clear. In my judgement, you need to know how much water vapor was initially in the balloon. If, in the initial state, the water vapor partial pressure were equal to its equilibrium saturation vapor pressure, how much water would there have been in the balloon?

Chet