HOW do I DO THIS??? URGENT?

Question

What is w (work) when 1.47 kg of H2O(liquid), initially at 25 degrees is converted into water vapour at 151 degrees against a constant external pressure of 1 atm.? Assume that the vapour behaves ideally and that the density of liquid water is 1 g/mL.


ok so I know I have to use this formula
W = -P ext (Vgas-Vliquid)

for v gas, we use v= nRT/P
but do i subtract temperatures or what? what about P ??
clue me in folks asap please!!!!!!!!

Answer 1

Well, at least you made sign wrong again--please pay very close attention to the definitions: ΔE = Q -W, where ΔE is the change of internal energy, Q the heat INTO the system, and W the work done by the system.
Also, this problem is vague in terms of temperature. It says "25 degrees" and "151 degrees" but does not specify if this is in Kalvin, Celsius or Fahrenheit. I would guess this is in Celsius.
Vliquid = (1.47 kg)/(1 g/mL) = 1.47L
Vgas = (1470/18.0)*0.08206*(273+151)/1.00 = 2.84x10^3 L.
Hence Vliquid can be ignored by comparing to Vgas.
W = P*Vgas = 2.84x10^3 atm•L = 288 kJ
since 1 J = 9.87 10-3 atm•L.