Don't understand the answer to this Chemistry question: about work and gas?
nliqu=nvap= 25gX(1molH2O/18g)
delta volume= Vvap-Vliq
Vvap=nvapRT/P
W=(Pext)(deltaV)=nvapRT
=(25/18)molx8.31J/mol-kX373K
=4.30 kJ
Just don't get how they got W to equal to nvapRT...
Answer:
OK...So are you OK with work being equal to P X Delta V?
Equating PXDeltaV to nvapRT is just applying the ideal gas law. They are assuming that the volume of the liquid water is insignificant compared to the volume of the water vapor, which is a very reasonable assumption. So, as the 25 grams of water is vaporized, it forms a very large volume of steam. As the steam is formed, it pushes against the piston, doing work.
Since work = P X Delta V, and since the ideal gas law says PV=nRT, then work = nRT.
That help?
This is based on an assumption.
Calculate the volumes of water and water vapor at the said temperatures.
The density of water = 1 gm/cc
So, the volume of water = 25 mL
The volume occupied by water vapour at that temperature = nRT/P = 42.5323 L = 42532361.11 mL
This is very large as compared to 25 mL. So we can assume delta V to be approximately the volume occupied by water vapour.
Hence the last expression is on the basis of the above assumption.
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