Water vapur drives a piston to accomplish work. What amount of work is done when 25.0 g of water at 373 K (100 degrees celsius) is vapourized to steam, pushing a piston against 1 atm of pressure?
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...
2007-07-16 02:33:31 · 2 answers · asked by Christine 1 in Science & Mathematics ➔ Chemistry
Also, why is work positive when it did work on the surroundings? Clearly energy was gain by the surroundings and not by the system....
2007-07-16 03:07:40 · update #1
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?
2007-07-16 02:50:49 · answer #1 · answered by hcbiochem 7 · 0⤊ 0⤋
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.
2007-07-16 02:48:50 · answer #2 · answered by Ajinkya N 5 · 0⤊ 0⤋