Calculate the work done in joules when 1.0 mole of water vaporizes at 1.0 atm and 100C. Assume that the volume of liquid water is negligible compared with that of steam at 100c, and ideal gas behavior
3 answers
-3.1x10^3J.
T =100+ 273= 373K
SINCE
PV=NRT
NOW,V=NRT/P
=1×.0821×373/1
=30.6L
W=-PV
=-1atm×30.6L
=101.3× 30.6
-3.1×10^3J
SINCE
PV=NRT
NOW,V=NRT/P
=1×.0821×373/1
=30.6L
W=-PV
=-1atm×30.6L
=101.3× 30.6
-3.1×10^3J
Although this problem is practically incorrect. In theory, the water vapor behaves as a perfect gas; therefore, we apply: PV = mRT
Work Done (approx.), W = Pdv = P(mRT/P) = mRT= 1kg-mol x 8.3145 KJ/Kg-mol-K x(100+273) = 3101.3085 KJ
Answer: Approx. 3101.31 KJ
Note that this answer is just an approximation
Work Done (approx.), W = Pdv = P(mRT/P) = mRT= 1kg-mol x 8.3145 KJ/Kg-mol-K x(100+273) = 3101.3085 KJ
Answer: Approx. 3101.31 KJ
Note that this answer is just an approximation
1 answer