one mole of liquid water is boiled at 100 °C is 40.670 mol-1 calculate the work done and change in internal energy
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Answer:
Volume of 0.5 mole of steam at 1 atm pressure
=
P
nRT
=
1.0
0.5×0.0821×373
=15.3L
Change in volume = Vol. of steam − Vol. of water
=15.3− negligible =15.3L
Work done by the system,
w=P
ext
×volumechange
=1×15.3=15.3litre−atm
=15.3×101.3J=1549.89J
w should be negative as the work has been done by the system on the surroundings
w=−1549.89J
Heat required to convert 0.5 mole of water at 100
o
C to steam
=0.5×40670J=20335J
According to the first law of thermodynamics
ΔU=q+w=20335−1549.89=18785.11J
Explanation:
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