calculate work done for the vaporization of 1 mole of water in calories(1atm,100°c)
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Answer:
Explanation:
Volume of water before vapourization,
1 mole of water = 18 g of water
Volume of water =
1g/ml
18 g
[∵V=
d
m
]
V
1
=18ml
Volume of water after vapourization,
V
2
=
P
nRT
=
1.0
1.0×0.0821×373
=30.6 litre
V
1
is negligible w.r.t. V
2
w=−P
ext
×ΔV=−(1.0)×(30.6)litre−atm
=(−30.6)×101.3J=−3098.3J
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