Question

calculate work done for the vaporization of 1 mole of water in calories(1atm,100°c)​

Answer 1

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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