Questions
While boiling 1 gm of water at pressure 1.013×105 N/m3, its volume 1471 cm3 from 1 cm3, then work done by the system is
a
148.911 J
b
150 J
c
130.24 J
d
120.57 J
detailed solution
Correct option is A
dQ=dU+dW=dU+pdvdW=pdv=1.013×105×(1470)×10−6=148.91 jouleTalk to our academic expert!
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