Question

When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20×106J/kg2.20×10

6
J/kg and the boiling point is 120∘C120
∘
C. At this pressure, 1.00 kg of water has a volume of 1.00×10−3m31.00×10
−3
m
3
, and 1.00 kg of steam has a volume of 0.824m30.824m
3
. (a) Compute the work done when 1.00 kg of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.

Answer 1

Step-by-step explanation:

a ) When 1 kg water is boiled at constant pressure of 1 atm , its volume increases by following volume

(.824 - .001 )m³

.823 m³

work done by steam = increase in volume x pressure

.823 x 10⁵ J

Heat added

= latent heat of vaporization x mass

= 2260000 J x 1

= 22.6 x 10⁵ J

Increase in internal energy of gas

= heat added - work done by gas

= (22.6 - .823) x 10⁵ J

= 21.777 x 10⁵ J .