One gram of water (volume = 1 cm 3 ) becomes 1671 cm 3 of steam when boiled at a pressure of one atmosphere. Latent heat of vaporization at this pressure is 539 cal/gm. Compute the work done. [1 atm = 1.013 x 10 5 Nm -2 ]
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(169.171 J)
Sol. Process is isobaric
W = P (V 2 – V 1 ) = 1.013 × 10 5 × (1671 – 1) × 10 –6
W = 169.171 J
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