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One gram of water (1 cm^3) becomes 1671 ...

One gram of water `(1 cm^3)` becomes `1671 cm^3` of steam when boiled at a constant pressure of 1 atm `(1.013xx10^5Pa)`. The heat of vaporization at this pressure is `L_v=2.256xx10^6J//kg`. Compute (a) the work done by the water when it vaporizes and (b) its increase in internal energy.

Text Solution

Verified by Experts

The correct Answer is:
`169.171 J`
Process is isobaric
`W = P(V_(2)-V_(1)) = 1.013 xx 10^(5) xx (1671 -1) xx 10^(-6)`
`W = 169.171 J`.
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