1g of a liquid is converted to vapour at 3 × 10 ^ 5 Pa pressure. If...

As pressure against which expansion is happening is constant, work done can be written as $= P\DeltaV$

$= \left(3 \times {10}^{5}\right) \times \left(1600 \times {10}^{- 6}\right)$

$= 480J$
As only $10$ of heat is used in work done.
Hence $\DeltaQ = \dfrac{W}{0 . 1} = 4800J$
The rest goes in increase in internal energy, which is $90$ of heat.
Change in internal energy $= 0 . 9 \times 4800 = 4320J$