1 g of a liquid is converted to vapour at 3 × 10 5 Pa pressure. If 10 % of the heat supplied is used for increasing the volume by 1600 cm 3 during this phase change, then the increase in internal energy in the process will be :

Step-by-step solution

As pressure against which expansion is happening is constant, work done can be written as = P Δ V = 3 × 10 5 × 1600 × 10 - 6 = 480   J As only 10 % of heat is used in work done. Hence Δ Q = W 0 . 1 = 4800   J The rest goes in increase in internal energy, which is 90 % of heat. Change in internal energy = 0 . 9 × 4800 = 4320   J