One gram 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

The work done during the phase change is given by W = P V . Substituting the given values, W = (3 10^5) (1600 10⁻⁶) = 480 J . It is given that 10 % of the heat supplied is used for increasing the volume, which means W = 0.1 Q . Therefore, the total heat supplied is Q = W 0.1 = 480 0.1 = 4800 J . According to the first law of thermodynamics, Q = U + W . The increase in internal energy is U = Q - W = 4800 - 480 = 4320 J . Answer: 4320 J