1 g of a liquid is converted to vapour at $$3 \times 10^5$$ 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:

The gas expands under a constant pressure of $$P = 3 \times 10^5$$ Pa and its volume increases by $$\Delta V = 1600$$ cm³, which equals $$1600 \times 10^{-6}$$ m³, with 10% of the heat supplied being used to perform expansion work.

The work done by the gas during expansion is given by the product of pressure and volume change:

$$W = P\Delta V = 3 \times 10^5 \times 1600 \times 10^{-6} = 480 \text{ J}$$

Since this expansion work corresponds to 10% of the total heat input $$Q$$, it follows that:

$$Q = \frac{480}{0.10} = 4800 \text{ J}$$

By applying the first law of thermodynamics, the increase in internal energy is:

$$\Delta U = Q - W = 4800 - 480 = 4320 \text{ J}$$

Therefore, the internal energy of the gas increases by 4320 J.