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 $$\text{cm}^3$$ during this phase change, then the increase in internal energy in the process will be :

The process involves a phase change from liquid to vapour at constant pressure. The heat supplied (Q) is used to increase the internal energy (ΔU) and to perform work against the external pressure (W). The first law of thermodynamics states:

$$Q = \Delta U + W$$

The work done by the system is given by $$W = P \Delta V$$, where P is the pressure and ΔV is the change in volume.

Given that 10% of the heat supplied is used for increasing the volume, this means the work done W is 10% of Q:

$$W = 0.1 Q$$

Substituting the expression for work:

$$P \Delta V = 0.1 Q$$

Now, plug in the known values:

Pressure, $$P = 3 \times 10^5$$ Pa
Change in volume, $$\Delta V = 1600 \text{cm}^3 = 1600 \times 10^{-6} \text{m}^3 = 0.0016 \text{m}^3$$

Calculate W:

$$W = P \Delta V = (3 \times 10^5) \times (0.0016) = 480 \text{J}$$

Using the relation $$W = 0.1 Q$$:

$$480 = 0.1 Q$$

Solve for Q:

$$Q = \frac{480}{0.1} = 4800 \text{J}$$

Now apply the first law to find ΔU:

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

Therefore, the increase in internal energy is 4320 J.

The correct option is A. 4320 J.