One mole of an ideal monatomic gas undergoes a cyclic process as...

For a complete cyclic process, the change in internal energy is zero, i.e., $\Delta U = 0$.

According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.

Therefore, the net heat supplied to the gas is equal to the net work done by the gas: $\Delta Q = \Delta W$.

The net work done $\Delta W$ is given by the area enclosed by the cycle on the $P-V$ diagram.

$\Delta W = \text{Area of rectangle } abcd$

$\Delta W = (P_a - P_d) \times (V_b - V_a)$

$\Delta W = (300 - 100) \times (5 - 2)$

$\Delta W = 200 \times 3 = 600 \text{ J}$

Thus, the total heat supplied to the gas is $600 \text{ J}$.

Answer: $600$ J