Initially, a gas is at a pressure of 100 kPa with a volume of 2.0...

The given thermodynamic cycle consists of three processes:

State 1 to State 2: Constant pressure expansion with $P_1 = 100 \text{ kPa}$, $V_1 = 2.0 \text{ m}^3$ to $V_2 = 2.5 \text{ m}^3$.

State 2 to State 3: Constant volume process with $V_2 = 2.5 \text{ m}^3$, $P_2 = 100 \text{ kPa}$ to $P_3 = 200 \text{ kPa}$.

State 3 to State 1: Linear return to the initial state.

These processes form a right-angled triangle on the $P-V$ diagram. For a complete cyclic process, the change in internal energy is zero ($\Delta U = 0$). According to the first law of thermodynamics, the net heat supplied or extracted is equal to the net work done, which corresponds to the area enclosed by the cycle on the $P-V$ diagram.

The magnitude of the work done is the area of the triangle:

$|W| = \dfrac{1}{2} \times \text{base} \times \text{height}$

$|W| = \dfrac{1}{2} \times (V_2 - V_1) \times (P_3 - P_2)$

$|W| = \dfrac{1}{2} \times (2.5 - 2.0) \text{ m}^3 \times (200 - 100) \times 10^3 \text{ Pa}$

$|W| = \dfrac{1}{2} \times 0.5 \times 100 \times 10^3 \text{ J}$

$|W| = 25000 \text{ J}$

Since the cycle is counter-clockwise, the net work is negative, meaning $25000 \text{ J}$ of heat was extracted from the gas.