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

The cyclic process consists of three steps:

Isobaric expansion from state A $(2.0 \text{ m}^3, 100 \text{ kPa})$ to state B $(2.5 \text{ m}^3, 100 \text{ kPa})$.

Isochoric process from state B to state C $(2.5 \text{ m}^3, 200 \text{ kPa})$.

Linear return from state C to state A.

Plotting these states on a $P-V$ diagram forms a right-angled triangle. The sequence $A \rightarrow B \rightarrow C \rightarrow A$ traces a counter-clockwise cycle.

The net work done in a counter-clockwise cycle on a $P-V$ diagram is negative ($W < 0$).

From the first law of thermodynamics for a complete cycle, $\Delta U = 0$, giving $Q = W$.

Since $W < 0$, $Q < 0$, indicating that net heat is extracted from the gas.