The work done by an ideal gas during closed cycle 1arrow 2 arrow 3...

The total work done by the gas is the algebraic sum of the work done in the two separate loops shown in the $P-V$ diagram.

From the indicated arrows, the lower loop is traversed in a clockwise direction. Thus, the work done in this loop is positive and equal to the area of the lower triangle.

$W_1 = +\dfrac{1}{2} \times \text{base} \times \text{height} = +\dfrac{1}{2} \times (5V_0 - V_0) \times (3P_0 - P_0)$

$W_1 = \dfrac{1}{2} \times 4V_0 \times 2P_0 = 4P_0V_0$

The upper loop is traversed in a counter-clockwise direction. Thus, the work done in this loop is negative and equal to the negative of the area of the upper triangle.

The upper and lower triangles are similar because their bases are parallel. The ratio of their heights is $\dfrac{4P_0 - 3P_0}{3P_0 - P_0} = \dfrac{P_0}{2P_0} = \dfrac{1}{2}$.

Therefore, the base of the upper triangle is half the base of the lower triangle, which is $\dfrac{1}{2} \times 4V_0 = 2V_0$.

$W_2 = -\dfrac{1}{2} \times \text{base} \times \text{height} = -\dfrac{1}{2} \times 2V_0 \times (4P_0 - 3P_0)$

$W_2 = -\dfrac{1}{2} \times 2V_0 \times P_0 = -P_0V_0$

The net work done during the complete cycle is $W = W_1 + W_2 = 4P_0V_0 - P_0V_0 = 3P_0V_0$.