Two moles of an ideal gas undergo a cyclic process A arrow B arrow...

The isobaric (constant pressure) segment of the cycle is the process $A \rightarrow B$, where the pressure remains constant at $1 \text{ atm}$.

During this process, the temperature increases from $T_A = 300 \text{ K}$ to $T_B = 600 \text{ K}$.

The work done by an ideal gas during an isobaric process is given by:
$W = P\Delta V = nR\Delta T$

Given $n = 2$ moles, the change in temperature is $\Delta T = 600 \text{ K} - 300 \text{ K} = 300 \text{ K}$.

Substituting the values:
$W = 2 \times R \times 300 = 600 R$

Calculating the area of the triangle on the $P-T$ graph ($75$) does not give the work done, and using $n=1$ gives the incorrect value of $300 R$.

Answer: $600 R$