One mole of an ideal gas in initial state A undergoes a cyclic...

From the given $V-T$ graph, the parameters for states A and B are:
State A: $T_A = T_0$, $V_A = V_0$, $P_A = P_0$
State B: $T_B = T_0$, $V_B = 4V_0$

Since $T_A = T_B = T_0$, the process AB is isothermal. For an ideal gas, internal energy depends only on temperature, which means $U_A = U_B$.

The work done by the gas in the isothermal process AB is:
$W_{AB} = nRT_0 \ln \left(\dfrac{V_B}{V_A}\right)<br><br>Given$n = 1$mole, and using the ideal gas equation at state A ($P_0 V_0 = RT_0$), we get:<br>$W_{AB} = P_0 V_0 \ln\left(\dfrac{4V_0}{V_0}\right) =$P_0 V_0 \ln(4)$

The line BC is straight but not specified to pass through the origin. If it did, both $P_C =$\dfrac{P_0}{4}$and T_C =$\dfrac{T_0}{4} would be true, leading to multiple correct options. Thus, only the work done in process AB is definitively correct.

Answer: Work done by the gas in process AB is P_0 V_0\ln(4)