Step-by-step Solution:

1. Understand the First Law of Thermodynamics: The first law of thermodynamics relates the change in internal energy ($\\\Delta U$), heat supplied to the system ($Q$), and work done by the system ($W$) as: $\Delta U = Q - W \quad \text{or} \quad Q = \Delta U + W$ Internal energy ($U$) is a state function, meaning its value depends only on the state of the system, not on the path taken to reach that state. Therefore, the change in internal energy between two states (e.g., from i to f) is the same regardless of the path (iaf or ibf).

2. Calculate the Internal Energy at the Final State ($U_f$) using Path iaf: We are given the following for path iaf:

   • Initial internal energy, $U_i = 100$ J.
   • Heat supplied, $Q_{iaf} = 500$ J. The total work done along path iaf is the sum of work done along ia and af: $W_{iaf} = W_{ia} + W_{af}$.
   • From the P-V diagram, the path ia is a vertical line, which represents an isochoric process (constant volume). For an isochoric process, the work done is zero. So, $W_{ia} = 0$.
   • The work done along path af is given as $W_{af} = 200$ J.
   • Therefore, the total work done is $W_{iaf} = 0 + 200 = 200$ J. Now, we apply the first law of thermodynamics to the path iaf: $\Delta U_{if} = Q_{iaf} - W_{iaf}$ $U_f - U_i = 500 \text{ J} - 200 \text{ J}$ $U_f - 100 \text{ J} = 300 \text{ J}$ $U_f = 300 \text{ J} + 100 \text{ J} = 400 \text{ J}$

3. Calculate the Heat Supplied along Path ib ($Q_{ib}$): Now we consider the process from state i to state b.

   • Initial state: $U_i = 100$ J.
   • Final state (for this segment): $U_b = 200$ J (given).
   • Change in internal energy: $\Delta U_{ib} = U_b - U_i = 200 \text{ J} - 100 \text{ J} = 100$ J.
   • Work done along path ib is given as $W_{ib} = 50$ J. Using the first law for path ib: $Q_{ib} = \Delta U_{ib} + W_{ib}$ $Q_{ib} = 100 \text{ J} + 50 \text{ J} = 150 \text{ J}$

4. Calculate the Heat Supplied along Path bf ($Q_{bf}$): Next, we consider the process from state b to state f.

   • Initial state (for this segment): $U_b = 200$ J.
   • Final state: $U_f = 400$ J (calculated in Step 2).
   • Change in internal energy: $\Delta U_{bf} = U_f - U_b = 400 \text{ J} - 200 \text{ J} = 200$ J.
   • Work done along path bf$ is given as $W_{bf} = 100$ J. (Note: Although the figure *schematically* shows $bf as an isochoric process, we must use the explicitly given numerical value for work done.) Using the first law for path bf: $Q_{bf} = \Delta U_{bf} + W_{bf}$ $Q_{bf} = 200 \text{ J} + 100 \text{ J} = 300 \text{ J}$

5. Calculate the Required Ratio: The problem asks for the ratio $Q_{bf} / Q_{ib}$. $\frac{Q_{bf}}{Q_{ib}} = \frac{300 \text{ J}}{150 \text{ J}} = 2$

The final answer is 2.