A thermodynamic system is taken from an initial state i with internal energy $U_{i} = 100 J$ to the final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the paths af, ib and bf are $W_{af} = 200 J, W_{ib} = 50 J$ and $W_{bf} = 100 J$ respectively. The heat supplied to the system along the path iaf, ib and bf are $Q_{iaf}, Q_{ib}$ and $Q_{bf}$ respectively. If the internal energy of the system in the state b is $U_{b} = 200 J$ and $Q_{iaf} = 500 J,$ the ratio $Q_{bf} / Q_{ib}$ is

Step 1: Identify the given parameters.

The given values for internal energy, work, and heat are listed below.

$$U_i = 100\text{ J}$$ $$W_{af} = 200\text{ J}$$ $$W_{ib} = 50\text{ J}$$ $$W_{bf} = 100\text{ J}$$ $$U_b = 200\text{ J}$$ $$Q_{iaf} = 500\text{ J}$$

Step 2: Determine the work done along path $ia$.

Since the path $i \to a$ is vertical on the $P\text{-}V$ diagram, the volume remains constant.

$$\Delta V = 0$$ $$W_{ia} = 0\text{ J}$$

Step 3: Calculate the total work done along path $iaf$.

Sum the work done along the segments $ia$ and $af$.

$$W_{iaf} = W_{ia} + W_{af}$$ $$W_{iaf} = 0 + 200 = 200\text{ J}$$

Step 4: Calculate the internal energy of the final state $f$.

Apply the first law of thermodynamics to the path $iaf$ to find the change in internal energy.

$$Q_{iaf} = \Delta U_{iaf} + W_{iaf}$$ $$\Delta U_{iaf} = 500 - 200 = 300\text{ J}$$ $$U_f = U_i + \Delta U_{iaf}$$ $$U_f = 100 + 300 = 400\text{ J}$$

Step 5: Calculate the heat supplied along path $ib$.

Apply the first law of thermodynamics to the path $ib$.

$$\Delta U_{ib} = U_b - U_i$$ $$\Delta U_{ib} = 200 - 100 = 100\text{ J}$$ $$Q_{ib} = \Delta U_{ib} + W_{ib}$$ $$Q_{ib} = 100 + 50 = 150\text{ J}$$

Step 6: Calculate the heat supplied along path $bf$.

Apply the first law of thermodynamics to the path $bf$.

$$\Delta U_{bf} = U_f - U_b$$ $$\Delta U_{bf} = 400 - 200 = 200\text{ J}$$ $$Q_{bf} = \Delta U_{bf} + W_{bf}$$ $$Q_{bf} = 200 + 100 = 300\text{ J}$$

Step 7: Calculate the ratio of the heats.

Divide the heat supplied along path $bf$ by the heat supplied along path $ib$.

$$\frac{Q_{bf}}{Q_{ib}} = \frac{300}{150}$$ $$\frac{Q_{bf}}{Q_{ib}} = 2$$

Answer: The ratio $\dfrac{Q_{bf}}{Q_{ib}}$ is $2$.