Ratio of Heat Supplied Along Thermodynamic Paths

To determine the ratio of the heat supplied $\frac{Q_{bf}}{Q_{ib}}$, we apply the First Law of Thermodynamics ($\Delta Q = \Delta U + W$) along the various segments of the thermodynamic process.

Step 1: Calculate the internal energy at the final state $f$ ($U_f$)

From the provided $P-V$ diagram, segment $i \to a$ is an isochoric process (constant volume), which means no work is done along this segment: $W_{ia} = 0$

Therefore, the total work done along the path $iaf$ is: $W_{iaf} = W_{ia} + W_{af} = 0 + 200\text{ J} = 200\text{ J}$

Using the First Law of Thermodynamics for the path $iaf$: $Q_{iaf} = \Delta U_{iaf} + W_{iaf} = (U_f - U_i) + W_{iaf}$

Given $Q_{iaf} = 500\text{ J}$, $U_i = 100\text{ J}$, and $W_{iaf} = 200\text{ J}$: $500 = (U_f - 100) + 200$ $U_f - 100 = 300$ $U_f = 400\text{ J}$

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Step 2: Calculate the heat absorbed along segment $ib$ ($Q_{ib}$)

For segment $ib$, the change in internal energy is: $\Delta U_{ib} = U_b - U_i = 200\text{ J} - 100\text{ J} = 100\text{ J}$

Given $W_{ib} = 50\text{ J}$, the heat supplied along segment $ib$ is: $Q_{ib} = \Delta U_{ib} + W_{ib} = 100\text{ J} + 50\text{ J} = 150\text{ J}$

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Step 3: Calculate the heat absorbed along segment $bf$ ($Q_{bf}$)

For segment $bf$, the change in internal energy is: $\Delta U_{bf} = U_f - U_b = 400\text{ J} - 200\text{ J} = 200\text{ J}$

Given $W_{bf} = 100\text{ J}$, the heat supplied along segment $bf$ is: $Q_{bf} = \Delta U_{bf} + W_{bf} = 200\text{ J} + 100\text{ J} = 300\text{ J}$

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Step 4: Compute the ratio $\frac{Q_{bf}}{Q_{ib}}$

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