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 $i a f$ and $i b f,$ as schematically shown in the figure. The work done by the system along the paths $a f, i b$ and $b f$ are $W_{a f} = 200 J, W_{i b} = 50 J$ and $W_{b f} = 100 J$ respectively. The heat supplied to the system along the path $i a f, i b$ and $b f$ are $Q_{i a f}, Q_{i b}$ and $Q_{b f}$ respectively. If the internal energy of the system in the state $b$ is $U_{b} = 200 J$ and $Q_{i a f} = 500 J,$ the ratio of $Q_{b f} / Q_{i b}$ is

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Detailed Solution

$\begin{array}{l} Δ Q_{i a f} = Δ U_{i a f} + W_{i a f} \\ \Rightarrow 500 = Δ U_{i a f} + 200 J \\ Δ U_{i a f} = 300 J; Δ U_{i b f} = 300 J; W_{i b f} = W_{i b} + W_{b f} = 50 + 100 = 150 J \\ Q_{i b f} = W_{i b f} + Δ U_{i b f} = 150 + 300 = 450 J \\ Δ Q_{i b f} = 4505 = Q_{i b} + Q_{b f} \\ U_{i} = 100 J; U_{b} = 200 J; Δ U_{i b} = 100 J; W_{i b} = 50 J \\ Q_{i b} = Δ_{u i b} + W_{i b} = 150 j 450 = Q_{i b} + Q_{b f} \Rightarrow Q_{b f} = 450 - 150 = 300 J \\ \Rightarrow \frac{Q_{b f}}{Q_{i b}} = \frac{300}{150} = 2 \end{array}$

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