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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 $Q_{b f} / Q_{i b}$ is

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answer is 2.

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Detailed Solution

$Δ Q_{i a f} = Δ U_{i a f} + W_{i a f} \Rightarrow 500 = Δ U_{i a f} + 200 J$

$Δ U_{i a f} = 300 J ∴ Δ U_{i b f} = 300 J$

$W_{i b f} = W_{i b} + W_{b f} = 50 + 100 = 150 J$

$Q_{i b f} = W_{i b f} + Δ U_{i b f} = 150 + 300 = 450$

$Δ Q_{i b f} = 450 J = Q_{i b} + Q_{b f}$

$U_{i} = 100 J U_{b} = 200 J Δ U_{i b} = 100 J W_{i b} = 50 J$

$∴ 450 = Q_{i b} + Q_{b f} \Rightarrow 450 = 150 J + Q_{b f}$

$Q_{b f} = 300 J \Rightarrow \frac{Q_{b f}}{Q_{i b}} = \frac{300}{150} = 2$

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