A thermodynamic system is taken from an initial state with internal energy to the final state along two different paths and , as schematically shown in the figure. [2014]

Question

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 $Q_{i a f} = 500 J,$ then the ratio $\frac{Q_{b f}}{Q_{i b}}$ is [2014]

Smart Achievers · Accepted Answer

(2)

Applying first law of thermodynamics to path $U_{i} = 100 J$

$Q_{i a f} = Δ U_{i a f} + W_{i a f}$

$500 = Δ U_{i a f} + 200$

$∴$ $Δ U_{i a f} = 300 J$

Now,

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

$Q_{i b} + Q_{b f} = 300 + 50 + 100 = 450 J ...(i)$

Also $Q_{i b} = Δ U_{i b} + W_{i b}$

$∴$ $Q_{i b} = 100 + 50 = 150 J ...(ii)$

From eq. (i) & (ii) $\frac{Q_{b f}}{Q_{i b}} = \frac{Q_{i b f} - Q_{i b}}{Q_{i b}} = \frac{300}{150} = 2$

Solution

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Solution

Ans. (2) Applying first law of thermodynamics to path iaf Qiaf=ΔUiaf+Wiaf 500=ΔUiaf+200 ∴ ΔUiaf=300 J Now, Qibf=ΔUibf+Wib+Wbf Qib+Qbf=300+50+100=450 J ...(i) Also Qib=ΔUib+Wib ∴ Qib=100+50=150 J ...(ii) From eq. (i) & (ii) QbfQib=Qibf-QibQib=300150=2

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