Heat And Thermodynamics Ques 162
- 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 $i b f$, 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 $Q _{b f} / Q _{i b} i$
(2014 Adv.)
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Answer:
Correct Answer: 162.(2)
Solution:
Formula:
- $W _{i b f}=W _{i b}+W _{b f}=50 J+100 J=150 J$
$$ \begin{aligned} W _{i a f} & =W _{i a}+W _{a f}=0+200 J=200 J \\ Q _{i a f} & =500 J \\ \Delta U _{i a f} & =Q _{i a f}-W _{i a f} \\ & =500 J-200 J \\ & =300 J=U _f-U _i \\ \text { So } \quad U _f & =U _{i a f}+U _i \\ & =300 J+100 J=400 J \\ \Delta U _{i b} & =U _b-U _i \\ & =200 J-100 J=100 J \\ Q _{i b} & =\Delta U _{i b}+W _{i b} \\ & =100 J+50 J=150 J \\ Q _{i b f} & =\Delta U _{i b f}+W _{i b f}=\Delta U _{i a f}+W _{i b f} \\ & =300 J+150 J=450 J \end{aligned} $$
So, the required ratio
$$ \begin{aligned} \frac{Q _{b f}}{Q _{i b}} & =\frac{Q _{i b f}-Q _{i b}}{Q _{i b}} \\ & =\frac{450-150}{150}=2 \end{aligned} $$
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