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Calculate the work done when `1.0` mol of water at `373K` vaporises against an atmosheric pressure of `1.0atm`. Assume ideal gas behaviour.A. `-6200 J`B. `-306 J`C. `-3100 J`D. `-1550 J` |
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Answer» Correct Answer - C Volume of `1` mol of water at `373 K`, `1` atm `(V_(1))=18cm^(3)=0.018 L` Volume of `1` mol of water in vapour state at `373 K`, `1 "atm"=(RT)/(P)` `=(0.0821xx373)/(1)=30.62 L` Now `w=-P(V_(2)-V_(1))` `=1(30.62-0.018)` `=-30.6 L-"atm"` `=-30.6xx101.3=-3100 J` |
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