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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. |
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Answer» The volume occupied by water is very small and thus the volume change is equal to the volume occupied by one gram mole of water vapour. `V=(nRT)/(P)=(1.0xx0.0821xx373)/(1.0)=31.0` litre `w=-P_(ext)xxDeltaV=-(1.0)xx(31.0)` litre-atm `=-(31.0)xx101.3J=-3140.3J` |
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