1.

Calculate the work done when `1.0` mol of water at `373K` vaporises against an atmosheric pressure of `1.0atm`. Assume ideal gas behaviour.

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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