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Calculate the work done when 1 mol of wa...

Calculate the work done when 1 mol of water vaporises at `100^(@)C` and 1 atm pressure. Assume water vapour behaves like an ideal gas.

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1 mol water, `100^(@)C,1atm to` 1 mol water vapour `100^(@)C,1atm`
Work done, `w=-P_(ex)(V_(2)-V_(1))`
`P_(ex)=`External pressure=1 atm, `V_(1) and V_(2)` are volumes of 1 mol of water and 1 mol of water vapour, respectively. The volume of 1 mol of water vapour is very much greater than that of 1 mol water. thus `(V_(2)gtgt V_(1))` and it makes `(V_(2)-V_(1))~~V_(2)`.
`therefore`Work done, `w=-P_(ex)V_(2)`
If water vapour behaves like an ideal gas, then
`P_(ex)V_(2)=RT" "[because" amount of water vapour=1mol"]`
`therefore w=-P_(ex)V_(2)=-RT`
Here, `T=(273+100)K=373K`
`therefore w=-RT=-8.314xx373=-3101.12J`
Therefore, amount of work done by the system=3101.12J.
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