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Question

Calculate the percentage of free volume available in one mole gaseous water at 1.0 atm and 100oC. Density of liquid H2O at 100oC is 0.958 g/mL (as nearest integer)

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Solution

For gaseous water PV=nRT
Thus, volume occupied by 1 mole gaseous water can be derived as (T=273+100=373K)
V=nRTP=1×0.0821×3731=30.62litre
Also, volume of 1 mole of liquid water = mass/ density
=180.958=1879mL=18.79×103litre
Thus, volume percentage occupied by water molecules in gaseous state
=18.79×10330.62×100=0.0614
Percentage of free volume=1000.0614=99.9386
so nearest integer value is 100.

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