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The volume occupied by 1.8 g of water vapour at $374^\circ C$ and 1 bar pressure will be: [Use R = 0.083 bar L K$^{-1}$ mol$^{-1}$]
Detailed Solution
$n=\frac{1.8}{18}=0.1$ mol, $T=374+273=647$ K. $V=\frac{nRT}{P}=\frac{0.1\times0.083\times647}{1}=5.37$ L.
