Answer
(a) The specific volume is volume per unit mass. Thus, the volume occupied by water vapour can be deteremined by the given expression :-
$V$ = $m$$v$ =$(5 Kg )$$( 0.2160$ $ m^{3}$/$kg$ ) = $1.08 $ $m^{3}$
(b) By using molecular weight of water from Table A-1 , moles can be given as below :-
$n$ = $\frac{m}{M}$ = ($ \frac{5 moles}{18.02 kg/kmol }$ )($ \frac{1000 moles}{1 kmol}$) $= 277.5$ $moles $
(c) By Avogardo's number to determine the number of molecules gives
Molecules = Avogardo's number $\times$ moles = ( $ 6.022 $$\times$$10^{23}$ ) ( $277.5$ ) = $1.671$ $\times$ $10^{26}$ molecules
Work Step by Step
(a) The specific volume is volume per unit mass. Thus, the volume occupied by water vapour can be deteremined by the given expression :-
$V$ = $m$$v$ =$(5 Kg )$$( 0.2160$ $ m^{3}$/$kg$ ) = $1.08 $ $m^{3}$
(b) By using molecular weight of water from Table A-1 , moles can be given as below :-
$n$ = $\frac{m}{M}$ = ($ \frac{5 moles}{18.02 kg/kmol }$ )($ \frac{1000 moles}{1 kmol}$) $= 277.5$ $moles $
(c) By Avogardo's number to determine the number of molecules gives
Molecules = Avogardo's number $\times$ moles = ( $ 6.022 $$\times$$10^{23}$ ) ( $277.5$ ) = $1.671$ $\times$ $10^{26}$ molecules
