At 100,^circtext{C} and 1 , text{atm} pressur | vedclass

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(C) The density of liquid water is $ho_l = 1.0 \, 	ext{g cm}^{-3}$.The density of water vapor is $ho_v = 0.0006 \, 	ext{g cm}^{-3} = 6 	imes 10

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(C) The density of liquid water is $ho_l = 1.0 \, 	ext{g cm}^{-3}$.The density of water vapor is $ho_v = 0.0006 \, 	ext{g cm}^{-3} = 6 	imes 10^{-4} \, 	ext{g cm}^{-3}$.In $1 \, 	ext{L}$ $(1000 \, 	ext{cm}^3)$ of steam,the mass of water vapor is $m = ho_v 	imes V = (6 	imes 10^{-4} \, 	ext{g cm}^{-3}) 	imes (1000 \, 	ext{cm}^3) = 0.6 \, 	ext{g}$.Since the mass of water remains constant during phase change,this $0.6 \, 	ext{g}$ of water occupies a volume in the liquid state given by $V_l = \frac{m}{ho_l} = \frac{0.6 \, 	ext{g}}{1.0 \, 	ext{g cm}^{-3}} = 0.6 \, 	ext{cm}^3$.

At $100\,^\circ\text{C}$ and $1 \, \text{atm}$ pressure,if the density of liquid water is $1.0 \, \text{g cm}^{-3}$ and the density of water vapor is $0.0006 \, \text{g cm}^{-3}$,then the volume occupied by water molecules in $1 \, \text{L}$ of steam is ............. $\text{cm}^3$.

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