Answer :
Answer:
[tex]m=1864.68\ g[/tex]
Explanation:
Given:
volume of air in the room, [tex]V=410\ m^3[/tex]
temperature of the room, [tex]T=25+273=298\ K[/tex]
Saturation water vapor pressure at any temperature T K is given as:
[tex]p_{sw}=\frac{e^{(77.3450 + 0.0057\times T - \frac{ 7235}{T} )}}{T^{8.2}}[/tex]
putting T=298 K we have
[tex]p_{sw}=3130\ Pa[/tex]
The no. of moles of water molecules that this volume of air can hold is:
Using Ideal gas law,
[tex]P.V=n.R.T[/tex]
[tex]n=\frac{P_{sw}.V}{R.T}[/tex]
[tex]n=\frac{3130\times 410}{8.314\times 298}[/tex]
[tex]n=518\ moles[/tex] is the maximum capacity of the given volume of air to hold the moisture.
Currently we have 80% of n, so the mass of 20% of n:
[tex]m=(20\%\ of\ n)\times M}[/tex]
where;
M= molecular mass of water
[tex]m=0.2\times 518\times 18[/tex]
[tex]m=1864.68\ g[/tex] is the mass of water that can vaporize further.