Question

If the humidity in a room of volume 410 m3 at 25 ∘C is 70%, what mass of water can still evaporate from an open pan? Express your answer using two significant figures.

Answer 1

Answer:

$m=1864.68\ g$

Explanation:

Given:

volume of air in the room, $V=410\ m^3$

temperature of the room, $T=25+273=298\ K$

Saturation water vapor pressure at any temperature T K is given as:

$p_{sw}=\frac{e^{(77.3450 + 0.0057\times T - \frac{ 7235}{T} )}}{T^{8.2}}$

putting T=298 K we have

$p_{sw}=3130\ Pa$

The no. of moles of water molecules that this volume of air can hold is:

Using Ideal gas law,

$P.V=n.R.T$

$n=\frac{P_{sw}.V}{R.T}$

$n=\frac{3130\times 410}{8.314\times 298}$

$n=518\ moles$ 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:

$m=(20\%\ of\ n)\times M}$

where;

M= molecular mass of water

$m=0.2\times 518\times 18$

$m=1864.68\ g$ is the mass of water that can vaporize further.