Answer
2.4 kg.
Work Step by Step
Table 13–3 gives the saturated vapor pressure at different temperatures. The saturated vapor pressure at $20^{\circ}C$ is $2.33\times10^3 Pa$. Use the ideal gas law to calculate the moles of water at both 95 percent and at 40 percent, and subtract the amounts to find the amount that must be removed.
Assume that the water vapor is an ideal gas and that the temperature remains constant.
$$PV=nRT$$
$$n_1-n_2=\frac{V}{RT}(P_1-P_2)$$
$$=\frac{ (105m^2)(2.4m)}{(8.314 J/(mol \cdot K))(273+20K)} (0.95-0.40)(2.33\times10^3Pa)=132.6mol$$
$$132.6mol\frac{0.018kg}{1mol}\approx 2.4kg$$
