91Ó°ÊÓ

Given,

The temperature ${\text{=26}}^{\text{o}}\text{C=299k}$.

$\begin{array}{c}\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{water}=4.2\mathrm{L}\\ 1\mathrm{L}=1000\mathrm{mL}\\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{water}=4.20\mathrm{L}×\frac{1000\mathrm{mL}}{1\mathrm{L}}\end{array}$

Density of ${\text{H}}_{\text{2}}\text{O=1.00g/mL}$

As,

$\text{NumberofMoles=}\frac{\text{Mass}}{\text{MolarMass}}$

And,

$\text{Density=}\frac{\text{Mass}}{\text{Volume}}$

$\text{²Ñ²¹²õ²õ=³Õ´Ç±ô³Ü³¾±ð׶ٱð²Ô²õ¾±³Ù²â}$

Therefore,

$\text{±·³Ü³¾²ú±ð°ù o´Ú â¶Ä‰M´Ç±ô±ð²õ=}\frac{\text{4.20³¢Ã—}\frac{\text{1000mL}}{\text{1L}}\text{×1.00²µ/³¾³¢}}{\text{18g/mole}}$

As,

Gas Constant, $\text{R=0.0821Latm/k/mole}$

Then, the pressure at temperature ${\text{=26}}^{\text{o}}\text{C=299k}$.

$\mathbf{PV}\mathbf{=}\mathbf{nRT}$

Therefore,

$\begin{array}{c}\mathrm{P}=\frac{\mathrm{n}×\mathrm{R}×\mathrm{T}}{}\\ \mathrm{P}=\frac{\left[\frac{4.20×\frac{1000\mathrm{mL}}{1\mathrm{L}}×1.00\mathrm{g}/\mathrm{mL}}{18.02\mathrm{g}/\mathrm{mL}}\right]}{256{\mathrm{m}}^3×\frac{1000\mathrm{L}}{1{\mathrm{m}}^3}}×(0.0821\mathrm{Latm}/\mathrm{mol}/\mathrm{k})×(299\mathrm{k})\\ \mathrm{P}=\frac{233.07}{}×24.55\\ \mathrm{P}=0.022\mathrm{atm}\end{array}$

The Absolute Humidity as the saturated air with water vapour.

$\begin{array}{c}\text{AbsoluteHumidity=}\frac{\text{Pressure}}{\text{³Ò²¹²õ C´Ç²õ²Ô³Ù²¹²Ô³Ù o´Ú W²¹³Ù±ð°ùװձ𳾱è±ð°ù²¹³Ù³Ü°ù±ð}}\\ \text{AbsoluteHumidity=}\frac{\text{0.022atm}}{\text{461.5×299°ì}}\\ {\text{´¡²ú²õ´Ç±ô³Ü³Ù±ð±á³Ü³¾¾±»å¾±³Ù²â=1.6×10}}^{\text{-7}}\end{array}$