Question

On a humid place the temperature and relative humidity of air are ${27}^{\circ }C$ and 80% respectively. What mass of water vapour must be added to saturate 1 $m^3$ of air? [${27}^{\circ }C,P_{sat}=3.6kPa$]

• A. 5.18 gram
• B. 4.61 gram
• C. 2.5 gram
• D. 3.6 gram

Answer 1

The correct option is A 5.18 gram

Relative humidity,
$\varphi =0.8$

$\varphi =\frac{p_v}{p_s}\Rightarrow p_v=0.8\times 3.6=2.88kPa$

Initial mass of water vapour

$m_1=\frac{2.88\times 1}{0.461\times 300}\times 1000=20.82gm$

$[\because R=\frac{\overline{R}}{M}=\frac{8.314}{18}=0.461]$

When air becomes saturated, the partial pressure of vapour= 3.6 kPa

Mass of water vapour
$m_2=\frac{3.6\times 1}{0.461\times 300}\times 1000=26gram$

$\therefore$ Mass of water vapour added= 26-20.82= 5.18 gram