On a winter day, the outside temperature is $0^{\circ} C$ and relative humidity $40 %$ . The air from outside comes into a room and is heated to $20^{\circ} C$ . What is the relative humidity in the room ? The saturation vapour pressure at $0^{\circ} C$ is $4.6 m m$ of mercury and at $20^{\circ} C$ it is $18 m m$ of mercury.

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Solution

Relative humidity $= 40 %$
$S V P = 4.6 m m$ of Hg
$0.4 = \frac{V P}{4.6} \Rightarrow V P = 0.4 \times 4.6 = 1.84$
$\frac{P_{1} V}{T_{1}} = \frac{P_{2} V}{T_{2}} \Rightarrow \frac{1.84}{273} = \frac{P_{2}}{293} \Rightarrow P_{2} = \frac{1.84}{273} \times 293$
Relative humidity at $20^{\circ} C$
$= \frac{V P}{S V P} = \frac{1.84 \times 293}{273 \times 10} = 0.109 = 10.9 %$

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On a winter day, the outside temperature is 0∘C and relative humidity 40%. The air from outside comes into a room and is heated to 20∘C. What is the relative humidity in the room ? The saturation vapour pressure at 0∘C is 4.6mm of mercury and at 20∘C it is 18mm of mercury.

Relative humidity =40%SVP=4.6mm of Hg 0.4=VP4.6⇒VP=0.4×4.6=1.84P1VT1=P2VT2⇒1.84273=P2293⇒P2=1.84273×293Relative humidity at 20∘C=VPSVP=1.84×293273×10=0.109=10.9%