The relative humidity of air is 80% at $27^{o} C$ . If the aqueous tension at the same temperature is 27 mm Hg, then the partial pressure of water vapour in the air will be:

21.6 mm

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27.5 mm

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25.2 mm

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23.9 mm

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Solution

The correct option is A 21.6 mm
% relative humidity $=$ partial pressure of water $\times (\frac{100}{a q u e o u s t e n s i o n})$

$⟹ 80 =$ partial pressure of water $\times (\frac{100}{27})$

Hence, the partial pressure of water is $21.6 m m$ .

Hence, the correct answer is option $A$ .

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The relative humidity of air is 80% at 27oC. If the aqueous tension at the same temperature is 27 mm Hg, then the partial pressure of water vapour in the air will be:

The correct option is A 21.6 mm% relative humidity = partial pressure of water ×(100aqueoustension)⟹80= partial pressure of water ×(10027)Hence, the partial pressure of water is 21.6 mm.Hence, the correct answer is option A.

The relative humidity of air is 80% at $27^{o} C$ . If the aqueous tension at the same temperature is 27 mm Hg, then the partial pressure of water vapour in the air will be:

The correct option is A 21.6 mm
% relative humidity $=$ partial pressure of water $\times (\frac{100}{a q u e o u s t e n s i o n})$

$⟹ 80 =$ partial pressure of water $\times (\frac{100}{27})$

Hence, the partial pressure of water is $21.6 m m$ .

Hence, the correct answer is option $A$ .