Suppose the inhaled air has partial pressure of water vapour of 25 mm Hg and exhaled air is nearly saturated at body temperature (310 K) with water vapour. The mass of water lost per day by a person a

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Suppose the inhaled air has partial pressure of water vapour of 25 mm Hg and exhaled air is nearly saturated at body temperature (310 K) with water vapour. The mass of water lost per day by a person assuming that the normal man breathes 10,000 litre per day. Saturated vapour pressure of water at 310 K is 45 mm Hg.

TestHub · Accepted Answer

ΔP = 45 - 25 = 20 mm = 20/760 atm. Moles of extra water vapour per day: n = ΔP·V/RT = (20/760 × 10000)/(0.0821 × 310) = 263.2/25.45 = 10.34 mol. Mass = 10.34 × 18 = 186.12 g. Confirms Ans: (d) 186.12 g.

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