Question

Suppose the inhaled air has partial pressure of water vapour of 5 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 breaths 10,000 litre per day. Saturated vapour pressure of water at 310 K is 45 mm Hg.

• A. 20.68 g
• B. 418.76 g
• C. 372.23 g
• D. 46.53 g

Answer 1

Answer: (C)

372.23 g

Answer 2

Calculate the difference in partial pressure of water vapour: $\Delta P = 45 - 5 = 40$ mm Hg. Convert $\Delta P$ to atm: $40/760$ atm. Use the ideal gas law $n = PV/RT$ for $V=10000$ L, $T=310$ K, and $R=0.0821 \, L \cdot atm/(mol \cdot K)$ to find moles of water vapour. Mass = moles $\times$ molar mass of water (18 g/mol).