Question

On a certain day, the vapour pressure is 24 mm of Hg for water vapour in air at 300 K. The saturated vapour pressure is 26.463 mm. How many moles of water vapour per litre of air would be required to saturate the air at this temperature?

• A. 0.1
• B. 0.0132
• C. 1.32 x $10^{-4}$
• D. 7.60

Answer 1

Answer: (C)

1.32 x $10^{-4}$

Answer 2

The pressure difference is $\Delta P = 26.463 \text{ mm Hg} - 24 \text{ mm Hg} = 2.463 \text{ mm Hg}$. Convert $\Delta P$ to atm: $\Delta P = \frac{2.463}{760} \text{ atm}$. Using the ideal gas law $PV = nRT$, we solve for $n$: $n = \frac{\Delta P \times V}{R \times T} = \frac{\left(\frac{2.463}{760} \text{ atm}\right) \times (1 \text{ L})}{(0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K})} \approx 1.303 \times 10^{-4}$ moles.