A box contains 0.90 g of liquid water in equilibrium with water vapour at 27°C. The equilibrium vapour pressure of water at 27°C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be __________ litre. [nearest integer]
(Given : R = 0.082 L atm K⁻¹ mol⁻¹)
(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

Solution

Chemistry ma’am · 3 steps

The last drop vanishes when 0.05 mol of vapour fills the box at exactly 32 Torr. Find that volume.

1. Pressure stays fixed. While any liquid is left, the vapour pressure stays at 32.0 Torr, however large the box gets. At the largest volume the box can have with liquid still present, all the water is vapour at this same pressure.
2. Moles and pressure. Moles of water: n = 0.90/18 = 0.05 mol. Pressure in atm, to match R: P = 32.0/760 atm = 0.0421 atm. Temperature: T = 27 + 273 = 300 K.

   V = nRT/P = (0.05 × 0.082 × 300)/(32/760) = 1.23 × 760/32 L

3. So. V = 1.23 × 23.75 = 29.2 L, which rounds to 29.

The common slip: 3. Putting T = 27 instead of 300 K gives about 2.6 L, which rounds to 3. Always convert to kelvin.

The idea to keep. With liquid present, the vapour pressure depends only on temperature, not on volume. All the liquid has turned to vapour when the whole sample, used as an ideal gas at that pressure, just fills the box.

Answer from the NTA final answer key. Solution by PrepWiser.