Question

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.)

Answer 1

Correct Answer: 29

To solve this problem, we need to determine the volume of the box when all the liquid water evaporates at equilibrium. To do this, we will use the Ideal Gas Law, \((PV = nRT)\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.

1. First, let's convert the temperature to Kelvin:
   \(T = 27°C + 273.15 = 300.15 \, \text{K}\)
2. Next, convert the given equilibrium pressure from Torr to Atmospheres, since \(R\) is provided in L atm K^-1mol^-1:
   \(P = 32.0 \, \text{Torr} \times \frac{1 \, \text{atm}}{760 \, \text{Torr}} = 0.0421 \, \text{atm}\)
3. Calculate the number of moles of water using its mass and molecular weight:
   Molecular weight of water \(= 18.0 \, \text{g/mol}\)
   \(n = \frac{0.90 \, \text{g}}{18.0 \, \text{g/mol}} = 0.05 \, \text{mol}\)
4. Apply the Ideal Gas Law to find the volume \(V\):
   \(PV = nRT \rightarrow V = \frac{nRT}{P}\)
   Substitute the known values:
   \(V = \frac{(0.05 \, \text{mol})(0.082 \, \text{L atm K}^{-1} \text{mol}^{-1})(300.15 \, \text{K})}{0.0421 \, \text{atm}}\)
   \(V = \frac{1.23063}{0.0421} = 29.23 \, \text{L}\)
5. Finally, round this to the nearest integer to get the required volume:
   \(V \approx 29 \, \text{L}\)

Thus, the volume of the box when all the liquid water evaporates is 29 litres, which fits the given range of 29 to 29.