Multiple choice

A container contains the mixture of water vapour and oxygen gas with total pressure 1.1 atm at a certain temperature. If the volume is made one third then the total pressure will be: [Assume aqueous tension of water at this temperature is 0.1 atm.]

  1. $3.3$ atm
  2. $3.1$ atm
  3. $0.43$ atm
  4. $\cfrac{1.1}{3}$ atm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total pressure = P_water_vapour + P_oxygen. Aqueous tension (P_water_vapour) is constant at 0.1 atm. P_oxygen = 1.1 - 0.1 = 1.0 atm. If volume is 1/3, P_oxygen becomes 1.0 * 3 = 3.0 atm. Total pressure = 3.0 + 0.1 = 3.1 atm.

AI explanation

The total pressure of 1.1 atm includes the aqueous tension of 0.1 atm, so the initial pressure of the oxygen gas alone is 1.0 atm. When the volume is reduced to one third, Boyle's law dictates the oxygen pressure triples to 3.0 atm, while the water vapour pressure remains constant at 0.1 atm. Adding these together gives a final total pressure of 3.1 atm.

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