Fisika Study Center

This problem sample below is taken from theoretical problems of first IphO that's conducted in Warsaw, Poland 1967, problem 4.

Problem 4
A closed vessel with volume V₀ = 10 l contains dry air in the normal conditions (t₀ = 0°C, p₀ = 1 atm). In some moment 3 g of water were added to the vessel and the system was warmed up to t = 100°C. Find the pressure in the vessel. Discuss assumption you made to solve the problem.

Solution
The water added to the vessel evaporates. Assume that the whole portion of water evaporated. Then the density of water vapor in 100°C should be 0.300 g/l. It is less than the density of saturated vapor at 100°C equal to 0.597 g/l. (The students were allowed to use physical tables.) So, at 100°C the vessel contains air and unsaturated water vapor only (without any liquid phase).
Now we assume that both air and unsaturated water vapor behave as ideal gases. In view of Dalton law, the total pressure p in the vessel at 100°C is equal to the sum of partial pressures of the air p_a and unsaturated water vapor p_v:

p = p_a + p_a

As the volume of the vessel is constant, we may apply the Gay-Lussac law to the air. We obtain:



The pressure of the water vapor may be found from the equation of state of the ideal gas:



where m denotes the mass of the vapor, μ - the molecular mass of the water and R – the universal gas constant. Thus,



and finally



Numerically:

p = (1.366 + 0.516) atm ≈ 1.88 atm.

Sources/Literatures :
-Waldemar Gorzkowski, Institute of Physics, Polish Academy of Sciences, Warsaw, Poland
-The Olympic home page www.jyu.fi/ipho