IPhO 1967 · Theory problem 4

Pressure in a closed vessel with air and evaporated water

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Comment: The Organizing Committee prepared three theoretical problems. Unfortunately, at the time of the 1st Olympiad the Romanian students from the last class had the entrance examinations at the universities. For that Romania sent a team consisting of students from younger classes. They were not familiar with electricity. To give them a chance the Organizers (under agreement of the International Board) added the fourth problem presented here. The students (not only from Romania) were allowed to chose three problems. The maximum possible scores for the problems were: 1st problem – 10 points, 2nd problem – 10 points, 3rd problem – 10 points and 4th problem – 6 points. The fourth problem was solved by 8 students. Only four of them solved the problem for 6 points.

A closed vessel with volume \(V_0 = 10\ \mathrm{l}\) contains dry air in the normal conditions (\(t_0 = 0^\circ\mathrm{C}\), \(p_0 = 1\ \mathrm{atm}\)). In some moment 3 g of water were added to the vessel and the system was warmed up to \(t = 100^\circ\mathrm{C}\). Find the pressure in the vessel. Discuss assumption you made to solve the problem.

Official English text of the problem.

Answers

  • Problem 4 (6 points): The final absolute pressure is \[ \boxed{p \simeq 1.88\ \mathrm{atm} \simeq 1.90\times10^5\ \mathrm{Pa}.} \] All \(3\ \mathrm{g}\) of water evaporate; the resulting vapour is unsaturated. This assumes a rigid vessel, no loss of air during water addition, ideal-gas behaviour, and thermal and phase equilibrium at \(100^\circ\mathrm{C}\).

Solution

Problem 4

Use absolute temperatures: \[ T_0=273.15\ \mathrm{K},\qquad T=373.15\ \mathrm{K}, \qquad V_0=1.00\times10^{-2}\ \mathrm{m^3}. \]

For an ideal mixture, Dalton’s law gives the total pressure as the sum of the air and water-vapour partial pressures: \[ p=p_{\mathrm a}+p_{\mathrm v}. \]

The amount of air is fixed. Applying the ideal-gas law initially and finally, \[ n_{\mathrm a}=\frac{p_0V_0}{RT_0}, \qquad p_{\mathrm a}=\frac{n_{\mathrm a}RT}{V_0} =p_0\frac{T}{T_0}. \] Thus \[ p_{\mathrm a} =1\ \mathrm{atm}\times\frac{373.15}{273.15} \simeq1.366\ \mathrm{atm}. \]

We must determine whether all the added water evaporates. If it does, its amount is \[ n_{\mathrm v}=\frac{m}{M} =\frac{3.0\ \mathrm{g}}{18.0\ \mathrm{g\,mol^{-1}}} =\frac16\ \mathrm{mol}. \] Its corresponding partial pressure would be \[ p_{\mathrm v}^{(\mathrm{all})} =\frac{mRT}{MV_0} =\frac{(1/6)(8.314)(373.15)}{1.00\times10^{-2}} \simeq5.17\times10^4\ \mathrm{Pa} \simeq0.510\ \mathrm{atm}. \]

At \(100^\circ\mathrm{C}\), water’s saturation vapour pressure is approximately \[ p_{\mathrm{sat}}=1\ \mathrm{atm}. \] Since \[ p_{\mathrm v}^{(\mathrm{all})}<p_{\mathrm{sat}}, \] even complete evaporation produces unsaturated vapour. Any liquid remaining would therefore continue to evaporate. Consequently, all the water evaporates.

Equivalently, the water mass needed just to saturate the vessel is \[ m_{\mathrm{sat}} =\frac{Mp_{\mathrm{sat}}V_0}{RT} \simeq5.88\ \mathrm{g}, \] which exceeds the available \(3\ \mathrm{g}\). Saturation is determined by the water’s partial pressure.

The final total pressure is therefore \[ \boxed{ p=p_0\frac{T}{T_0}+\frac{mRT}{MV_0} \simeq1.366+0.510 =1.876\ \mathrm{atm} \simeq1.90\times10^5\ \mathrm{Pa}. } \]

The assumptions underlying this result are:

  • Constant vessel volume: thermal expansion and pressure-induced deformation of the vessel are negligible.
  • Conservation of air: adding the water does not allow air to escape, and the vessel remains sealed.
  • Ideal-gas behaviour: air and water vapour obey the ideal-gas law and Dalton’s law. These are reasonable approximations at the stated temperature and pressures.
  • Equilibrium: sufficient heat and time are supplied for the entire contents to reach \(100^\circ\mathrm{C}\) and for evaporation to finish. The latent heat of evaporation affects the required heat input, but not the final pressure at the specified temperature.

The solution was checked against the official solution.