Sealed flask: evaporation, vapour pressure and a non-ideal gas
A rigid 2.0 L glass flask at 20 °C contains dry air at 1.00×10⁵ Pa plus a few millilitres of liquid ethanol. The flask is sealed and left until equilibrium. The saturated vapour pressure of ethanol at 20 °C is 5.9 kPa and its molar mass is 46 g/mol. A student then heats the flask to 60 °C in a water bath. At 60 °C the saturated vapour pressure of ethanol is 47 kPa. The student claims: 'Because the flask is rigid and sealed, the total pressure inside must be the same at 60 °C as at 20 °C, since no gas can escape.' A second student argues the pressure will simply scale as 333/293.
1. Calculate the total pressure in the flask at 20 °C at equilibrium and explain the role of the liquid ethanol.
Read the worked answer
At equilibrium the vapour is saturated, so its partial pressure equals the saturated vapour pressure p_vap = 5.9 kPa, independent of how much liquid is present (as long as some liquid remains).
The dry air occupies the full 2.0 L and contributes its own partial pressure. Since the flask is sealed and rigid, the amount of air is fixed and its partial pressure at 20 °C is the stated 1.00×10⁵ Pa.
By Dalton's law: p_total = p_air + p_vap = 100 kPa + 5.9 kPa = 105.9 kPa ≈ 1.06×10⁵ Pa.
The liquid ethanol acts as a reservoir: it supplies vapour until the partial pressure reaches p_sat, then stops evaporating (dynamic equilibrium: evaporation rate = condensation rate). The liquid does not 'add pressure' beyond p_sat.
2. Evaluate both students' claims for the 60 °C state. Compute the total pressure at 60 °C, stating clearly any assumption about whether liquid ethanol remains.
Read the worked answer
Both claims are wrong.
First student (pressure unchanged): false. The flask is rigid, but the temperature rises and more liquid evaporates, increasing the number of gas molecules. Pressure must rise.
Second student (simple T scaling, p₂ = p₁·T₂/T₁): this applies only to a fixed amount of ideal gas. Here (a) the amount of ethanol vapour changes because more evaporates, and (b) the vapour pressure is set by saturation, not by the ideal-gas law with fixed n. So simple scaling is invalid for the ethanol component.
Correct treatment:
- Dry air: fixed amount, rigid volume → p_air(60 °C) = 100 kPa × (333/293) = 113.7 kPa.
- Ethanol vapour: saturated at 60 °C → p_vap = 47 kPa (provided liquid remains).
Total: p = 113.7 + 47 = 160.7 kPa ≈ 1.61×10⁵ Pa.
Assumption: enough liquid ethanol is present that it is not fully evaporated at 60 °C. This must be checked (next task).
3. The flask contains only 'a few millilitres' of ethanol. Show quantitatively whether the saturation assumption at 60 °C is safe, and describe what happens to the pressure if it is not.
Read the worked answer
Estimate the vapour needed to saturate 2.0 L at 60 °C. Using the ideal-gas law for the vapour: n = p V / (R T) = (47×10³ × 2.0×10⁻³) / (8.31 × 333) = 94 / 2767 ≈ 0.034 mol.
Mass of ethanol: m = n M = 0.034 × 46 g/mol ≈ 1.56 g, i.e. about 2.0 mL of liquid ethanol (density ≈ 0.79 g/mL).
So 'a few millilitres' (say 3–5 mL) is just enough - but if the flask held only ~2 mL or less, all the liquid would evaporate before saturation, and the vapour would no longer be saturated. In that case the ethanol would behave as a fixed amount of gas and its partial pressure would scale with T from the actual evaporated amount, giving a lower pressure than 47 kPa.
Conclusion: the 160.7 kPa answer is an upper bound valid only if ≥ ~2 mL of liquid remains at 60 °C. A careful answer states this condition explicitly, since the problem deliberately makes the liquid amount marginal.