Q1(h). Sealed sample: ethane, neon or a mixture
A sealed container contains a gaseous sample at 300 K consisting of either pure ethane, or pure neon, or a mixture of the two. The pressure inside the container at this temperature is 1.00 atm. When the container is cooled to 150 K, the pressure is 0.37 atm. What is the composition of the sample; pure ethane, pure neon or a mixture of both? Explain your answer. Given: Vapour pressure of C₂H₆ at 150 K is 0.10 atm; critical temperature of neon = 44 K.
Answer in brief
A mixture: neon 0.54 atm and ethane 0.46 atm at 300 K (mole fractions 0.54 and 0.46).
Key points
- At constant volume a gas that stays a gas obeys p ∝ T, so halving T should halve the pressure.
- Pure neon cannot condense (Tc = 44 K), so it would give 0.50 atm at 150 K; observed 0.37 atm, so rejected.
- Pure ethane would condense and the pressure would stay at its vapour pressure, 0.10 atm; observed 0.37 atm, so rejected.
- Mixture: ethane contributes only 0.10 atm, neon the rest (Dalton’s law), pNe = 0.27 atm at 150 K, i.e. 0.54 atm at 300 K; pC₂H₆ = 0.46 atm.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversDalton's law of partial pressurescritical temperaturevapour pressuregas mixturecondensation
All parts of Question 1
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