23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2014
Question 2 of 7: Vapour Pressure of Gasoline in a Sealed Can
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2014 — 04-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A, 15 marks), one of Q3–Q4 (Part B, 25 marks) and two of Q5–Q7 (Part C, 30 marks each); four questions total 100 marks. All seven are solved below for completeness. Property data are stated explicitly in each Given block; units follow the paper (mixed SI and US/older conventions).
Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material & energy balances, single-phase systems, combustion and recycle; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — VLE and excess-property (Margules) models, compressor work, and reaction equilibrium; supporting property data from Perry's Chemical Engineers' Handbook (9th ed.) and the NIST Chemistry WebBook (Antoine constants, C₀p polynomials, standard enthalpies and Gibbs energies of formation).
Question 2: Vapour Pressure of Gasoline in a Sealed Can (Part A, 15 marks)
Given. A liquid gasoline whose volatile constituents (liquid mole fractions) are 11% i-butane, 10% i-pentane, 12% n-hexane, 7% n-heptane; the balancing 60 mol% is non-volatile ($P^{\text{sat}}\approx0$) and contributes no vapour. Antoine constants ($\log_{10}P^{\text{sat}}[\text{mm Hg}] = A - B/(T[^{\circ}\text{C}]+C)$):
Species
$A$
$B$
$C$
$P^{\text{sat}}(38^{\circ}\text{C})$
i-butane
6.74808
882.80
240.00
3737 mm Hg
i-pentane
6.78967
1020.01
233.10
1064 mm Hg
n-hexane
6.87776
1171.53
224.37
259 mm Hg
n-heptane
6.89385
1264.37
216.64
85 mm Hg
Find. (a) whether the can deforms at 38 °C; (b) the temperature at which the gasoline’s vapour first drives the gauge pressure to the 200 mm Hg limit.
Approach. By Raoult’s law each volatile species exerts a partial pressure $p_i = x_i P_i^{\text{sat}}(T)$ in the head-space; their sum is the pressure the gasoline adds above the trapped air, i.e. the gauge pressure.
Check (modelling assumption): the can is taken to be sealed at atmospheric pressure, so the trapped air holds ~760 mm Hg and the gauge reading equals the gasoline’s added vapour pressure $\sum x_i P_i^{\text{sat}}$. Non-volatile components are retained in the liquid mole-fraction basis (they dilute but do not evaporate).
Partial pressures at 38 °C (Raoult). Multiplying each saturation pressure by its liquid mole fraction:
$$p_{\text{iC4}}=0.11(3737)=411,\quad p_{\text{iC5}}=0.10(1064)=106,$$
$$p_{\text{nC6}}=0.12(259)=31,\quad p_{\text{nC7}}=0.07(85)=6\ \text{mm Hg}.$$
Total gasoline vapour pressure. Summing the volatile contributions,
$$P_{\text{vap}}(38^{\circ}\text{C}) = \sum_i x_i P_i^{\text{sat}} = \boxed{554\ \text{mm Hg (gauge)}}.$$
Since $554 \gg 200$ mm Hg, the answer to (a) is yes — the can will deform.
Threshold temperature for part (b). Set the gauge sum to the rated 200 mm Hg and solve $\sum_i x_i P_i^{\text{sat}}(T)=200$ for $T$. Because i-butane dominates and its $P^{\text{sat}}$ falls steeply on cooling, the root is
$$T \approx \boxed{5.9^{\circ}\text{C}}.$$
Interpret. The can only stays within its 200 mm Hg rating below about 6 °C; at any ordinary ambient temperature — and certainly at 38 °C — the volatile i-butane fraction alone (411 mm Hg) already exceeds the limit.
