The tutor implicates the Clausius-Clapeyron equation to explain why combustion is more difficult at lower temperatures.
To burn, a fuel must evaporate.1
In colder temperatures, fuel has less tendency to evaporate. For two specific temperatures (T1 and T2), the Clausius-Clapeyron equation can give a comparison of the vapor pressures (p2 vs p1) of a liquid with molar enthalpy of evaporation ΔHvap:
ln(p2/p1) = ΔHvap(T2 – T1)/(R*T2*T1), where
R = 8.314 J/(K*mol), the idea gas constant.
T1, T2 are absolute temperatures.
Note, from my post from Feb 20, that ΔHvap for gasoline is approximately 38.1kJ/mol.
Example: Compare the vapor pressure of gasoline at 12°C (285K) vs -5°C (268K).
Solution: Let T2 be 268K (-5°C). Then
ln(p2/p1) = 38100(268 – 285)/(268*285) = -1.02
Taking the antilog gives
p2/p1 = 0.36 = 36%
Apparently, gasoline is about 36% as likely to evaporate at -5°C as at 12°C. Not much surprise, really, that a car can be more difficult to start in colder temperatures.
Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.
White, J. Edmund. Physical Chemistry, Harcourt Brace Jovanovich College Outline Series. San Diego: Harcourt Brace Jovanovich, 1987.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.