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 ΔH_vap:

ln(p2/p1) = ΔH_vap(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 ΔH_vap 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.

Source:

¹Only Vapors Burn 1947 Chemistry of Fire US War Department

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.