To burn, a fuel must evaporate.1<\/sup><\/p>\n In colder temperatures, fuel has less tendency to evaporate. For two specific temperatures (T1<\/sub> and T2<\/sub>), the Clausius-Clapeyron equation can give a comparison of the vapor pressures (p2<\/sub> vs p1<\/sub>) of a liquid with molar enthalpy of evaporation \u0394Hvap<\/sub>:<\/p>\n ln(p2\/p1) = \u0394Hvap<\/sub>(T2<\/sub> – T1<\/sub>)\/(R*T2<\/sub>*T1<\/sub>), where<\/p>\n R = 8.314 J\/(K*mol), the idea gas constant. <\/p>\n T1<\/sub>, T2<\/sub> are absolute temperatures.<\/p>\n Note, from my post from Feb 20,<\/a> that \u0394Hvap<\/sub> for gasoline is approximately 38.1kJ\/mol.<\/p>\n Example: Compare the vapor pressure of gasoline at 12\u00b0C (285K) vs -5\u00b0C (268K).<\/strong><\/p>\n Solution: Let T2<\/sub> be 268K (-5\u00b0C). Then<\/p>\n ln(p2\/p1) = 38100(268 – 285)\/(268*285) = -1.02<\/p>\n Taking the antilog gives<\/p>\n p2\/p1 = 0.36 = 36%<\/p>\n Apparently, gasoline is about 36% as likely to evaporate at -5\u00b0C as at 12\u00b0C. Not much surprise, really, that a car can be more difficult to start in colder temperatures.<\/p>\n Source:<\/p>\n 1<\/sup>Only Vapors Burn 1947 Chemistry of Fire US War Department<\/a><\/p>\n Mortimer, Charles E. Chemistry<\/u>, sixth ed. Belmont: Wadsworth, 1986.<\/p>\n White, J. Edmund. Physical Chemistry<\/u>, Harcourt Brace Jovanovich College Outline Series. San Diego: Harcourt Brace Jovanovich, 1987.<\/p>\n