Portable lanterns and stoves used for camping often use a mixture of hydrocarbons for fuel. One such hydrocarbon is an alkane called n-pentane. These lanterns and stoves are often difficult to light on a cold day because the fuel has a low vapor pressure at low temperatures. Determine the vapor pressure of n-pentane on a night when the temperature is 0.0°C. The enthalpy of vaporization of n-pentane is 27.6 kJ/mol, and its boiling point is 36.0°C.

Question

Quizplus · Accepted Answer

To determine the vapor pressure of n-pentane at 0.0°C, we need to use the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔHvap/R) x ((1/T2)-(1/T1))
where P1 is the vapor pressure at the boiling point (36.0°C) and T1 is the boiling point in Kelvin (309.2 K). P2 is the vapor pressure at 0.0°C and T2 is the temperature in Kelvin (273.2 K). ΔHvap is the enthalpy of vaporization of n-pentane (27.6 kJ/mol) and R is the gas constant (8.314 J/mol*K).

Substituting the values, we get:
ln(P2/760) = -(27.6x10³)/(8.314) x ((1/273.2)-(1/309.2))
ln(P2/760) = -2.184
P2/760 = e^-2.184
P2 = 760 x e^-2.184
P2 = 184 torr (rounded to the nearest whole number)

Therefore, the vapor pressure of n-pentane at 0.0°C is 184 torr.

Portable Lanterns and Stoves Used for Camping Often Use a Mixture