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^oC. 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)[(1/T2) - (1/T1)]

where P1 and T1 are the normal boiling point values (1 atm and 36.0°C, respectively), ΔHvap is the enthalpy of vaporization (27.6 kJ/mol), R is the gas constant (8.314 J/mol·K), T2 is the given temperature in Kelvin (273.15 K + 0.0°C = 273.15 K), and P2 is the vapor pressure we want to find.

Plugging in the values and solving for P2:

ln(P2/1 atm) = -(27.6 kJ/mol / 8.314 J/mol·K) [(1/273.15 K) - (1/309.15 K)]

ln(P2/1 atm) = -3.3252

P2/1 atm = e^(-3.3252)

P2 = 0.000184 atm

Finally, we convert the pressure to torr:

0.000184 atm x 760 torr/atm = 0.1398 torr ≈ 184 torr.

Portable Lanterns and Stoves Used for Camping Often Use a Mixture