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 $\circ$ C. The enthalpy of vaporization of n-pentane is 27.6 kJ/mol, and its boiling point is 36.0 $\circ$ C.
A)228 torr
B)367 torr
C)184 torr
D)479 torr
E)209 torr

Question

Quizplus · Accepted Answer

First, we need to convert the temperature from Celsius to Kelvin: 
0°C + 273.15 = 273.15 K

Next, we can use the Clausius-Clapeyron equation to calculate the vapor pressure:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

where:
P1 = vapor pressure at the boiling point (in atm or torr)
P2 = vapor pressure at the given temperature (in atm or torr)
ΔHvap = enthalpy of vaporization (in J/mol)
R = gas constant (8.314 J/mol*K)
T1 = boiling point temperature (in K)
T2 = given temperature (in K)

Converting the given temperature to Kelvin, we have:

T2 = 273.15 K

Substituting the given values and solving for P2:

ln(P2/1) = -(27600 J/mol) / (8.314 J/mol*K) * (1/273.15 K - 1/309.15 K)
ln(P2/1) = -(-5.105)
ln(P2/1) = 5.105
P2/1 = e^(5.105)
P2 = e^(5.105) * 1

P2 = 165.9 torr

Therefore, the vapor pressure of n-pentane at 0°C (273.15 K) is 165.9 torr, which is closest to choice C (184 torr).

Portable Lanterns and Stoves Used for Camping Often Use a Mixture