Gasoline is a mixture of hydrocarbons, a major component of which is octane,CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressure of 13.95 torr at 25 degrees celcius and a vapor pressure of 144.78 torr at 75 degrees celcius. Use these data and the equation in part (a) to calculate the heat of vaporization of octane.
To calculate the heat of vaporization of octane, we can use the Clausius-Clapeyron equation, which relates the vapor pressures of a substance at two different temperatures to its heat of vaporization. The equation is as follows:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively
ΔHvap is the heat of vaporization
R is the ideal gas constant (8.314 J/(mol·K))
T1 and T2 are the temperatures in Kelvin
Now, let's calculate the heat of vaporization of octane:
Given:
Octane vapor pressure at 25°C (298K) = 13.95 torr
Octane vapor pressure at 75°C (348K) = 144.78 torr
First, we need to convert the vapor pressures from torr to atm:
1 atm = 760 torr
P1 = 13.95 torr / 760 torr/atm = 0.0183 atm
P2 = 144.78 torr / 760 torr/atm = 0.1904 atm
Next, we plug the values into the Clausius-Clapeyron equation:
ln(0.1904/0.0183) = -(ΔHvap/8.314) * (1/348 - 1/298)
Simplifying the equation:
ln(10.4) = -(ΔHvap/8.314) * (0.002865 - 0.003356)
Dividing both sides of the equation by -8.314:
ln(10.4) / -8.314 = ΔHvap * (0.000491)
Rearranging the equation:
ΔHvap = (ln(10.4) / -8.314) / 0.000491
Calculating ΔHvap:
ΔHvap = -8283.55 J/mol
Therefore, the heat of vaporization of octane is approximately -8283.55 J/mol.
To calculate the heat of vaporization of octane, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔHvap/R)((1/T2) - (1/T1))
Where:
P1 = Vapor pressure at temperature T1
P2 = Vapor pressure at temperature T2
ΔHvap = Heat of vaporization
R = Universal gas constant (8.314 J/(mol·K))
Let's plug in the given values and solve for ΔHvap:
P1 = 13.95 torr
P2 = 144.78 torr
T1 = 25 degrees Celsius = 25 + 273.15 Kelvin
T2 = 75 degrees Celsius = 75 + 273.15 Kelvin
R = 8.314 J/(mol·K)
ln(144.78/13.95) = -(ΔHvap/8.314)((1/(75+273.15)) - (1/(25+273.15)))
First, let's calculate the natural logarithm on the left side:
ln(144.78/13.95) ≈ 1.903
Now, let's simplify the right side of the equation:
(1/(75+273.15)) - (1/(25+273.15)) ≈ 0.0038
(-ΔHvap/8.314) * 0.0038 = 1.903
Now we can solve for ΔHvap:
-ΔHvap * 0.0038 = 1.903 * 8.314
-ΔHvap ≈ 15.803
Finally, we can find the heat of vaporization by multiplying both sides by -1:
ΔHvap ≈ -15.803 J/mol
Therefore, the heat of vaporization of octane is approximately -15.803 J/mol.