Gasoline is primarily a mixture of hydrocarbons and is sold with an octane rating that is based on a comparison with the combustion properties of isooctane. Gasoline usually contains an isomer of isooctane called tetramethylbutane (C8H18), which has an enthalpy of vaporization of 43.3 kJ/mol and a boiling point of 106.5°C. Determine the vapor pressure of tetramethylbutane on a very hot summer day when the temperature is 38°C.
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To determine the vapor pressure of tetramethylbutane at a given temperature, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 is the vapor pressure at temperature T1
P2 is the vapor pressure at temperature T2
ΔHvap is the enthalpy of vaporization
R is the ideal gas constant (8.314 J/(mol·K))
T1 is the initial temperature
T2 is the final temperature
Let's calculate the vapor pressure of tetramethylbutane at 38°C (38°C = 311 K), assuming the initial temperature (T1) is the boiling point of tetramethylbutane (106.5°C = 379.65 K):
ln(P2/P1) = (43.3 kJ/mol / (8.314 J/(mol·K))) * (1/379.65 K - 1/311 K)
Simplifying:
ln(P2/P1) = 5202.77 * (0.002634 - 0.003215)
ln(P2/P1) = 5202.77 * (-0.000581)
ln(P2/P1) = -3.027
Now, let's solve for P2/P1:
P2/P1 = e^(-3.027)
P2/P1 ≈ 0.0489
Finally, we can calculate the vapor pressure at 38°C (P2) by assuming the vapor pressure at the boiling point (P1) is 1 atm:
P2 = P1 * (P2/P1)
P2 = 1 atm * 0.0489
P2 ≈ 0.0489 atm
Therefore, the vapor pressure of tetramethylbutane on a hot summer day when the temperature is 38°C is approximately 0.0489 atm.
To determine the vapor pressure of tetramethylbutane (C8H18) at a given temperature, you can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance to its enthalpy of vaporization and temperature.
The Clausius-Clapeyron equation is given as follows:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
where:
P1 is the initial vapor pressure at temperature T1
P2 is the vapor pressure at temperature T2
ΔHvap is the enthalpy of vaporization
R is the gas constant (8.314 J/(mol·K))
T1 and T2 are the temperatures in Kelvin
First, we need to convert the given temperatures from Celsius to Kelvin. The conversion formula is:
T(K) = T(C) + 273.15
Therefore, the given temperatures are:
T1 = 106.5°C + 273.15 = 379.65 K
T2 = 38°C + 273.15 = 311.15 K
Next, we can substitute the values into the Clausius-Clapeyron equation:
ln(P2/P1) = (43.3 kJ/mol / (8.314 J/(mol·K))) * (1/379.65 K - 1/311.15 K)
Now, we need to solve for ln(P2/P1):
ln(P2/P1) = (5.217 * 10^-2) * (0.002634 - 0.003212)
Evaluating the expression:
ln(P2/P1) = (5.217 * 10^-2) * (-5.78 * 10^-4)
ln(P2/P1) = -3.01 * 10^-6
Since ln(P2/P1) is an extremely small value, we can approximate it to zero:
0 ≈ -3.01 * 10^-6
This implies that P2/P1 is approximately equal to 1.
Now we can solve for P2, which is the vapor pressure of tetramethylbutane at 38°C:
P2/P1 = 1
P2/1 atm = 1
Therefore, the vapor pressure of tetramethylbutane on a very hot summer day when the temperature is 38°C is approximately 1 atmosphere (atm).