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Answer :
Final Answer:
The vapor pressure of isooctane (C8H18) at a temperature of 38.0°C is approximately 25.27 kPa.
Explanation:
To determine the vapor pressure of isooctane at 38.0°C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its enthalpy of vaporization, temperature, and its normal boiling point. The equation is as follows:
ln(P2/P1) = (-ΔHvap / R) * (1/T2 - 1/T1)
Where:
P1 is the vapor pressure at the normal boiling point (known).
P2 is the vapor pressure at the desired temperature (unknown).
ΔHvap is the enthalpy of vaporization (given as 35.8 kJ/mol).
R is the universal gas constant (8.314 J/(mol*K)).
T1 is the normal boiling point temperature (known as 98.2°C, which needs to be converted to Kelvin).
T2 is the desired temperature (38.0°C, which also needs to be converted to Kelvin).
Converting temperatures to Kelvin:
T1 = 98.2°C + 273.15 = 371.35 K
T2 = 38.0°C + 273.15 = 311.15 K
Now we can plug these values into the equation:
ln(P2/P1) = (-35.8 * 10^3 J/mol / (8.314 J/(mol*K))) * (1/311.15 K - 1/371.35 K)
Solving for ln(P2/P1):
ln(P2/P1) = -6986.87
Now, we can find P2/P1:
P2/P1 = e^(-6986.87)
P2 = P1 * e^(-6986.87)
P1 is the vapor pressure at the normal boiling point, which can be found in reference tables (for isooctane, it's around 29.9 kPa). Therefore:
P2 ≈ 29.9 kPa * e^(-6986.87)
P2 ≈ 25.27 kPa
So, the vapor pressure of isooctane at 38.0°C is approximately 25.27 kPa.
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