Final answer:
The vapor pressure of Benzene at its normal boiling point of 80°C can be determined using the Clausius-Clapeyron equation. By substituting the known values into the equation and solving for the unknown, we find that the vapor pressure of Benzene at 80°C is 101.3 kPa.
Step-by-step explanation:
The vapor pressure of Benzene at its normal boiling point of 80°C can be determined using the Clausius-Clapeyron equation. The equation is: ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1), where P1 is the known vapor pressure at T1, P2 is the unknown vapor pressure at T2, ΔHvap is the enthalpy of vaporization, R is the ideal gas constant, T1 is the known temperature, and T2 is the unknown temperature.
In this case, we know that T1 = 80°C = 353.3 K and P1 = 101.3 kPa. We want to find P2 at T2 = 80°C. Substituting the values into the equation, we get:
ln(P2/101.3) = -(30.8/8.314) * (1/353.3 - 1/353.3)
Simplifying the equation gives:
ln(P2/101.3) = 0
Therefore, P2/101.3 = e^0 = 1
So, the vapor pressure of Benzene at 80°C is 101.3 kPa.