The vapor pressure of benzene, C6H6, is 60.0 mm Hg at 13.5°C. The molar heat of vaporization of benzene is 30.8 kJ/mol and R = 8.31 10-3 kJ/mol·K. What is the vapor pressure of benzene at 70.0°C?
Sam see your post about 5, 6, or 7 pages back. I showed how to do this.
To find the vapor pressure of benzene at 70.0°C, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 = vapor pressure at temperature 1 (13.5°C)
P2 = vapor pressure at temperature 2 (70.0°C)
ΔHvap = molar heat of vaporization of benzene (30.8 kJ/mol)
R = universal gas constant (8.31 10-3 kJ/mol·K)
T1 = temperature 1 in Kelvin (13.5°C + 273.15)
T2 = temperature 2 in Kelvin (70.0°C + 273.15)
Let's calculate:
T1 = 13.5 + 273.15 = 286.65 K
T2 = 70.0 + 273.15 = 343.15 K
ln(P2/60.0) = (30.8/8.31 10-3) * (1/286.65 - 1/343.15)
Now we solve for P2:
ln(P2/60.0) = 3700.6 * (0.003487 - 0.002916)
ln(P2/60.0) = 3700.6 * 0.000571
ln(P2/60.0) = 2.114278
Apply the natural exponential function (e) to both sides of the equation:
P2/60.0 = e^(2.114278)
Now solve for P2:
P2 = 60.0 * e^(2.114278)
Using a calculator, we find that P2 ≈ 93.028 mm Hg.
Therefore, the vapor pressure of benzene at 70.0°C is approximately 93.028 mm Hg.