Benzene is a hydrocarbon that is commonly
used as a commercial solvent. However, it is
carcinogenic; i.e., accumulations in the body
can cause cancer. What is the vapor pressure
of benzene at 29�C? The normal boiling point
of benzene is 80.0�C and its molar heat of
vaporization is 30.8 kJ/mol.
Answer in units of torr
To find the vapor pressure of benzene at 29ºC, we can use the Clausius-Clapeyron equation, which relates vapor pressure to temperature and the molar heat of vaporization.
The Clausius-Clapeyron equation is:
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 molar heat of vaporization (30.8 kJ/mol for benzene)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T1 and T2 are the temperatures in Kelvin.
First, we need to convert the temperatures given to Kelvin:
29ºC + 273.15 = 302.15 K (T2)
80.0ºC + 273.15 = 353.15 K (T1)
Now we can plug the values into the equation:
ln(P2/P1) = -(30.8 kJ/mol) / (0.0821 L·atm/(mol·K)) * (1/302.15 K - 1/353.15 K)
Simplifying:
ln(P2/P1) = -375.9 / (0.0821) * (-0.003313)
ln(P2/P1) ≈ 16.73
To find P2/P1, we need to raise e (the base of natural logarithm) to the power of both sides:
P2/P1 = e^(16.73)
Using a calculator, we find:
P2/P1 ≈ 587555.95
Since we are looking for the vapor pressure of benzene at 29ºC (P2), we can rearrange the equation:
P2 = P1 * (P2/P1)
Finally, we can substitute the known values:
P2 = 1 atm * 587555.95
P2 ≈ 587555.95 torr
Therefore, the vapor pressure of benzene at 29ºC is approximately 587555.95 torr.