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 determine the vapor pressure of benzene at 29°C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and heat of vaporization.
The Clausius-Clapeyron equation is as follows: ln(P1/P2) = (-ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.
ΔHvap is the molar heat of vaporization.
R is the ideal gas constant (8.314 J/(mol·K)).
T1 and T2 are the temperatures in Kelvin.
First, we need to convert the given temperatures to Kelvin. To convert from Celsius to Kelvin, we add 273.15.
Temperature (T1) = 29°C + 273.15 = 302.15 K
Temperature (T2) = 80.0°C + 273.15 = 353.15 K
Now we can calculate the vapor pressure at 29°C using the given data.
ln(P1/P2) = (-ΔHvap/R) * (1/T1 - 1/T2)
ΔHvap = 30.8 kJ/mol = 30.8 * 10^3 J/mol
R = 8.314 J/(mol·K)
Substituting the values into the equation:
ln(P1/P2) = (-30.8 * 10^3 J/mol / (8.314 J/(mol·K))) * (1/302.15 K - 1/353.15 K)
ln(P1/P2) = (-3.708 * 10^3) * (0.003307 - 0.002835)
ln(P1/P2) = -3.708 * 10^3 * 0.000472
ln(P1/P2) = -1.7505
Now we can solve for P1/P2 by taking the exponential of both sides:
P1/P2 = e^(-1.7505)
P1/P2 = 0.1739
Lastly, we can solve for P1 by multiplying P2 by the ratio we just obtained:
P1 = P2 * (P1/P2)
P2 represents the vapor pressure at the boiling point, which can be looked up in a reference source. For benzene, the vapor pressure at its normal boiling point of 80.0°C is approximately 760 torr.
P1 = 760 torr * 0.1739 = 131.95 torr
Therefore, the vapor pressure of benzene at 29°C is approximately 131.95 torr.