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.

Clausius-Clapeyron Equation will handle this, look it up. You are given vapor pressure at 80C (101.3kPa), the heat of vaporization, and the two temps. Solve for VP at 29C

To find the vapor pressure of benzene at 29°C, we can use the Clausius-Clapeyron equation:

ln(P1/P2) = (-ΔHvap/R) * (1/T1 - 1/T2)

Where:
P1 = vapor pressure at the first temperature (unknown)
P2 = vapor pressure at the second temperature (normal boiling point)
ΔHvap = molar heat of vaporization
R = ideal gas constant (8.314 J/(mol*K))
T1 = first temperature (29°C + 273.15 = 302.15 K)
T2 = second temperature (80°C + 273.15 = 353.15 K)

Now, let's substitute the values into the equation and solve for P1:

ln(P1/760 torr) = (-30.8 kJ/mol / (8.314 J/(mol*K))) * (1/302.15 K - 1/353.15 K)

Simplifying the right side:

ln(P1/760 torr) = (-3.709 mol / K) * (0.001745 K)

ln(P1/760 torr) ≈ -0.00646

Now, to find P1, we can take the exponential of both sides:

P1/760 torr ≈ e^(-0.00646)

P1 ≈ 760 torr * e^(-0.00646)

Using a scientific calculator or an online calculator, we find:

P1 ≈ 747.78 torr

Therefore, the vapor pressure of benzene at 29°C is approximately 747.78 torr.