Benzene has a heat of vaporization of 30.72kJ/mol and a normal boiling point of 80.1 ∘C.

At what temperature does benzene boil when the external pressure is 480torr ?

Use the Clausius-Clapeyron equation. Don't forget to substitute dHvap in J/mol.

To determine the boiling temperature of benzene at an external pressure of 480 torr, we need to assess the relationship between boiling point, vapor pressure, and atmospheric pressure.

1. First, convert the external pressure from torr to atm by dividing by the conversion factor of 760 torr per 1 atm.

480 torr ÷ 760 torr/atm = 0.632 atm

2. Next, we need to establish the relationship between vapor pressure and boiling point using the Clausius-Clapeyron equation:

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

where:
P₁ = initial pressure (vapor pressure at T₁)
P₂ = final pressure (external pressure at T₂)
ΔH_vap = heat of vaporization
R = ideal gas constant (8.314 J/(mol∙K))
T₁ = initial temperature (boiling point)
T₂ = final temperature (unknown boiling point)

3. Rearrange the equation to solve for T₂ (final temperature at the given pressure):

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

Given that ΔH_vap (heat of vaporization) for benzene is 30.72 kJ/mol and its normal boiling point (T₁) is 80.1 ∘C, we need to convert it to Kelvin by adding 273.15:

T₁ = 80.1 ∘C + 273.15 = 353.25 K

4. Substitute the known values into the equation and solve for T₂:

T₂ = (30.72 kJ/mol / (8.314 J/(mol∙K))) * (1 / (353.25 K - ln(0.632/1)) )

T₂ ≈ 373.10 K

Therefore, benzene will boil at approximately 373.10 K (or 99.95 ∘C) when the external pressure is 480 torr.