Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling point of 80.1. At what temperature does benzene boil when the external pressure is 480 torr?

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Sra

To determine the boiling point of benzene at a specific pressure, we can use the Clausius-Clapeyron equation, which relates the boiling point of a substance to its heat of vaporization and the external pressure. The equation is as follows:

ln(P1/P2) = ΔHvap/R * (1/T2 - 1/T1)

Where:
P1 is the initial pressure (760 torr, since it is the standard pressure)
P2 is the final pressure (480 torr in this case)
ΔHvap is the heat of vaporization of benzene (30.72 kJ/mol)
R is the ideal gas constant (8.314 J/(mol*K))
T1 is the initial temperature (the given normal boiling point of benzene, 80.1 °C converted to Kelvin)
T2 is the unknown boiling point temperature we're trying to find

First, we need to convert the given heat of vaporization from kJ/mol to J/mol:
ΔHvap = 30.72 kJ/mol * 1000 J/kJ = 30720 J/mol

Now let's plug in the values into the Clausius-Clapeyron equation and solve for T2:

ln(760/480) = (30720 J/mol)/(8.314 J/(mol*K)) * (1/T2 - 1/353.25 K)

Simplifying:

ln(760/480) = 3701.46 * (1/T2 - 0.00283)

Dividing both sides by 3701.46:

ln(760/480)/ 3701.46 = 1/T2 - 0.00283

Rearranging the equation:

1/T2 = ln(760/480)/ 3701.46 + 0.00283

Calculating the right-hand side:

1/T2 = 0.193 + 0.00283 = 0.19583

Taking the reciprocal to isolate T2:

T2 = 1/0.19583 = 5.108 K

So, the boiling point of benzene at an external pressure of 480 torr is approximately 5.108 K.