Benzene has a heat of vaporization of 30.72kJ/mol and a normal boiling point of 80.1 degrees C. At what temperature does benzene boil when the external pressure is 470 torr?

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

To determine the boiling point of benzene at a specific external pressure, we can use the Clausius-Clapeyron equation:

ln(P1/P2) = -ΔHvap/R * (1/T2 - 1/T1)

Where:
P1 = pressure at the known boiling point (at 760 torr)
P2 = pressure at the desired boiling point (470 torr)
ΔHvap = heat of vaporization of benzene (30.72 kJ/mol)
R = ideal gas constant (8.314 J/(mol·K) or 0.0821 L·atm/(mol·K))
T1 = known boiling point temperature (80.1 degrees C + 273.15 K)
T2 = desired boiling point temperature (unknown)

Now let's calculate T2, the desired boiling point temperature:

ln(760/470) = -30.72 * 10^3 / (0.0821) * (1/T2 - 1/(80.1 + 273.15))

ln(1.617) = -30.72 * 10^3 / 0.0821 * (1/T2 - 1/353.25)

Taking the antilog of both sides:

1.617 = (1/T2 - 1/353.25) * (30.72 * 10^3 / 0.0821)

1.617 = (1/T2 - 0.0028295) * 373501.81467

1.617 = (1/T2 - 0.0028295) * 373501.81467

1.617 = (373501.81467/T2 - 1.05597)

Solving for T2:

1.617 + 1.05597 = 373501.81467 / T2

T2 = 373501.81467 / 2.67297

T2 ≈ 139,747.48 K

Converting T2 from Kelvin to degrees Celsius:

T2 ≈ 139,747.48 - 273.15

T2 ≈ 139,474.33 degrees C

Therefore, benzene will boil at approximately 139,474.33 degrees Celsius when the external pressure is 470 torr.

To determine the boiling point of benzene at a given external pressure, we need to use the Clausius-Clapeyron equation:

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

Where:
P₁ and P₂ are the initial and final pressures respectively.
ΔH_vap is the heat of vaporization of the substance.
R is the ideal gas constant.
T₁ and T₂ are the initial and final temperatures respectively.

First, let's convert the external pressure from torr to atm:
1 atm = 760 torr

So, we have:
External pressure = 470 torr / 760 torr/atm = 0.618 atm

Now we can substitute the given values into the equation:
ln(0.618 atm / 1 atm) = -(30.72 kJ/mol / R)((1/T₂) - (1/353 K))

Note: We convert 80.1 degrees C to Kelvin by adding 273.15:
T₁ = 80.1°C + 273.15 = 353 K

Now we need to rearrange the equation to solve for T₂:
(1/T₂) - (1/353 K) = -(ln(0.618 atm / 1 atm) x R / 30.72 kJ/mol)

We know that R is the ideal gas constant with a value of 8.314 J/(mol·K).

Let's substitute the values and solve for T₂:
(1/T₂) - (1/353 K) = -(ln(0.618) x (8.314 J/(mol·K)) / (30.72 kJ/mol)

Now, we can calculate T₂ by solving the equation above for (1/T₂), then taking the reciprocal to get the boiling temperature in Kelvin.

After finding T₂, convert it back to degrees Celsius by subtracting 273.15.

Note: It's important to use consistent units throughout the calculation to avoid errors.

Can't you use the Clausius-Clapeyron equation?