Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling point of 80.1 degrees celcius.
At what temperature does Benzene boil when the external pressure is 440 torr?
Use the Clausius-Clapeyron equation.
ln(P2/P1) = -dHvap/R * (1/T2- 1/T1)
At the normal boiling point T1, the vapor pressure P1 is 760 torr. A lower temperature T2 will lower the vapor pressure to 440 torr.
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To determine the boiling point of Benzene at a given external pressure, you need to use the Clausius-Clapeyron equation:
ln(P₂/P₁) = ΔH_vap/R * (1/T₁ - 1/T₂)
where:
P₁ and T₁ are the pressure and temperature at the known boiling point,
P₂ is the given external pressure,
T₂ is the unknown boiling point temperature,
ΔH_vap is the heat of vaporization of Benzene, and
R is the ideal gas constant (0.008314 J/(mol·K)).
First, convert the given external pressure from torr to atm:
1 atm = 760 torr
440 torr x (1 atm/760 torr) = 0.579 atm
Now, rearrange the equation to solve for T₂:
ln(P₂/P₁) = ΔH_vap/R * (1/T₁ - 1/T₂)
ln(0.579/1) = (30.72 kJ/mol / (0.008314 J/(mol·K))) * (1/(80.1 + 273.15) - 1/T₂)
Simplify the equation:
ln(0.579) = (3696.7 K / 0.008314) * (1/353.25 - 1/T₂)
Simplify further:
ln(0.579) = 443122.377 * (0.0028314 - 1/T₂)
Now, solve for T₂:
443122.377 * (0.0028314 - 1/T₂) = ln(0.579)
0.0028314 - 1/T₂ = ln(0.579) / 443122.377
-1/T₂ = ln(0.579) / 443122.377 - 0.0028314
T₂ = -1 / (-ln(0.579) / 443122.377 + 0.0028314)
T₂ ≈ 48.28 degrees Celsius
Therefore, Benzene boils at approximately 48.28 degrees Celsius when the external pressure is 440 torr.
To determine the boiling point of Benzene at a given external pressure, we can use the Clausius-Clapeyron equation. This equation relates the boiling point of a substance to its heat of vaporization and the external pressure. The equation is written as:
ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 = initial pressure (given as the normal boiling point pressure)
P2 = final pressure (440 Torr in this case)
ΔHvap = heat of vaporization
R = gas constant (8.314 J/(mol*K))
T1 = initial temperature (given as the normal boiling point temperature)
T2 = final temperature (desired boiling point temperature)
By rearranging the equation, we can solve for T2:
T2 = (ΔHvap / (R * ln(P1/P2))) + T1
Let's plug in the values and calculate:
ΔHvap = 30.72 kJ/mol (given)
R = 8.314 J/(mol*K)
P1 = pressure at normal boiling point (use 101.325 kPa or 760 Torr)
P2 = 440 Torr (given)
T1 = 80.1 °C (given)
Converting units:
ΔHvap = 30.72 kJ/mol * 1000 J/kJ = 30720 J/mol
R = 8.314 J/(mol*K)
P1 = 760 Torr = 760/760 = 1 atm
T1 = 80.1 °C + 273.15 = 353.25 K
Now we can plug these values into the equation and calculate T2:
T2 = (30720 J/mol / (8.314 J/(mol*K) * ln(1/440))) + 353.25 K
T2 ≈ 353.25 K + 6.920 K
So, at an external pressure of 440 Torr, Benzene will boil at approximately 360.17 K (or 87.02 °C).