The normal boiling point of liquid butanol is 391 K. Assuming that its molar heat of vaporization is constant at 45.9 kJ/mol, the boiling point of C4H9OH when the external pressure is 0.647 atm is
To determine the boiling point of C4H9OH (butanol) when the external pressure is 0.647 atm, we need to use the Clausius-Clapeyron equation. This equation relates the boiling point of a substance to its molar heat of vaporization, the gas constant, and the difference in vapor pressure at two different temperatures.
The Clausius-Clapeyron equation is given as:
ln(P1/P2) = (∆Hvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.
∆Hvap is the molar heat of vaporization.
R is the gas constant (8.314 J/(mol·K)).
We can rearrange this equation to solve for the boiling point (T2) when the external pressure (P2) is known:
T2 = (∆Hvap/R) * (1/(ln(P1/P2)) + 1/T1)
Given:
T1 = 391 K (normal boiling point of butanol)
P2 = 0.647 atm (external pressure)
∆Hvap = 45.9 kJ/mol (molar heat of vaporization)
R = 8.314 J/(mol·K)
We need to convert the molar heat of vaporization from kJ/mol to J/mol by multiplying by 1000:
∆Hvap = 45.9 kJ/mol * 1000 J/1 kJ = 45,900 J/mol
Substituting the given values into the equation:
T2 = (45,900 J/mol / 8.314 J/(mol·K)) * (1/ln(P1/P2) + 1/391 K)
Now we can calculate the boiling point (T2):
T2 = (45,900 / 8.314) * (1/ln(1 / 0.647) + 1/391)
Using a calculator, we can evaluate this equation to find the boiling point of C4H9OH when the external pressure is 0.647 atm.
To find the boiling point of C4H9OH (butanol) at an external pressure of 0.647 atm, we need to use the Clausius-Clapeyron equation:
ln(P₁/P₂) = (∆H_vap/R) * (1/T₁ - 1/T₂)
Where:
P₁ = Normal boiling point pressure (1 atm)
P₂ = External pressure (0.647 atm)
∆H_vap = Molar heat of vaporization (45.9 kJ/mol)
R = Ideal gas constant (8.314 J/(mol·K))
T₁ = Normal boiling point temperature (in Kelvin)
T₂ = Unknown boiling point temperature (in Kelvin)
First, we convert the molar heat of vaporization to J/mol:
∆H_vap = 45.9 kJ/mol × 1000 J/kJ = 45,900 J/mol
Next, substitute the known values into the equation:
ln(1/0.647) = (45,900 J/mol / (8.314 J/(mol·K))) * (1/391 K - 1/T₂)
Simplifying:
ln(1.546) = 5539.37 * (0.002559 - 1/T₂)
Now, isolate T₂:
T₂ = 1 / (1/391 K - (ln(1.546) / 5539.37))
Calculate:
T₂ ≈ 360.16 K
Therefore, the boiling point of C4H9OH (butanol) at an external pressure of 0.647 atm is approximately 360.16 K.