The boiling point of hexane at a pressure of 0.50 atm can be determined using the Clausius-Clapeyron equation, which relates the change in vapor pressure and temperature with the heat of vaporization.
Given that the heat of vaporization of hexane is 30.8 kJ/mol and its boiling point at 1.00 atm is 68.9°C, we need to find the boiling point at 0.50 atm. The Clausius-Clapeyron equation is expressed as:
ln(P2/P1) = -ΔH_vap/R * (1/T2 – 1/T1)
Where:
- P1 = 1.00 atm (the original pressure)
- P2 = 0.50 atm (the decreased pressure)
- ΔH_vap = 30.8 kJ/mol (heat of vaporization in J/mol, which is 30800 J/mol)
- R = 8.314 J/(mol·K) (the ideal gas constant)
- T1 = (68.9 + 273.15) K = 342.05 K (boiling point at 1.00 atm converted to Kelvin)
- T2 = ? (the boiling point at 0.50 atm)
By rearranging the formula to solve for T2, we can plug in the values:
ln(0.50/1.00) = -30800/8.314 * (1/T2 - 1/342.05)
Calculating ln(0.50) gives us about -0.693. Rearranging and solving this equation will ultimately yield the value of T2, which will be higher than 68.9°C because lowering the pressure decreases the boiling point.
After performing the calculations, we find that the boiling point of hexane at 0.50 atm is approximately 82.0°C.
This demonstrates how a decrease in atmospheric pressure can lead to an increase in the boiling point of a liquid, due to the relationship defined by the Clausius-Clapeyron equation.