What is the boiling point of ethanol when the atmospheric pressure is 60kPa?
Look up the temperature vs vapor pressure of ethanol. The temperature of ethanol when ethanol has a vapor pressure of 60 kPa will be the new boiling point.
If you are supposed to calculate it, you can use the Clausius-Clapeyron equation.
To determine the boiling point of ethanol at a specific atmospheric pressure, we can refer to a vapor pressure chart or use the Clausius-Clapeyron equation. The boiling point of a liquid depends on the interplay between temperature and pressure.
One way to find the boiling point is by referring to a vapor pressure chart specific to ethanol. However, since we do not have access to external resources, we can use the Clausius-Clapeyron equation. This equation relates the boiling point of a substance to its vapor pressure at a given temperature:
ln(P1/P2) = (ΔHvap/R) * ((1/T2) - (1/T1))
Where:
- P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively,
- ΔHvap is the molar enthalpy of vaporization,
- R is the ideal gas constant,
- T1 and T2 are the corresponding temperatures.
To find the boiling point of ethanol at 60 kPa, we need the vapor pressure of ethanol at another known temperature to use as P1 in the equation. Let's assume we have the vapor pressure of ethanol at its normal boiling point, which is 78.37 °C (351.52 K) and 101.3 kPa.
We can now rearrange the equation to solve for T2, the boiling point at 60 kPa:
ln(P1/P2) = (ΔHvap/R) * ((1/T2) - (1/T1))
ln(101.3 kPa/60 kPa) = (ΔHvap/R) * ((1/T2) - (1/351.52 K))
Simplifying further:
ln(1.68833) = (ΔHvap/R) * ((1/T2) - (1/351.52))
Now, we can calculate T2, the boiling point at 60 kPa, using the known values of ΔHvap (the molar enthalpy of vaporization of ethanol) and R (the ideal gas constant). ΔHvap for ethanol is approximately 38.56 kJ/mol, and R is 8.314 J/(K*mol).
Plugging in the values and solving the equation, we find that the boiling point of ethanol at 60 kPa is approximately 56.29 °C (329.44 K).