what is the delta Hvap of a liquid that has a vapor pressure of 612 torr at 77.6 degrees C and a boiling point of 94.9 degrees C at 1 atm
To find the delta Hvap (the enthalpy of vaporization) of a liquid, we need to use the Clausius-Clapeyron equation. This equation relates the vapor pressure of a substance at different temperatures to its enthalpy of vaporization.
Here's the step-by-step process to find the delta Hvap:
1. Begin by converting the given temperature values from Celsius to Kelvin. To do this, add 273.15 to each temperature:
- Given boiling point: T1 = 94.9°C + 273.15 = 368.05 K
- Given temperature: T2 = 77.6°C + 273.15 = 350.75 K
2. Next, convert the given vapor pressure from torr to atm. Divide the pressure by 760 torr/atm:
- Given vapor pressure: P1 = 612 torr / 760 torr/atm ≈ 0.805 atm
3. Plug the values into the Clausius-Clapeyron equation:
ln(P1/P2) = (-ΔHvap/R) * (1/T1 - 1/T2)
where P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively, ΔHvap is the enthalpy of vaporization, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T1 and T2 are the temperatures in Kelvin.
4. Rearrange the equation to solve for ΔHvap:
ΔHvap = -(R * (T2 - T1)) / (ln(P1/P2))
5. Substitute the known values into the equation:
ΔHvap = -(0.0821 L·atm/(mol·K) * (350.75 K - 368.05 K)) / ln(0.805 atm / 1 atm)
6. Perform the calculations:
ΔHvap = -(0.0821 L·atm/(mol·K) * (-17.3 K)) / ln(0.805) ≈ 44.83 kJ/mol
So, the delta Hvap of the liquid is approximately 44.83 kJ/mol.