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 (enthalpy of vaporization) of a liquid, you need to know its boiling point and vapor pressure.
1. Convert the given temperatures from degrees Celsius to Kelvin:
Boiling point = 94.9 degrees C + 273.15 = 368.05 K
Temperature = 77.6 degrees C + 273.15 = 350.75 K
2. Using the boiling point and vapor pressure, we can calculate the enthalpy of vaporization (delta Hvap) using the Clausius-Clapeyron equation:
ln(P2/P1) = - delta Hvap / R * (1/T2 - 1/T1)
Where:
P1 = vapor pressure at the lower temperature
P2 = vapor pressure at the higher temperature
T1 = lower temperature
T2 = higher temperature
R = ideal gas constant (0.0821 L * atm / (mol * K))
Rearranging the equation, we get:
delta Hvap = -R * (T2 - T1) / ln(P2/P1)
3. Plug in the values we have:
P1 = 612 torr
P2 = 1 atm
T1 = 350.75 K
T2 = 368.05 K
R = 0.0821 L * atm / (mol * K)
delta Hvap = -0.0821 L * atm / (mol * K) * (368.05 K - 350.75 K) / ln(1 atm / 612 torr)
4. Calculate the natural logarithm (ln) of the pressure ratio:
ln(1 atm / 612 torr) = ln(1.63399) ≈ 0.49164
5. Plugging the values into the equation:
delta Hvap ≈ -0.0821 L * atm / (mol * K) * (368.05 K - 350.75 K) / 0.49164
6. Calculate the final result:
delta Hvap ≈ -9.95144 L * atm / mol
So, the delta Hvap of the liquid is approximately -9.95144 L * atm / mol.