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
To find the delta Hvap (enthalpy of vaporization) of a liquid, you can use the Clausius-Clapeyron equation. This equation relates the vapor pressure and temperature of a substance to its enthalpy of vaporization.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = -(delta Hvap/R) *(1/T2 - 1/T1)
Where:
P1 is the vapor pressure at temperature T1 (in Kelvin).
P2 is the vapor pressure at temperature T2 (in Kelvin).
delta Hvap is the enthalpy of vaporization (in J/mol).
R is the gas constant (8.314 J/mol K).
T1 and T2 are the corresponding temperatures in Kelvin.
First, you need to convert the temperatures from degrees Celsius to Kelvin by adding 273.15.
T1 = 77.6°C + 273.15 = 350.75 K
T2 = 94.9°C + 273.15 = 368.05 K
Next, convert the vapor pressure from torr to atm by dividing by 760.
P1 = 612 torr / 760 = 0.805 atm
P2 = Since the boiling point is given, the vapor pressure at the boiling point is 1 atm.
Now, we can substitute these values into the Clausius-Clapeyron equation:
ln(1/0.805) = -(delta Hvap/8.314) * (1/368.05 - 1/350.75)
Solving this equation will give you the delta Hvap of the liquid.