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 calculate the delta Hvap (the molar heat of vaporization) of a liquid, you need to know its boiling point and vapor pressure at a certain temperature. In this case, we are given the vapor pressure of the liquid at 77.6 degrees C and its boiling point at 1 atm.

The formula to calculate the delta Hvap is:

delta Hvap = -R * T * ln(P1/P2)

Where:
- R is the ideal gas constant (8.314 J/(mol*K))
- T is the temperature in Kelvin
- P1 is the vapor pressure at the desired temperature
- P2 is the vapor pressure at the boiling point at 1 atm

First, we need to convert the temperatures from degrees Celsius to Kelvin, which is done by adding 273.15 to each value:

77.6 degrees C + 273.15 = 350.75 K
94.9 degrees C + 273.15 = 368.05 K

Next, we substitute the values into the formula:

delta Hvap = -8.314 J/(mol*K) * (350.75 K) * ln(612 torr/1 atm)

Now, we need to convert the vapor pressure from torr to atm:

612 torr ÷ 760 torr/atm = 0.8053 atm

delta Hvap = -8.314 J/(mol*K) * (350.75 K) * ln(0.8053)

Now, you can solve the equation by calculating the natural logarithm of 0.8053 and then applying the rest of the multiplication and division:

delta Hvap ≈ -8.314 J/(mol*K) * (350.75 K) * (-0.217)

Calculating this expression will give you the value of delta Hvap in joules per mole (J/mol).