Question: Chloroform, CHCl3, a volatile liquid was once used an anesthetic but has ben replaced by safer compounds. chloroform has a normal boiling point at 61.7 degree C and has a heat of vaporization of 31.4 kj/mol. compute its vapor pressure at 36.2 degree C?

Is the equation ln(p)= -Delta Hvap/RT + C. So than you plug in the numbers and get -31.4 kj/mole/ 8.31* (61.7+273.15) + 36.2 degree C. Is this how you solve for this problem. Did i plug in correctly?

Close.

I would use the full form of the Clausius-Clapeyron equation, i.e.,
ln(p2/p1) = etc.

Chloroform, CHCl3 a volatile liquid was once used as an anesthetic

but has been replaced by safer compounds. It boils at 61.7 C, and
has a heat of vaporization of 31.4 kJ/mol. What is its vapour pressure
at 25

To solve for the vapor pressure of chloroform at 36.2 degrees Celsius, you can use the Clausius-Clapeyron equation, which is commonly written as:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

In this equation:
- P2 is the vapor pressure at the desired temperature (36.2 degrees Celsius).
- P1 is the known vapor pressure at the normal boiling point (61.7 degrees Celsius).
- ΔHvap is the heat of vaporization of chloroform (31.4 kJ/mol).
- R is the gas constant (8.314 J/(mol*K)).
- T2 and T1 are the respective temperatures in Kelvin.

First, convert the temperatures from degrees Celsius to Kelvin:
- 36.2 degrees Celsius + 273.15 = 309.35 K
- 61.7 degrees Celsius + 273.15 = 334.85 K

Next, substitute the values into the equation:
ln(P2/1 atm) = -31.4 kJ/mol / (8.314 J/(mol*K)) * (1/309.35 K - 1/334.85 K)

Note: 1 atm is typically used as the reference pressure, so P1 is equal to 1 atm.

Now you can solve for ln(P2/1 atm) by simplifying the equation and plugging in the values correctly.

I got 89.1 mmHg.