Question

An unknown liquid has a heat of vaporization of 32.45 kJ/mole. If the normal boiling point is 84, what is vapor pressure (in torr) of this liquid at room temperature of 25 degrees C? HINT: Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).

Answer 1

To solve this problem, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)

where P1 is the vapor pressure at temperature T1, P2 is the vapor pressure at temperature T2, ΔHvap is the heat of vaporization, R is the gas constant (8.31 J/mol*K), and ln is the natural logarithm.

First, we need to convert the heat of vaporization from kJ/mol to J/mol:

ΔHvap = 32.45 kJ/mol * 1000 J/kJ = 32,450 J/mol

Next, we need to convert the boiling point from Celsius to Kelvin:

T1 = 84 + 273.15 = 357.15 K

We are given that the room temperature is 25 degrees Celsius, so we need to convert this to Kelvin as well:

T2 = 25 + 273.15 = 298.15 K

We are looking for the vapor pressure of the liquid at room temperature, so we can set P2 = ? and P1 = 1 atm = 760 torr. Substituting all of these values into the Clausius-Clapeyron equation, we get:

ln(P2/760 torr) = (32,450 J/mol / 8.31 J/mol*K) * (1/357.15 K - 1/298.15 K)

Simplifying and solving for P2, we get:

P2 = 760 torr * e^[(32,450 J/mol / 8.31 J/mol*K) * (1/357.15 K - 1/298.15 K)]

P2 = 0.000047 torr

Therefore, the vapor pressure of the liquid at room temperature is approximately 0.000047 torr.