a liquid has an enthalpy of vaporization of 30.8 kg.mol. at 273 K it has a vapor pressure of 102 mmHg. What is the normal boiling point of this liquid
To determine the normal boiling point of a liquid, we need to find the temperature at which its vapor pressure reaches atmospheric pressure (760 mmHg).
First, convert the given vapor pressure from mmHg to atm:
1 atm = 760 mmHg
Therefore, the vapor pressure is 102 mmHg / 760 mmHg/atm = 0.134 atm.
Next, we need to calculate the heat required to vaporize one mole (kg.mol = kilogram-mole) of the liquid. The enthalpy of vaporization (ΔHvap) is given as 30.8 kg.mol.
Now, let's write down the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 = vapor pressure at temperature T1 (273 K)
P2 = vapor pressure at the boiling point (760 mmHg or 1 atm)
T1 = initial temperature (273 K)
T2 = boiling point (to be determined)
R = ideal gas constant = 0.0821 L.atm/(mol.K)
Rearrange the equation to solve for T2:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Simplify:
ln(P2/P1) = (-ΔHvap/R) * (T1 - T2) / (T1 * T2)
Multiply both sides by (T1 * T2):
ln(P2/P1) * (T1 * T2) = (-ΔHvap/R) * (T1 - T2)
Distribute:
(T1 * T2) * ln(P2/P1) = -ΔHvap * (T1 - T2) / R
Multiply both sides by -R:
-R * (T1 * T2) * ln(P2/P1) = ΔHvap * (T1 - T2)
Expand:
-RT1T2 * ln(P2/P1) = ΔHvapT1 - ΔHvapT2
Rearrange and combine terms:
ΔHvapT2 - ΔHvapT1 = -RT1T2 * ln(P2/P1)
Move the T2 term to the left side:
ΔHvapT2 + RT1T2 * ln(P2/P1) = ΔHvapT1
Factor out T2:
T2(ΔHvap + RT1 * ln(P2/P1)) = ΔHvapT1
Divide both sides by (ΔHvap + RT1 * ln(P2/P1)):
T2 = (ΔHvapT1) / (ΔHvap + RT1 * ln(P2/P1))
Now, plug in the given values:
ΔHvap = 30.8 kg.mol
T1 = 273 K
P1 = 0.134 atm
R = 0.0821 L.atm/(mol.K)
P2 = 1 atm (as it reaches atmospheric pressure)
Calculating:
T2 = (30.8 kg.mol * 273 K) / (30.8 kg.mol + 0.0821 L.atm/(mol.K) * 273 K * ln(1 atm / 0.134 atm))
Using a calculator, we can find that T2 ≈ 335.5 K
Therefore, the normal boiling point of this liquid is approximately 335.5 K.