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 find the normal boiling point of a liquid, we need to determine the temperature at which its vapor pressure is equal to the standard pressure, which is usually taken as 1 atmosphere or 760 mmHg.
In this case, we are given that the liquid's vapor pressure at 273 K is 102 mmHg. Therefore, we need to calculate the temperature at which the vapor pressure becomes 760 mmHg.
To find the boiling point, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at two different temperatures to the enthalpy of vaporization:
ln(P2/P1) = (-ΔHvap/R)((1/T2) - (1/T1))
Where:
P1 = vapor pressure at temperature T1
P2 = vapor pressure at temperature T2
ΔHvap = enthalpy of vaporization
R = ideal gas constant (8.314 J/(mol·K))
Rearranging the equation to solve for T2, we have:
T2 = (1/((ln(P2/P1)/(-ΔHvap/R)) + (1/T1)))
Plugging in the values:
P1 = 102 mmHg
P2 = 760 mmHg
ΔHvap = 30.8 kg·mol
R = 8.314 J/(mol·K)
T1 = 273 K
T2 = (1/((ln(760/102)/(-30.8*10^3/(8.314 J/(mol·K))) + (1/273))))
Calculating this equation will give us the normal boiling point of the liquid.