Question

7. (5 pts) A liquid with an enthalpy of vaporization, Hv, of 52.80 KJ/mole has a normal boiling point at 38.38C. Its vapor pressure at 24.86C is:
(a) 166.09 Torr (b) 233.91 Torr (c) 301.4 Torr (d) 374.5 Torr

Answer 1

If vapor pressure of a substance is known at some temperature then vapor pressure of that substance can be known at some other temperature also by using Clausius-Clapeyron equation provided enthalpy of vaporization (ΔHv) is given.

By using Clausius-Clapeyron equation, we can write

ln (P₂/P₁) = -ΔHv/ (R) [1/T₂ - 1/T₁] ----------(1)

Given:

It is said that, a liquid has normal boiling point at Temperature (T₁) = 38.38⁰C = (38.38 + 273) =311.38 K

At normal condition, vapor pressure (P₁) = 1 atm = 760 torr

At temp. (T₂) = 24.86⁰C = (24.86 + 273) = 297.86 K, Vapor pressure (P₂) = ?

Enthalpy of vaporization, ΔHvap = 52.80 kJ/mol = 52800 j/mol and R = gas constant = 8.314J/mol-K

Putting all the values in equation (1), we get

ln (P₂/P₁) = (-52800) / (8.314) [ (1/297.86) - (1/311.38) ]

ln (P₂/P₁) = (-6350.7338) [0.0033572 - 0.0032115]

ln (P₂/P₁) = (-6350.7338) [0.0001457]

P₂/P₁ = e^(-0.9253) = 0.39641124

P₂ = 760^0.39641124 torr = 301.27526 torr almost equal to 301.4 torr

Thus, vapor pressure at 24.86⁰C will be 301.4 torr

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