P_{1} = 1 atm (pressure of normal boiling point)

T_{1} = -42.0 °C + 273.15 = 231.15 K

T_{2} = 30.0 °C + 273.15 = 303.15 K

$\overline{){\mathbf{ln}}{\mathbf{}}\frac{{\mathbf{P}}_{\mathbf{2}}}{{\mathbf{P}}_{\mathbf{1}}}{\mathbf{}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{\u2206}{\mathbf{H}}_{\mathbf{vap}}}{\mathbf{R}}{\mathbf{[}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}{\mathbf{-}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{]}}}$

Propane has a normal boiling point of -42.0^{o}C and a heat of vaporization (H_{vap}) of 19.04 kJ/mol. What is the vapor pressure of propane at 30.0 ^{o}C?

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