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

**Convert kJ/mol to J/mol:**

$\mathbf{\u2206}{\mathbf{H}}_{\mathbf{vap}}\mathbf{=}\mathbf{29}\mathbf{.}\mathbf{1}\mathbf{}\frac{\overline{)\mathbf{kJ}}}{\mathbf{mol}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{J}}{\mathbf{1}\mathbf{}\overline{)\mathbf{kJ}}}$

**ΔH _{vap} = 29100 J/mol**

**Solving for ****P _{2}:**

T_{1} = 25.0°C + 273.15 = **298.15 K**

T_{2} = 95.0°C + 273.15 = **368.15 K**

Diethyl ether has a Δ*H*°_{vap} of 29.1 kJ/mol and a vapor pressure of 0.703 atm at 25.0°C. What is its vapor pressure at 95.0°C?

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