Questions: Diethyl ether is a volatile, highly flammable organic liquid that is used mainly as a solvent. The vapor pressure of diethyl ether is 401 mmHg at 18°C. Calculate its vapor pressure at 32°C.

Solution Steps

We need to calculate the vapor pressure of diethyl ether at \(32^{\circ} \mathrm{C}\) given its vapor pressure at \(18^{\circ} \mathrm{C}\) is 401 mmHg. This can be done using the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature.

The Clausius-Clapeyron equation is given by:

\[ \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

where:

• \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) respectively,
• \(\Delta H_{\text{vap}}\) is the enthalpy of vaporization,
• \(R\) is the universal gas constant (\(8.314 \, \text{J/mol K}\)),
• \(T_1\) and \(T_2\) are the absolute temperatures in Kelvin.

Convert the given temperatures from Celsius to Kelvin:

\[ T_1 = 18 + 273.15 = 291.15 \, \text{K} \] \[ T_2 = 32 + 273.15 = 305.15 \, \text{K} \]

We need the enthalpy of vaporization \(\Delta H_{\text{vap}}\) for diethyl ether to proceed. Assuming \(\Delta H_{\text{vap}} = 29.1 \, \text{kJ/mol}\) (a typical value for diethyl ether), convert it to J/mol:

\[ \Delta H_{\text{vap}} = 29.1 \times 10^3 \, \text{J/mol} \]

Substitute the known values into the Clausius-Clapeyron equation:

\[ \ln \left( \frac{P_2}{401} \right) = -\frac{29.1 \times 10^3}{8.314} \left( \frac{1}{305.15} - \frac{1}{291.15} \right) \]

Calculate the right-hand side of the equation:

\[ \ln \left( \frac{P_2}{401} \right) = -\frac{29.1 \times 10^3}{8.314} \left( \frac{1}{305.15} - \frac{1}{291.15} \right) \approx 0.4525 \]

Exponentiate both sides to solve for \(P_2\):

\[ \frac{P_2}{401} = e^{0.4525} \]

\[ P_2 = 401 \times e^{0.4525} \approx 401 \times 1.5723 \approx 630.4 \, \text{mmHg} \]

Final Answer