Let's solve the problem **step-by-step** using the **Clausius-Clapeyron equation**. --- ## **Step 1: Write the Clausius-Clapeyron Equation** \[ \ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] Where: - \( P_1 \) = vapor pressure at \( T_1 \) (boiling point, so \( P_1 = 1\,atm \)) - \( T_1 \) = boiling point in Kelvin - \( P_2 \) = vapor pressure at \( T_2 \) - \( T_2 \) = 20.°C in Kelvin - \( \Delta H_{vap} \) = enthalpy of vaporization (convert to J/mol) - \( R \) = gas constant (\( 8.314 \, J/(mol\,K) \)) --- ## **Step 2: Convert Temperatures to Kelvin** \[ T_1 = 34.6^\circ C + 273.15 = 307.75\,K \] \[ T_2 = 20.^\circ C + 273.15 = 293.15\,K \] --- ## **Step 3: Convert \(\Delta H_{vap}\) to J/mol** \[ \Delta H_{vap} = 27.53\,kJ/mol = 27530\,J/mol \] --- ## **Step 4: Plug Values into the Equation** \[ \ln \left(\frac{P_2}{1\,atm}\right) = -\frac{27530}{8.314} \left(\frac{1}{293.15} - \frac{1}{307.75}\right) \] --- ## **Step 5: Calculate the Exponent** First, calculate the temperature difference: \[ \frac{1}{293.15} - \frac{1}{307.75} = .003413 - .003249 = .000164\, K^{-1} \] Now calculate \(\frac{27530}{8.314}\): \[ \frac{27530}{8.314} = 3311.6 \] So exponent is: \[ -3311.6 \times .000164 = -.5431 \] --- ## **Step 6: Solve for \( P_2 \)** \[ \ln \left(\frac{P_2}{1}\right) = -.5431 \] \[ \frac{P_2}{1} = e^{-.5431} \] \[ P_2 = e^{-.5431} = .581 \] --- ## **Final Answer** \[ \boxed{P_2 = .581 \,\text{atm}} \] **The vapor pressure of diethyl ether at \(20.^\circ C\) is approximately \(.58\, \text{atm}\) (rounded to 2 significant figures).**