The normal boiling point for ethyl alcohol is \(78.4^{\circ} \mathrm{C} .5^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) is \(282.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\). At what temperature is the vapor pressure of ethyl alcohol \(357 \mathrm{~mm} \mathrm{Hg} ?\)

Since we are given the molar heat capacity at constant volume for C2H5OH(g), we can find the enthalpy of vaporization at T1 = 78.4°C (the given normal boiling point) using the following equation: \(\Delta H_{vap} = C_{v}(T_{2} - T_{1}) \text{(Assumption: constant molar heat capacity)}\) We have the molar heat capacity and T1, but we need to find T2. At the boiling point, the vapor pressure of the liquid is equal to the atmospheric pressure. Assuming standard conditions, the atmospheric pressure is 1 atm, which can be converted to mm Hg: \(1 \ \text{atm} = 760 \ \text{mm Hg}\) Since we know the vapor pressure is 357 mm Hg, we can find T2 by solving for it using the Clausius-Clapeyron equation. However, we first need the enthalpy of vaporization.

To use the Clausius-Clapeyron equation and find the enthalpy of vaporization, we need to convert the given Celsius temperatures to Kelvin: \(T_{1(K)} = T_{1(°C)} + 273.15\) \(T_{1(K)} = 78.4 + 273.15 = 351.55 \ \text{K}\)

Now we can substitute the conversion factor from Step 2 into the first equation for enthalpy of vaporization: \(\Delta H_{vap} = C_{v}(T_{2} - T_{1})\) \(\Delta H_{vap} = 282.7 \ \text{J/mol K} \cdot (760 - 351.55) \ \text{K}\) \(\Delta H_{vap} \approx 115189.485 \ \text{J/mol}\)

With all necessary values to use the Clausius-Clapeyron equation, we can solve for the temperature at which the vapor pressure of ethyl alcohol is 357 mm Hg: \(ln\frac{P_2}{P_1}=-\frac{\Delta H_{vap}}{R}\left(\frac{1}{T2}-\frac{1}{T1}\right)\) \(ln\frac{357}{760}=-\frac{115189.485}{8.314}\left(\frac{1}{T2}-\frac{1}{351.55}\right)\) Now, solve for the temperature T2: \(T2 \approx 339.55 \ \text{K}\)

Finally, we'll convert the temperature back to Celsius for the final answer: \(T_{2(°C)} = T_{2(K)} - 273.15\) \(T_{2(°C)} = 339.55 - 273.15\) \(T_{2(°C)} \approx 66.4 ^{\circ}\text{C}\) The temperature at which the vapor pressure of ethyl alcohol is 357 mm Hg is approximately 66.4°C.