Question

At sea level, the barometric pressure is \(760 \mathrm{~mm}\) Hg. Calculate the difference in the normal boiling point for water boiled in an open pot and water boiled in a pressure cooker set for 1.75 atm. \(\left(\Delta H_{\text {vap }}\right.\) for water \(\left.=40.7 \mathrm{~kJ} / \mathrm{mol}\right)\)

Short answer

The difference in the normal boiling point for water boiled in an open pot and water boiled in a pressure cooker set for 1.75 atm is approximately 21.71°C.

Step-by-step solution

Convert the given pressure and enthalpy values.

First, let's convert the given pressure and enthalpy values into appropriate units. Normal pressure at sea level: \(760 \mathrm{~mm} \mathrm{Hg}\). We need to convert this pressure to atmospheres (atm): \(760 \mathrm{~mm} \mathrm{Hg} \cdot \frac{1 \mathrm{~atm}}{760 \mathrm{~mm} \mathrm{Hg}} = 1 \mathrm{~atm}\) Enthalpy of vaporization of water is given as \(40.7 \mathrm{~kJ} / \mathrm{mol}\). We need to convert this value to Joules per mole: \(40.7 \mathrm{~kJ} / \mathrm{mol} \cdot \frac{1000 \mathrm{~J}}{1 \mathrm{~kJ}} = 40700 \mathrm{~J} / \mathrm{mol}\) Now we have the necessary values: \(P_1 = 1 \mathrm{~atm}\) \(P_2 = 1.75 \mathrm{~atm}\) \(ΔH_\text{vap}= 40700 \mathrm{~J/\mathrm{mol}}\)

Determine the boiling point of water in an open pot.

We know that the normal boiling temperature of water at sea level is 100°C. Thus, we have the initial temperature \(T_1 = 373.15 \mathrm{K}\) as we must use Kelvin.

Find the boiling point of water in a pressure cooker using the Clausius-Clapeyron equation.

Applying the Clausius-Clapeyron equation, we get: \(ln\frac{P_2}{P_1}=-\frac{ΔH_\text{vap}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\) Inserting values: \(ln\frac{1.75}{1}=-\frac{40700}{8.314}\left(\frac{1}{T_2}-\frac{1}{373.15}\right)\) Now, we need to solve for \(T_2\) (boiling point in the pressure cooker): \(ln(1.75) = \frac{-40700}{8.314} \cdot \left(\frac{1}{T_2} - \frac{1}{373.15}\right)\) Rearrange and solve for \(T_2\): \(T_2 = \frac{373.15}{1 - \frac{8.314 \cdot ln(1.75)}{-40700}}\) \(T_2 \approx 394.86 \mathrm{~K}\) Now we have the boiling point of water in the pressure cooker, which is approximately \(394.86 \mathrm{~K}\) or \(121.71 \mathrm{°C}\).

Calculate the difference in boiling points between the open pot and pressure cooker.

Finally, we'll find the difference in boiling points between the open pot and pressure cooker: Boiling point difference = \(\left| T_\text{pressure cooker} - T_\text{open pot}\right|\) Boiling point difference = \(\left| 394.86 \mathrm{~K} - 373.15 \mathrm{~K}\right|\) Boiling point difference \(\approx 21.71 \mathrm{°C}\) Therefore, the difference in the normal boiling point for water boiled in an open pot and water boiled in a pressure cooker set for \(1.75\) atm is approximately \(21.71 \mathrm{°C}\).