Certainly! Let’s go through this step by step using the Clausius-Clapeyron equation. Our task is to determine the change in boiling point of water with changes in atmospheric pressure.
Given:
1. Heat of vaporization ([tex]\(\Delta H_v\)[/tex]) = 539.7 [tex]\(\text{cal/g}\)[/tex]
2. Molar mass of water = 18 [tex]\(\text{g/mol}\)[/tex]
3. Molar volume of liquid water ([tex]\(V_l\)[/tex]) = 80.78 [tex]\(\text{cm}^3\)[/tex]
4. Molar volume of steam ([tex]\(V_v\)[/tex]) = 30.199 [tex]\(\text{dm}^3\)[/tex] = 30199 [tex]\(\text{cm}^3\)[/tex]
5. Boiling point is 100 °C (373.15 K) at 1 atm pressure ([tex]\(P_1=1\)[/tex]).
To find the change in boiling point [tex]\(\Delta T_b\)[/tex] with a small change in pressure, we use the Clausius-Clapeyron equation:
[tex]\[ \frac{dP}{dT} = \frac{\Delta H_v}{T \Delta V} \][/tex]
Where:
- [tex]\( \Delta H_v \)[/tex] is the molar enthalpy (heat) of vaporization
- T is the absolute temperature
- [tex]\( \Delta V = V_v - V_l \)[/tex]
First, convert heat of vaporization to a molar basis because [tex]\( \Delta H_v \)[/tex] is given in terms of per gram.
[tex]\[ \Delta H_v = 539.7 \, \text{cal/g} \times 18 \, \text{g/mol} = 9714.6 \, \text{cal/mol} \][/tex]
We also need this in joules per mole because SI units are easier to work with:
[tex]\[ \Delta H_v = 9714.6 \, \text{cal/mol} \times 4.184 \, \frac{\text{J}}{\text{cal}} = 40663.1 \, \text{J/mol} \][/tex]
Next, calculate [tex]\( \Delta V\)[/tex]:
[tex]\[ \Delta V = V_v - V_l = 30199 \, \text{cm}^3 - 80.78 \, \text{cm}^3 = 30118.22 \, \text{cm}^3 \][/tex]
The Clausius-Clapeyron relation can be used as an approximation for small changes around an equilibrium at constant temperatures:
[tex]\[ \Delta T_b = \frac{T (\Delta V)}{\Delta H_v} \Delta P \][/tex]
Now to find the change in boiling point per millimeter change in pressure:
1 atm pressure = 101325 Pa, [tex]\(1 \, \text{atm} \approx 760 \, \text{mm Hg}\)[/tex]
So, for a 1 mm Hg change:
[tex]\[ \Delta P = \frac{101325 \, \text{Pa}}{760 \, \text{mm Hg}} \approx 133.3 \, \text{Pa} \][/tex]
Using the Clausius-Clapeyron equation:
[tex]\[ \Delta T_b = \frac{373.15 \times 30118.22 \, \text{cm}^3}{40663.1 \, \text{J/mol}} \times 133.3 \, \text{Pa} \][/tex]
[tex]\[ = \frac{373.15 \times 30118.22 \times 133.3}{40663.1} \][/tex]
Converting [tex]\( 30118.22 \times 133.3 \)[/tex] to K (since J/g conversion is kept for temperatures):
[tex]\[ = \frac{373.15 \times 4014792.676}{40663.1}\][/tex]
Since units of calorimetry and vapor are simplified:
[tex]\[ \Delta T_b \approx 36.2 \, \text{mK per mm Hg} \][/tex]
This detailed step-by-step clearly shows the relationship between pressure changes and boiling points using the Clausius-Clapeyron equation.
