To find the enthalpy of vaporization ([tex]\(\Delta H_{\text{vap}}\)[/tex]) of the liquid A, we can use the Clausius-Clapeyron equation, which relates the vapor pressures at two different temperatures to [tex]\(\Delta H_{\text{vap}}\)[/tex]. The Clausius-Clapeyron equation is given by:
[tex]\[ \ln \left( \frac{P_2}{P_1} \right) = - \frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \][/tex]
where:
- [tex]\(P_1\)[/tex] and [tex]\(P_2\)[/tex] are the vapor pressures at temperatures [tex]\(T_1\)[/tex] and [tex]\(T_2\)[/tex], respectively.
- [tex]\(R\)[/tex] is the universal gas constant, [tex]\(8.314 \, \text{J/mol·K}\)[/tex].
- [tex]\(\Delta H_{\text{vap}}\)[/tex] is the enthalpy of vaporization.
- [tex]\(T_1\)[/tex] and [tex]\(T_2\)[/tex] should be in Kelvin.
Here are the given values:
- [tex]\(P_1 = 175 \, \text{torr}\)[/tex]
- [tex]\(P_2 = 760 \, \text{torr}\)[/tex]
- [tex]\(T_1 = 15.5^\circ \text{C}\)[/tex]
- [tex]\(T_2 = 75.5^\circ \text{C}\)[/tex]
First, we need to convert the temperatures from Celsius to Kelvin:
[tex]\[ T_1 = 15.5 + 273.15 = 288.65 \, \text{K} \][/tex]
[tex]\[ T_2 = 75.5 + 273.15 = 348.65 \, \text{K} \][/tex]
Next, we can rearrange the Clausius-Clapeyron equation to solve for [tex]\(\Delta H_{\text{vap}}\)[/tex]:
[tex]\[ \Delta H_{\text{vap}} = -R \cdot \frac{\ln \left( \frac{P_2}{P_1} \right)}{\left( \frac{1}{T_2} - \frac{1}{T_1} \right)} \][/tex]
Now, we can plug in all the known values:
- [tex]\(R = 8.314 \, \text{J/mol·K}\)[/tex]
- [tex]\(P_1 = 175 \, \text{torr}\)[/tex]
- [tex]\(P_2 = 760 \, \text{torr}\)[/tex]
- [tex]\(T_1 = 288.65 \, \text{K}\)[/tex]
- [tex]\(T_2 = 348.65 \, \text{K}\)[/tex]
First, calculate the natural logarithm [tex]\(\ln \left( \frac{P_2}{P_1} \right)\)[/tex]:
[tex]\[ \ln \left( \frac{760}{175} \right) \][/tex]
Then, find the temperature difference term [tex]\(\frac{1}{T_2} - \frac{1}{T_1}\)[/tex]:
[tex]\[ \frac{1}{348.65} - \frac{1}{288.65} \][/tex]
Combining the values, the result for [tex]\(\Delta H_{\text{vap}}\)[/tex] (in J/mol) becomes:
[tex]\[ \Delta H_{\text{vap}} = -8.314 \cdot \frac{\ln \left( \frac{760}{175} \right)}{\left( \frac{1}{348.65} - \frac{1}{288.65} \right)} \][/tex]
Finally, to convert the answer from J/mol to kJ/mol:
[tex]\[ \Delta H_{\text{vap}} \text{ (in } \text{kJ/mol)} = \frac{\Delta H_{\text{vap}} \text{ (in } \text{J/mol)}}{1000} \][/tex]
Thus, after calculation, the enthalpy of vaporization ([tex]\(\Delta H_{\text{vap}}\)[/tex]) for liquid A is approximately:
[tex]\[ 20.478755054362544 \, \text{kJ/mol} \][/tex]
So, [tex]\(\Delta H_{\text{vap}}\)[/tex] is [tex]\(20.48 \, \text{kJ/mol}\)[/tex] (rounded to two decimal places).
