To determine the temperature at which the vapor pressure of the liquid is 0.245 atm, given that its normal boiling point is 282°C and the enthalpy of vaporization (ΔHvap) is 28.5 kJ/mol, we can use the Clausius-Clapeyron equation:
[tex]\[
\ln \left( \frac{P2}{P1} \right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T2} - \frac{1}{T1} \right)
\][/tex]
where:
- [tex]\( P1 \)[/tex] is the vapor pressure at the normal boiling point, which is 1 atm,
- [tex]\( P2 \)[/tex] is the target vapor pressure, 0.245 atm,
- [tex]\( T1 \)[/tex] is the normal boiling point in Kelvin,
- [tex]\( T2 \)[/tex] is the temperature we are solving for in Kelvin,
- [tex]\(\Delta H_{\text{vap}}\)[/tex] is the enthalpy of vaporization in J/mol,
- [tex]\( R \)[/tex] is the gas constant, 8.314 J/(mol·K).
First, we need to convert the given values to consistent units:
1. Convert the normal boiling point from Celsius to Kelvin:
[tex]\[
T1 = 282^\circ C + 273.15 = 555.15 \text{ K}
\][/tex]
2. Convert the enthalpy of vaporization from kJ/mol to J/mol:
[tex]\[
\Delta H_{\text{vap}} = 28.5 \text{ kJ/mol} \times 1000 = 28500 \text{ J/mol}
\][/tex]
We can now rearrange the Clausius-Clapeyron equation to solve for [tex]\( T2 \)[/tex]:
[tex]\[
\ln \left( \frac{P2}{P1} \right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T2} - \frac{1}{T1} \right)
\][/tex]
[tex]\[
\frac{1}{T2} = \frac{1}{T1} - \frac{R}{\Delta H_{\text{vap}}} \ln \left( \frac{P2}{P1} \right)
\][/tex]
Plugging in the values:
[tex]\[
\ln \left( \frac{0.245}{1} \right) = \ln (0.245) = -1.4076
\][/tex]
[tex]\[
\frac{1}{T2} = \frac{1}{555.15} - \frac{8.314}{28500} (-1.4076)
\][/tex]
[tex]\[
\frac{1}{T2} = 0.001801 - \left( \frac{8.314 \times 1.4076}{28500} \right)
\][/tex]
[tex]\[
\frac{1}{T2} = 0.001801 + 0.000411 = 0.002212
\][/tex]
[tex]\[
T2 = \frac{1}{0.002212} \approx 451.17 \text{ K}
\][/tex]
Finally, convert [tex]\( T2 \)[/tex] from Kelvin back to Celsius:
[tex]\[
T2 \text{ (in °C)} = 451.17 \text{ K} - 273.15 = 178.02^\circ C
\][/tex]
Rounding to two decimal places, the temperature at which the vapor pressure is 0.245 atm is:
[tex]\[
179.01^\circ C
\][/tex]
