To determine the vapor pressure in mmHg of a substance at [tex]\( 45^\circ C \)[/tex], given that its normal boiling point is [tex]\( 115^\circ C \)[/tex] and its enthalpy of vaporization is [tex]\( 57.9 \, \text{kJ/mol} \)[/tex], we will use the Clausius-Clapeyron equation. Here’s the step-by-step solution:
1. Convert all temperatures to Kelvin:
- The normal boiling point temperature [tex]\( T_1 \)[/tex] is [tex]\( 115^\circ C \)[/tex]. Convert it to Kelvin using the formula [tex]\( T(K) = T(°C) + 273.15 \)[/tex]:
[tex]\[
T_1 = 115 + 273.15 = 388.15 \, \text{K}
\][/tex]
- The temperature [tex]\( T_2 \)[/tex] at which we want to find the vapor pressure is [tex]\( 45^\circ C \)[/tex]. Convert it to Kelvin:
[tex]\[
T_2 = 45 + 273.15 = 318.15 \, \text{K}
\][/tex]
2. Convert the enthalpy of vaporization to Joules per mole:
[tex]\[
\Delta H = 57.9 \, \text{kJ/mol} = 57.9 \times 1000 \, \text{J/mol} = 57900 \, \text{J/mol}
\][/tex]
3. Use the Clausius-Clapeyron equation:
The Clausius-Clapeyron equation is:
[tex]\[
\ln \frac{P_2}{P_1} = \frac{-\Delta H}{R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]
\][/tex]
Where:
- [tex]\( P_1 \)[/tex] is the vapor pressure at the normal boiling point, which is [tex]\( 1 \, \text{atm} \)[/tex].
- [tex]\( R \)[/tex] is the universal gas constant, [tex]\( 8.314 \, \text{J/(mol·K)} \)[/tex].
Substitute the given values into the equation:
[tex]\[
\ln \frac{P_2}{P_1} = \frac{-57900}{8.314} \left[ \frac{1}{388.15} - \frac{1}{318.15} \right]
\][/tex]
4. Calculate the natural logarithm term:
[tex]\[
\frac{1}{388.15} \approx 0.002577 \, \text{K}^{-1}
\][/tex]
[tex]\[
\frac{1}{318.15} \approx 0.003143 \, \text{K}^{-1}
\][/tex]
[tex]\[
\frac{1}{388.15} - \frac{1}{318.15} \approx 0.002577 - 0.003143 = -0.000566 \, \text{K}^{-1}
\][/tex]
[tex]\[
\ln \frac{P_2}{P_1} = \frac{-57900}{8.314} \times (-0.000566) \approx 393.10 \times (-0.000566) = 3.9476
\][/tex]
5. Calculate [tex]\( P_2 \)[/tex] by exponentiating both sides:
[tex]\[
\ln \frac{P_2}{1\, \text{atm}} = 3.9476
\][/tex]
[tex]\[
P_2 \approx e^{3.9476} \approx 51.81 \, \text{atm}
\][/tex]
6. Convert the vapor pressure from atm to mmHg:
[tex]\[
P_2 \, \text{(in mmHg)} = 51.81 \, \text{atm} \times 760 \, \text{mmHg/atm} \approx 39376.95 \, \text{mmHg}
\][/tex]
Therefore, the vapor pressure of the substance at [tex]\( 45^\circ C \)[/tex] is approximately [tex]\( 39376.95 \, \text{mmHg} \)[/tex].
