QuestionJuly 6, 2025

The normal boiling point of mercury (Hg) is 356.7^circ C What is the vapor pressure of mercury at 344.5^circ C in atm? (Delta H_(vap)=58.51kJ/mol)

The normal boiling point of mercury (Hg) is 356.7^circ C What is the vapor pressure of mercury at 344.5^circ C in atm? (Delta H_(vap)=58.51kJ/mol)
The normal boiling point of mercury (Hg) is 356.7^circ C
What is the vapor pressure of mercury at 344.5^circ C in
atm? (Delta H_(vap)=58.51kJ/mol)

Solution
4.0(336 votes)

Answer

P_2 \approx 0.739 \, \text{atm} Explanation 1. Use Clausius-Clapeyron Equation The Clausius-Clapeyron equation is given by **\ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)**. Here, P_1 is the vapor pressure at the normal boiling point (1 atm), T_1 = 356.7 + 273.15 K, T_2 = 344.5 + 273.15 K, and R = 8.314 \, \text{J/mol K}. 2. Convert Temperatures to Kelvin T_1 = 356.7 + 273.15 = 629.85 \, \text{K}; T_2 = 344.5 + 273.15 = 617.65 \, \text{K}. 3. Calculate \ln(P_2) Substitute values into the equation: \ln \left(\frac{P_2}{1}\right) = -\frac{58510}{8.314} \left(\frac{1}{617.65} - \frac{1}{629.85}\right). 4. Solve for P_2 Compute the right side: \ln(P_2) = -\frac{58510}{8.314} \times (-0.0000204). Calculate \ln(P_2) and then exponentiate to find P_2.

Explanation

1. Use Clausius-Clapeyron Equation<br /> The Clausius-Clapeyron equation is given by **$\ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$**. Here, $P_1$ is the vapor pressure at the normal boiling point (1 atm), $T_1 = 356.7 + 273.15$ K, $T_2 = 344.5 + 273.15$ K, and $R = 8.314 \, \text{J/mol K}$.<br /><br />2. Convert Temperatures to Kelvin<br /> $T_1 = 356.7 + 273.15 = 629.85 \, \text{K}$; $T_2 = 344.5 + 273.15 = 617.65 \, \text{K}$.<br /><br />3. Calculate $\ln(P_2)$<br /> Substitute values into the equation: $\ln \left(\frac{P_2}{1}\right) = -\frac{58510}{8.314} \left(\frac{1}{617.65} - \frac{1}{629.85}\right)$.<br /><br />4. Solve for $P_2$<br /> Compute the right side: $\ln(P_2) = -\frac{58510}{8.314} \times (-0.0000204)$. Calculate $\ln(P_2)$ and then exponentiate to find $P_2$.
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