The enthalpy of vaporization for methanol is 35.2 kJ/mol. Methanol has a vapor pressure of 760 torr at 64.7 °C. Using the Clausius-Clapeyron equation, what is the vapor pressure for methanol at 34.2 °C? Give your answer in torr, to the first decimal point.
The enthalpy of vaporization for methanol is 35.2 kJ/mol. Methanol has a vapor pressure of 760 torr at 64.7 °C. Using the Clausius-Clapeyron equation, what is the vapor pressure for methanol at 34.2 °C? Give your answer in torr, to the first decimal point.
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Question
![**Problem Statement:**
The enthalpy of vaporization for methanol is 35.2 kJ/mol. Methanol has a vapor pressure of 760 torr at 64.7°C. Using the Clausius-Clapeyron equation, what is the vapor pressure for methanol at 34.2°C? Give your answer in torr, to the first decimal point.
**Solution Explanation:**
To solve this problem, we will use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at different temperatures to its enthalpy of vaporization. It is expressed as:
\[
\ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
\]
Where:
- \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) respectively.
- \(\Delta H_{vap}\) is the enthalpy of vaporization.
- \(R\) is the universal gas constant (8.314 J/mol·K).
- Temperatures \(T_1\) and \(T_2\) must be in Kelvin.
**Steps to solve:**
1. Convert the temperatures from Celsius to Kelvin:
- \(T_1 = 64.7°C + 273.15 = 337.85\,K\)
- \(T_2 = 34.2°C + 273.15 = 307.35\,K\)
2. Rearrange the equation to solve for \(P_2\):
\[
\ln(P_2) = \ln(P_1) - \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
\]
3. Substitute the given values:
\[
\ln(P_2) = \ln(760) - \frac{35200}{8.314} \left(\frac{1}{307.35} - \frac{1}{337.85}\right)
\]
4. Calculate:
- Calculate the change in reciprocal temperatures.
- Compute the expression inside the exponential function.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39d99037-ab62-46d9-8eef-e84624aeb8d9%2F60e97e61-c986-4646-be69-d11088ed267a%2F10ij0fo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
The enthalpy of vaporization for methanol is 35.2 kJ/mol. Methanol has a vapor pressure of 760 torr at 64.7°C. Using the Clausius-Clapeyron equation, what is the vapor pressure for methanol at 34.2°C? Give your answer in torr, to the first decimal point.
**Solution Explanation:**
To solve this problem, we will use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at different temperatures to its enthalpy of vaporization. It is expressed as:
\[
\ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
\]
Where:
- \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) respectively.
- \(\Delta H_{vap}\) is the enthalpy of vaporization.
- \(R\) is the universal gas constant (8.314 J/mol·K).
- Temperatures \(T_1\) and \(T_2\) must be in Kelvin.
**Steps to solve:**
1. Convert the temperatures from Celsius to Kelvin:
- \(T_1 = 64.7°C + 273.15 = 337.85\,K\)
- \(T_2 = 34.2°C + 273.15 = 307.35\,K\)
2. Rearrange the equation to solve for \(P_2\):
\[
\ln(P_2) = \ln(P_1) - \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
\]
3. Substitute the given values:
\[
\ln(P_2) = \ln(760) - \frac{35200}{8.314} \left(\frac{1}{307.35} - \frac{1}{337.85}\right)
\]
4. Calculate:
- Calculate the change in reciprocal temperatures.
- Compute the expression inside the exponential function.
Expert Solution
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