An unknown liquid has a heat of vaporization of 31.32 kJ/mole. If the normal boiling point is 88, what is vapor pressure (in torr) of this liquid at room temperature of 25 degrees C? HINT: Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).
An unknown liquid has a heat of vaporization of 31.32 kJ/mole. If the normal boiling point is 88, what is vapor pressure (in torr) of this liquid at room temperature of 25 degrees C? HINT: Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).
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Question
![**Determining the Vapor Pressure of an Unknown Liquid**
**Problem Statement:**
An unknown liquid has a heat of vaporization of 31.32 kJ/mole. If the normal boiling point is 88°C, what is the vapor pressure (in torr) of this liquid at room temperature of 25°C?
**Hint:**
Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).
**Explanation:**
To solve this problem, we use the Clausius-Clapeyron equation which relates the vapor pressure of a substance at two different temperatures.
The Clausius-Clapeyron equation can be written as:
\[ \ln \left( \frac{P_2}{P_1} \right) = - \frac{\Delta H_\text{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Where:
- \( P_1 \) is the initial vapor pressure (760 mm Hg at the boiling point).
- \( P_2 \) is the vapor pressure at the new temperature.
- \( \Delta H_\text{vap} \) is the heat of vaporization (31.32 kJ/mole).
- \( R \) is the ideal gas constant (8.314 J/mol·K).
- \( T_1 \) is the initial temperature in Kelvin (88°C = 361 K).
- \( T_2 \) is the new temperature in Kelvin (25°C = 298 K).
Solve for \( P_2 \):
1. Convert the heat of vaporization from kJ/mole to J/mole:
\[ \Delta H_\text{vap} = 31.32 \text{ kJ/mole} \times 1000 \text{ J/kJ} = 31320 \text{ J/mole} \]
2. Substitute the values into the Clausius-Clapeyron equation:
\[ \ln \left( \frac{P_2}{760 \text{ mm Hg}} \right) = - \frac{31320 \text{ J/mole}}{8.314 \text{ J/mol·K}} \left( \frac{1}{298 \text{ K}} - \frac{1}{361 \text{ K}} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6746fa6f-1856-443b-addc-ecae5d9da9b9%2Fb4fd2384-c0f6-46da-aa4a-4452fa0eb8f5%2Fdyl3e8e.jpeg&w=3840&q=75)
Transcribed Image Text:**Determining the Vapor Pressure of an Unknown Liquid**
**Problem Statement:**
An unknown liquid has a heat of vaporization of 31.32 kJ/mole. If the normal boiling point is 88°C, what is the vapor pressure (in torr) of this liquid at room temperature of 25°C?
**Hint:**
Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).
**Explanation:**
To solve this problem, we use the Clausius-Clapeyron equation which relates the vapor pressure of a substance at two different temperatures.
The Clausius-Clapeyron equation can be written as:
\[ \ln \left( \frac{P_2}{P_1} \right) = - \frac{\Delta H_\text{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Where:
- \( P_1 \) is the initial vapor pressure (760 mm Hg at the boiling point).
- \( P_2 \) is the vapor pressure at the new temperature.
- \( \Delta H_\text{vap} \) is the heat of vaporization (31.32 kJ/mole).
- \( R \) is the ideal gas constant (8.314 J/mol·K).
- \( T_1 \) is the initial temperature in Kelvin (88°C = 361 K).
- \( T_2 \) is the new temperature in Kelvin (25°C = 298 K).
Solve for \( P_2 \):
1. Convert the heat of vaporization from kJ/mole to J/mole:
\[ \Delta H_\text{vap} = 31.32 \text{ kJ/mole} \times 1000 \text{ J/kJ} = 31320 \text{ J/mole} \]
2. Substitute the values into the Clausius-Clapeyron equation:
\[ \ln \left( \frac{P_2}{760 \text{ mm Hg}} \right) = - \frac{31320 \text{ J/mole}}{8.314 \text{ J/mol·K}} \left( \frac{1}{298 \text{ K}} - \frac{1}{361 \text{ K}} \
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