C The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.415 atm? (AHvap = 28.5 kJ/mol) BUD ! 1 Bo Q A N @ 2 F2 W S > #3 80 F3 E D $ 4 a F4 R F % 5 tv F5 T MacBook Air A 6 G e F6 Y & 7 H F7 U 8 DII F8 J 7 +/- AC 9 D
C The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.415 atm? (AHvap = 28.5 kJ/mol) BUD ! 1 Bo Q A N @ 2 F2 W S > #3 80 F3 E D $ 4 a F4 R F % 5 tv F5 T MacBook Air A 6 G e F6 Y & 7 H F7 U 8 DII F8 J 7 +/- AC 9 D
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter9: Liquids And Solids
Section: Chapter Questions
Problem 2QAP
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![The image presents a question about the physical properties of a liquid, specifically related to its boiling point and vapor pressure. Below is the transcribed text, formatted appropriately for an educational website:
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### Boiling Point and Vapor Pressure Calculation
**Problem Statement:**
The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.415 atm? (ΔHvap = 28.5 kJ/mol)
---
In order to solve this problem, you would typically use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature. The equation is given by:
\[ \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where:
- \( P_1 \) is the vapor pressure at the boiling point (which is 1 atm at 282 °C).
- \( P_2 \) is the new vapor pressure (0.415 atm).
- \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization (28.5 kJ/mol).
- \( R \) is the gas constant (8.314 J/mol·K).
- \( T_1 \) is the boiling point in Kelvin.
- \( T_2 \) is the temperature we need to find.
To proceed, we need to convert temperatures from Celsius to Kelvin:
\[ T_1 = 282 + 273.15 = 555.15 \text{ K} \]
By substituting known values into the Clausius-Clapeyron equation, you can solve for \( T_2 \).
Note: This is a higher-level chemistry problem typically covered in undergraduate physical chemistry courses.
---
This formatted text provides clarity and context, suitable for enhancing learning on an educational platform.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57980898-f1ea-4baf-a646-fad54b56b73d%2Ff399c3a1-8745-4204-99af-1906ffbf52b4%2Fkrb2gxo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image presents a question about the physical properties of a liquid, specifically related to its boiling point and vapor pressure. Below is the transcribed text, formatted appropriately for an educational website:
---
### Boiling Point and Vapor Pressure Calculation
**Problem Statement:**
The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.415 atm? (ΔHvap = 28.5 kJ/mol)
---
In order to solve this problem, you would typically use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature. The equation is given by:
\[ \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where:
- \( P_1 \) is the vapor pressure at the boiling point (which is 1 atm at 282 °C).
- \( P_2 \) is the new vapor pressure (0.415 atm).
- \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization (28.5 kJ/mol).
- \( R \) is the gas constant (8.314 J/mol·K).
- \( T_1 \) is the boiling point in Kelvin.
- \( T_2 \) is the temperature we need to find.
To proceed, we need to convert temperatures from Celsius to Kelvin:
\[ T_1 = 282 + 273.15 = 555.15 \text{ K} \]
By substituting known values into the Clausius-Clapeyron equation, you can solve for \( T_2 \).
Note: This is a higher-level chemistry problem typically covered in undergraduate physical chemistry courses.
---
This formatted text provides clarity and context, suitable for enhancing learning on an educational platform.
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