The enthalpy of vaporization for acetone is 32.0 kJ/mol. The normal boiling point (@ 1.00 atm) for acetone is 56.5 C. What is the temperature of acetone at 560 torr?
The enthalpy of vaporization for acetone is 32.0 kJ/mol. The normal boiling point (@ 1.00 atm) for acetone is 56.5 C. What is the temperature of acetone at 560 torr?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The enthalpy of vaporization for acetone is 32.0 kJ/mol. The normal boiling point (@ 1.00 atm) for acetone is 56.5 C. What is the temperature of acetone at 560 torr? Can you explain using the claussius equation i provided below, m horrible at math so plz help me understand.
![**Clausius-Clapeyron Equation**
The Clausius-Clapeyron equation is presented as follows:
\[
\ln \left( \frac{P_1}{P_2} \right) = \frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)
\]
where:
- \( \ln \left( \frac{P_1}{P_2} \right) \) is the natural logarithm of the ratio of pressures \( P_1 \) and \( P_2 \).
- \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization.
- \( R \) is the universal gas constant, with a value of \( 8.3145 \, \frac{J}{\text{mol} \cdot K} \).
- \( T_1 \) and \( T_2 \) are the temperatures in Kelvin.
This equation is used to describe the phase transition between two phases of matter, such as liquid and gas, and can be applied to estimate the vapor pressure of a liquid at a given temperature or the enthalpy of vaporization.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08bb2a94-da17-43f2-90ab-4ad444df37b3%2F83638440-d6ad-4722-83fd-889096eff3ab%2Fym5tczw_processed.png&w=3840&q=75)
Transcribed Image Text:**Clausius-Clapeyron Equation**
The Clausius-Clapeyron equation is presented as follows:
\[
\ln \left( \frac{P_1}{P_2} \right) = \frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)
\]
where:
- \( \ln \left( \frac{P_1}{P_2} \right) \) is the natural logarithm of the ratio of pressures \( P_1 \) and \( P_2 \).
- \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization.
- \( R \) is the universal gas constant, with a value of \( 8.3145 \, \frac{J}{\text{mol} \cdot K} \).
- \( T_1 \) and \( T_2 \) are the temperatures in Kelvin.
This equation is used to describe the phase transition between two phases of matter, such as liquid and gas, and can be applied to estimate the vapor pressure of a liquid at a given temperature or the enthalpy of vaporization.
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