5. Using the trendline on the graph, calculate the enthalpy of vaporization (AHvap) of water. ( R = 8.314 x 10* kJ/mol-K). Vapor Pressure of Water 6.20 6.00 5.80 y = -4948.2x + 19.925 5.60 5.40 5.20 5.00 4.80 4.60 4.40 0.0028 0.00285 0.0029 0.00295 0.003 0.00305 0.0031 1/T (K-1) Ln P
5. Using the trendline on the graph, calculate the enthalpy of vaporization (AHvap) of water. ( R = 8.314 x 10* kJ/mol-K). Vapor Pressure of Water 6.20 6.00 5.80 y = -4948.2x + 19.925 5.60 5.40 5.20 5.00 4.80 4.60 4.40 0.0028 0.00285 0.0029 0.00295 0.003 0.00305 0.0031 1/T (K-1) Ln P
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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Clausius-Clapeyron equation
LnP = -deltaH vap /RT+ C
Transcribed Image Text:**Appendix C: Vapor Pressure of Water**
This section explores the vapor pressure of water as a function of temperature. The relationship between temperature in degrees Celsius (°C) and vapor pressure in Torr is provided as both a table and a graph.
**Graph Explanation:**
The graph illustrates how vapor pressure increases with temperature. The x-axis represents the temperature in degrees Celsius, ranging from 0°C to 37°C, and the y-axis represents the vapor pressure in Torr. The curve steepens as temperature rises, indicating a rapid increase in vapor pressure with higher temperatures.
**Data Table:**
| Temperature (°C) | Pressure (Torr) |
|------------------|-----------------|
| 0 | 4.6 |
| 5 | 6.5 |
| 10 | 9.2 |
| 11 | 9.8 |
| 12 | 10.5 |
| 13 | 11.2 |
| 14 | 12.0 |
| 15 | 12.5 |
| 16 | 13.6 |
| 17 | 14.5 |
| 18 | 15.5 |
| 19 | 16.5 |
| 20 | 17.5 |
| 21 | 18.6 |
| 22 | 19.8 |
| 23 | 21.0 |
| 24 | 22.3 |
| 25 | 23.8 |
| 26 | 25.2 |
| 27 | 26.7 |
| 28 | 28.3 |
| 29 | 30.0 |
| 30 | 31.8 |
| 31 | 33.7 |
| 32 | 35.7 |
| 33 | 37.7 |
| 34 | 39.9 |
| 35 | 42.2 |
| 37* | 47.1 |
| 100 | 760 |
*\*Body temperature*
The table shows a gradual increase in vapor pressure as the temperature increases from 0°C to 100°C, where it reaches
![### Transcription and Explanation for Educational Use
**Problem Statement:**
1. Using the trendline on the graph, calculate the enthalpy of vaporization (\( \Delta H_{\text{vap}} \)) of water. ( \( R = 8.314 \times 10^{-3} \) kJ/mol·K).
**Graph Details:**
- **Title:** Vapor Pressure of Water
- **Axes:**
- **Y-axis:** \( \ln P \) (natural log of vapor pressure)
- **X-axis:** \( 1/T \) (K\(^{-1}\)), where \( T \) is the temperature in Kelvin
- **Trendline Equation:**
- \( y = -4948.2x + 19.925 \)
**Explanation:**
The graph illustrates the relationship between the natural logarithm of the vapor pressure of water and the inverse of temperature. The trendline, represented by the equation \( y = -4948.2x + 19.925 \), describes this relationship linearly.
**Application:**
- To determine the enthalpy of vaporization (\( \Delta H_{\text{vap}} \)), we utilize the slope of the trendline. According to the Clausius-Clapeyron equation, the slope (\(-\Delta H_{\text{vap}}/R\)) of the line is equal to \(-4948.2\).
- By rearranging, we can solve for \( \Delta H_{\text{vap}} \):
\[
\Delta H_{\text{vap}} = -(\text{slope} \times R) = 4948.2 \times 8.314 \times 10^{-3}
\]
- Calculating this product will give the enthalpy of vaporization of water in kJ/mol.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F159596e9-0ab1-4ebf-b731-c1d379bc4b68%2F916abb0c-6b1f-4fdf-a409-938f8bb434ea%2Fib1mr5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Transcription and Explanation for Educational Use
**Problem Statement:**
1. Using the trendline on the graph, calculate the enthalpy of vaporization (\( \Delta H_{\text{vap}} \)) of water. ( \( R = 8.314 \times 10^{-3} \) kJ/mol·K).
**Graph Details:**
- **Title:** Vapor Pressure of Water
- **Axes:**
- **Y-axis:** \( \ln P \) (natural log of vapor pressure)
- **X-axis:** \( 1/T \) (K\(^{-1}\)), where \( T \) is the temperature in Kelvin
- **Trendline Equation:**
- \( y = -4948.2x + 19.925 \)
**Explanation:**
The graph illustrates the relationship between the natural logarithm of the vapor pressure of water and the inverse of temperature. The trendline, represented by the equation \( y = -4948.2x + 19.925 \), describes this relationship linearly.
**Application:**
- To determine the enthalpy of vaporization (\( \Delta H_{\text{vap}} \)), we utilize the slope of the trendline. According to the Clausius-Clapeyron equation, the slope (\(-\Delta H_{\text{vap}}/R\)) of the line is equal to \(-4948.2\).
- By rearranging, we can solve for \( \Delta H_{\text{vap}} \):
\[
\Delta H_{\text{vap}} = -(\text{slope} \times R) = 4948.2 \times 8.314 \times 10^{-3}
\]
- Calculating this product will give the enthalpy of vaporization of water in kJ/mol.
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