Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I faced a lot of computational problems. Please help me keep my transplants tidy. And my teacher gave me a comment, but I don't know how to approach it.
![**Educational Explanation of Van der Waals Equation Calculations**
---
**Van der Waals Equation Calculation:**
The Van der Waals equation is used to calculate the pressure of a real gas by incorporating molecular interactions and volumes, which the ideal gas law does not consider.
**Equation:**
\[ P = \frac{nRT}{V-nb} - \frac{n^2a}{V^2} \]
1. **Parameters:**
- \( P \) = Pressure
- \( n \) = Number of moles
- \( R \) = Universal gas constant
- \( T \) = Temperature in Kelvin
- \( V \) = Volume
- \( a \) and \( b \) = Van der Waals constants for specific gases
**Calculation Process:**
\[ n = 2 \, \text{g CH}_4 / \text{mol} \]
\[ R = 0.082 \, \text{L atm} / (\text{mol K}) \]
\[ T = 273K \times 10^2 \]
\[ \frac{1}{mol} \times 0.023 \, \text{L/mol} \]
\[ V = n^2 L - n2g \times \frac{1}{mol} \times 0.023 \, \text{L/mol} \]
**Pressure Calculation:**
\[ P = 1,900 \, \text{atm} \]
\[ \frac{n^2g \times \frac{1}{mol}}{(\text{lg CH}_4)^2} \times \frac{k \times 3 \times (2 \, \text{atm})}{\text{50mC}^2} \]
\[ = 1,900 \, \text{atm} \]
**Conversion to mmHg:**
\[ P = 1,900 \, \text{atm} \times \frac{760 \, \text{mmHg}}{1 \, \text{atm}} = 1,444,000 \, \text{mmHg} \]
**Interpretation:**
- The pressure calculated is higher because the Van der Waals equation considers the interaction between molecules, which cannot be ignored at high pressure.
**Note:**
Both calculations have errors, but the reasoning process is clear. One missing element is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ae764b7-d237-4053-a9e9-5b016ede3169%2F44dd3606-e223-45ea-9b4c-bf01a4344c98%2Frmgi86_processed.png&w=3840&q=75)
Transcribed Image Text:**Educational Explanation of Van der Waals Equation Calculations**
---
**Van der Waals Equation Calculation:**
The Van der Waals equation is used to calculate the pressure of a real gas by incorporating molecular interactions and volumes, which the ideal gas law does not consider.
**Equation:**
\[ P = \frac{nRT}{V-nb} - \frac{n^2a}{V^2} \]
1. **Parameters:**
- \( P \) = Pressure
- \( n \) = Number of moles
- \( R \) = Universal gas constant
- \( T \) = Temperature in Kelvin
- \( V \) = Volume
- \( a \) and \( b \) = Van der Waals constants for specific gases
**Calculation Process:**
\[ n = 2 \, \text{g CH}_4 / \text{mol} \]
\[ R = 0.082 \, \text{L atm} / (\text{mol K}) \]
\[ T = 273K \times 10^2 \]
\[ \frac{1}{mol} \times 0.023 \, \text{L/mol} \]
\[ V = n^2 L - n2g \times \frac{1}{mol} \times 0.023 \, \text{L/mol} \]
**Pressure Calculation:**
\[ P = 1,900 \, \text{atm} \]
\[ \frac{n^2g \times \frac{1}{mol}}{(\text{lg CH}_4)^2} \times \frac{k \times 3 \times (2 \, \text{atm})}{\text{50mC}^2} \]
\[ = 1,900 \, \text{atm} \]
**Conversion to mmHg:**
\[ P = 1,900 \, \text{atm} \times \frac{760 \, \text{mmHg}}{1 \, \text{atm}} = 1,444,000 \, \text{mmHg} \]
**Interpretation:**
- The pressure calculated is higher because the Van der Waals equation considers the interaction between molecules, which cannot be ignored at high pressure.
**Note:**
Both calculations have errors, but the reasoning process is clear. One missing element is
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