Part C 13.0 moles of gas are in a 8.00 L tank at 23.6°C. Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a = 2.300 L?. atm/mol and b=0.0430 L/mol. Express your answer with the appropriate units. > View Available Hint(s) HA ? pressure difference = 32.4 atm %3D
Part C 13.0 moles of gas are in a 8.00 L tank at 23.6°C. Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a = 2.300 L?. atm/mol and b=0.0430 L/mol. Express your answer with the appropriate units. > View Available Hint(s) HA ? pressure difference = 32.4 atm %3D
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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Part c show all work

Transcribed Image Text:### Van der Waals Real Gas Calculation
**Problem Statement:**
Calculate the difference in pressure between methane and an ideal gas under the given conditions.
**Given:**
- **Number of moles (n):** 13.0 moles
- **Volume (V):** 8.00 L
- **Temperature (T):** 23.6 °C
- **Van der Waals constants for methane:**
- \( a = 2.300 \, \text{L}^2 \cdot \text{atm/mol}^2 \)
- \( b = 0.0430 \, \text{L/mol} \)
**Task:**
Express your answer with the appropriate units.
**Attempted Solution:**
- **Pressure Difference:** 32.4 atm
- **Result:** Incorrect; 3 attempts remaining
This exercise requires calculating the pressure difference using the Van der Waals equation for real gases and comparing it with the ideal gas law result. The answer input was found to be incorrect, indicating a need to revisit the calculations.
Expert Solution

Step 1
Real gasses have attraction so they have less pressure than ideal gasses
Step by step
Solved in 2 steps with 1 images

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