The van der Waals equation of state for a real gas is n²a P+ y2 nb| = nRT At what pressure will 1.00 mole of CH, be in a 10.0 L container at 298 Kassuming CH, is a real gas. (van der Waals constants for CH4 are a = 2.253 L² atm mol 2, b = 0.04278 L mol-1) O 2.43 atm O 2.28 atm O 2.51 atm O 24.5 atm O 0.440 atm
The van der Waals equation of state for a real gas is n²a P+ y2 nb| = nRT At what pressure will 1.00 mole of CH, be in a 10.0 L container at 298 Kassuming CH, is a real gas. (van der Waals constants for CH4 are a = 2.253 L² atm mol 2, b = 0.04278 L mol-1) O 2.43 atm O 2.28 atm O 2.51 atm O 24.5 atm O 0.440 atm
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 van der Waals equation of state for a real gas is
n²a
P+
部
- nb| = nRT
At what pressure will 1.00 mole of CH, be in a 10.0 L container at 298 K assuming CH4 is a real gas.
(van der Waals constants for CH4 are a = 2.253 L2 atm mol-2, b = 0.04278 L mol-1)
O 2.43 atm
O 2.28 atm
O 2.51 atm
O 24.5 atm
O 0.440 atm](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9db4a9f0-7ea2-4c88-8b5d-c7bdc3d3b6d8%2F63e94eaf-9338-40b2-b9b5-fca21f6de571%2Ff8cneh2_processed.png&w=3840&q=75)
Transcribed Image Text:The van der Waals equation of state for a real gas is
n²a
P+
部
- nb| = nRT
At what pressure will 1.00 mole of CH, be in a 10.0 L container at 298 K assuming CH4 is a real gas.
(van der Waals constants for CH4 are a = 2.253 L2 atm mol-2, b = 0.04278 L mol-1)
O 2.43 atm
O 2.28 atm
O 2.51 atm
O 24.5 atm
O 0.440 atm
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