Hi everyone, i'm having troubles with this chem problem.?
If you could, please give me step by step solution that way i understand it better thank you!
12.0 moles of gas are in a 3.00 L tank at 24.4 ∘C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2⋅atm/mol2 and b=0.0430 L/mol.
- pisgahchemistLv 74 years ago
Real vs ideal gases......
For many situations we can treat gases as ideal gases, and use the ideal gas equation: PV=nRT
Gases often deviate from what is predicted by the ideal gas equation. We refer to them as "real" gases. The pressure of a real gas can be determined from the van der Waals equation and the empirically determined constants for that gas. The van der Waals constants for methane are a=2.300L2⋅atm/mol2 and b=0.0430 L/mol.
Methane as an ideal gas
P = nRT / V
P = 12.0 mol x 0.08206 Latm/molK x 297.55K / 3.00L
P = 97.7 atm ......... to three significant digits
Methane as a real gas
P = nRT/(V-nb) - (an²/V²)
P = 12.0 mol x 0.08206Latm/molK x 297.55K / (3.00 - 12.0(0.0430))L - (2.300 x 12.0² / 3.00²)atm
P = 81.5 atm
The difference is 16.2 atm. The real pressure is less than the ideal pressure because of the attraction between molecules of methane.
Pressure is proportional to the number of collisions between the gas molecules and the walls of the container. Two factors are instrumental in causing the pressure to deviate from ideality:
1. Gas molecules are not points, but have finite volume and take up space. For low volumes the space occupied by the molecules themselves reduces the overall volume and causes the pressure to be greater than what the ideal gas equation predicts because the apparent number of collisions between the gas molecules and the walls of the container are greater.
2. Gas molecules are attracted to each other by intermolecular forces. The attraction serves to decrease the apparent pressure because the number of collisions between the gas molecules and the walls of the container will be reduced.