According to the ideal gas law, a 1.008 mol sample of oxygen gas in a 1.684 L container at 269.7 K should exert a pressure of 13.25 atm. What is the percent difference between the L² atm pressure calculated using the van der Waals' equation and the ideal pressure? For O₂ gas, a = 1.360 and b 0.03183 L mol mol² Percent difference = Percent difference = % Pideal Pvan der Waals Pideal +Pran der Waals 2 x 100
Ideal and Real Gases
Ideal gases obey conditions of the general gas laws under all states of pressure and temperature. Ideal gases are also named perfect gases. The attributes of ideal gases are as follows,
Gas Laws
Gas laws describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. The very first observations about the physical properties of gases was made by Robert Boyle in 1662. Later discoveries were made by Charles, Gay-Lussac, Avogadro, and others. Eventually, these observations were combined to produce the ideal gas law.
Gaseous State
It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed forms of matter studied are solids, gases and liquids. The best example of a substance that is present in different states is water. It is solid ice, gaseous vapor or steam and liquid water depending on the temperature and pressure conditions. This is due to the difference in the intermolecular forces and distances. The occurrence of three different phases is due to the difference in the two major forces, the force which tends to tightly hold molecules i.e., forces of attraction and the disruptive forces obtained from the thermal energy of molecules.
![### Ideal and Real Gases: Van der Waals Equation Application
According to the ideal gas law, a 1.008 mol sample of oxygen gas in a 1.684 L container at 269.7 K should exert a pressure of 13.25 atm. What is the percent difference between the pressure calculated using the van der Waals equation and the ideal pressure? For \( \text{O}_2 \) gas, \( a = 1.360 \, \frac{L^2 \cdot atm}{mol^2} \) and \( b = 0.03183 \, \frac{L}{mol} \).
#### Formula for Percent Difference
\[
\text{Percent difference} = \left( \frac{|P_{ideal} - P_{van \, der \, Waals}|}{\left( \frac{P_{ideal} + P_{van \, der \, Waals}}{2} \right)} \right) \times 100
\]
#### Calculation Box
Percent difference = \([ \text{Enter Value Here} ]\%\)
In this task, you will need to use the given constants for \( \text{O}_2 \) gas and calculate the pressure using both methods to find the percent difference.
This involves comparing the pressures obtained from the ideal gas law with those calculated using the Van der Waals equation, accounting for real gas behavior due to molecular interactions and volume.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0f1975b-ad07-4857-b626-1368fa1e7067%2Fb3c69f88-7f4e-4827-afee-8ded5a5bc0bc%2Fedcaxr_processed.png&w=3840&q=75)
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