Water vapor is a polar gas molecule. Because of this, it has fairly strong interactions with itself in the gas phase that can make it behave less like an ideal gas that other gases under similar conditions. What is the difference between the calculated pressures found using the ideal gas law and those found using the van der Waals gas equation for 1 mole of water vapor at −100°C in a 1.00 L container?(van der Waals constants for water are a = 5.464 L² atm mol⁻², b = 0.03049 L mol⁻¹)

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The Answer of Water vapor is a polar gas molecule. Because of this, it has fairly strong interactions with itself in the gas phase that can make it behave less like an ideal gas that other gases under similar conditions. What is the difference between the calculated pressures found using the ideal gas law and those found using the van der Waals gas equation for 1 mole of water vapor at −100°C in a 1.00 L container?(van der Waals constants for water are a = 5.464 L² atm mol⁻², b = 0.03049 L mol⁻¹)

Water Vapor Is a Polar Gas Molecule