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The van der Waals equation is a modification of the ideal gas law that accounts for the non-ideal behavior of real gases. It introduces parameters to correct for intermolecular forces and the volume occupied by gas molecules. This topic appears in exams to test your understanding of how real gases deviate from ideal behavior and the conditions under which these deviations are significant.
This topic is frequently tested in chemistry, physics, and engineering exams, particularly in advanced high school and undergraduate courses. It typically carries 10-15% of the total marks and tests your ability to apply theoretical concepts to real-world scenarios, understand molecular interactions, and perform calculations accurately.
The van der Waals equation is given by: [ (P + \frac{a}{V_m^2})(V_m - b) = RT ] - Primary Rule: This equation modifies the ideal gas law to account for real gas behavior. - Sub-rules: - (a) corrects for intermolecular forces. - (b) corrects for the volume occupied by gas molecules. - (V_m) is the molar volume. - Mnemonic: Remember "a for attraction, b for bulk."
Intermediate
Question: Calculate the molar volume of 1 mole of CO? at 298 K and 1 atm using the van der Waals equation. Given (a = 3.59 \text{ L}^2 \text{ atm/mol}^2) and (b = 0.0427 \text{ L/mol}).
Step-by-Step:1. Substitute the given values into the van der Waals equation.2. Solve for (V_m).
Answer: (V_m \approx 24.5 \text{ L/mol})
Question: Determine the compressibility factor (Z) for methane (CH?) at 300 K and 10 atm. Given (a = 2.25 \text{ L}^2 \text{ atm/mol}^2) and (b = 0.0428 \text{ L/mol}).
Step-by-Step:1. Use the van der Waals equation to find (V_m).2. Calculate (Z) using (Z = \frac{PV_m}{RT}).
Answer: (Z \approx 0.95)
Question: Explain why real gases deviate from ideal behavior at high pressures and low temperatures. Use the van der Waals equation to support your explanation.
Step-by-Step:1. Discuss the significance of (a) and (b) in the van der Waals equation.2. Explain how intermolecular forces and molecular volume become significant at high pressures and low temperatures.3. Show how the van der Waals equation accounts for these deviations.
Answer: Real gases deviate due to intermolecular forces and molecular volume, which are accounted for by the van der Waals constants (a) and (b).
Question: What is the molar volume of 1 mole of N? at 300 K and 1 atm using the van der Waals equation? Given (a = 1.39 \text{ L}^2 \text{ atm/mol}^2) and (b = 0.0391 \text{ L/mol}).
Options: A) 24.8 L/mol B) 25.2 L/mol C) 24.5 L/mol D) 23.9 L/mol
Correct Answer: C) 24.5 L/mol
Explanation: Substitute the given values into the van der Waals equation and solve for (V_m).
Why the Distractors Are Tempting: - A) and B) are close to the ideal gas value. - D) is a common rounding error.
Question: The compressibility factor (Z) for a real gas is found to be 0.85 at a certain temperature and pressure. What does this indicate?
Options: A) The gas is behaving ideally. B) The gas is under high pressure. C) The gas is at low temperature. D) The gas is at high temperature.
Correct Answer: B) The gas is under high pressure.
Explanation: A (Z) value significantly less than 1 indicates non-ideal behavior, often due to high pressure.
Why the Distractors Are Tempting: - A) and D) suggest ideal behavior. - C) is a common misconception about temperature effects.
Question: Which of the following is a correct statement about the van der Waals constants (a) and (b)?
Options: A) (a) accounts for molecular volume. B) (b) accounts for intermolecular forces. C) Both (a) and (b) are zero for ideal gases. D) (a) and (b) are always positive for real gases.
Correct Answer: D) (a) and (b) are always positive for real gases.
Explanation: (a) and (b) are positive constants that account for intermolecular forces and molecular volume, respectively.
Why the Distractors Are Tempting: - A) and B) mix up the roles of (a) and (b). - C) suggests ideal gases have non-zero (a) and (b).
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