Gas Laws Quiz
Grade 11 · Science · Quiz
10 multiple-choice questions testing mastery of gas behavior, pressure-volume-temperature relationships, and the Ideal Gas Law.
Questions
Which law states that pressure and volume are inversely proportional at constant temperature?
- Charles's Law
- Gay-Lussac's Law
- Boyle's Law
- Dalton's Law
Show answer
Answer: Boyle's Law
Boyle's Law: P₁V₁ = P₂V₂ at constant T and n.
A gas at 300 K and 1.5 atm is heated to 450 K in a rigid container. What is the new pressure?
- 1.0 atm
- 1.5 atm
- 2.0 atm
- 2.25 atm
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Answer: 2.25 atm
Gay-Lussac's Law: 1.5/300 = P₂/450 → P₂ = 2.25 atm.
What temperature scale MUST be used in all gas law calculations?
- Celsius
- Fahrenheit
- Kelvin
- Rankine
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Answer: Kelvin
Kelvin is the absolute scale used in gas laws: K = °C + 273.15.
A 10.0 L gas at 2.0 atm and 300 K is compressed to 5.0 L and cooled to 150 K. What is the new pressure?
- 1.0 atm
- 2.0 atm
- 4.0 atm
- 8.0 atm
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Answer: 2.0 atm
Combined Gas Law: (2.0×10)/300 = (P₂×5.0)/150 → P₂ = 2.0 atm.
What is the value of R when using units of L, atm, mol, and K?
- 0.0821 L·atm/mol·K
- 8.314 J/mol·K
- 1.987 cal/mol·K
- 22.4 L/mol
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Answer: 0.0821 L·atm/mol·K
R = 0.0821 L·atm/mol·K when P is in atm and V is in liters.
A mixture contains O₂ at 0.40 atm, N₂ at 0.50 atm, CO₂ at 0.10 atm. Total pressure?
- 0.50 atm
- 0.90 atm
- 1.00 atm
- 1.10 atm
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Answer: 1.00 atm
Dalton's Law: 0.40 + 0.50 + 0.10 = 1.00 atm.
At STP, what volume does 0.500 mol of an ideal gas occupy?
- 11.2 L
- 22.4 L
- 44.8 L
- 5.6 L
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Answer: 11.2 L
0.500 mol × 22.4 L/mol = 11.2 L.
Which condition makes a real gas behave most like an ideal gas?
- High pressure, low temperature
- Low pressure, low temperature
- High pressure, high temperature
- Low pressure, high temperature
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Answer: Low pressure, high temperature
At low pressure molecules are far apart (minimal intermolecular forces); at high temperature kinetic energy dominates — both ideal gas assumptions.
A balloon at 25°C has volume 2.0 L. Moved outside to −25°C at constant pressure. New volume?
- 1.0 L
- 1.67 L
- 2.40 L
- 3.0 L
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Answer: 1.67 L
T₁=298 K, T₂=248 K. V₂ = 2.0 × (248/298) ≈ 1.67 L.
Using PV = nRT, what is the pressure of 2.0 mol of gas in 10.0 L at 400 K?
- 3.28 atm
- 6.57 atm
- 8.21 atm
- 13.14 atm
Show answer
Answer: 6.57 atm
P = nRT/V = (2.0 × 0.0821 × 400)/10.0 = 65.68/10.0 = 6.57 atm.
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