Gas laws practice questions.

A fixed mass of an ideal gas occupies 2.0 × 10⁻³ m³ at a pressure of 100 kPa. It is compressed at constant temperature until its volume is 0.50 × 10⁻³ m³. What is the new pressure?

• 25 kPa
• 200 kPa
• 400 kPa
• 800 kPa

Answer: 400 kPa

At constant temperature p₁V₁ = p₂V₂, so p₂ = 100 × (2.0/0.50) = 400 kPa. Choosing 25 kPa comes from inverting the volume ratio.

An ideal gas is heated from 27 °C to 327 °C. What is the ratio of the average translational kinetic energy of its molecules at the higher temperature to that at the lower temperature?

• 12
• 1.4
• 4.0
• 2.0

Answer: 2.0

Average kinetic energy is proportional to the absolute temperature: 600 K / 300 K = 2.0. Using Celsius temperatures (327/27 ≈ 12) is the classic error.

Which of the following is NOT an assumption of the kinetic model of an ideal gas?

• All molecules move with the same speed
• Collisions between molecules and the container walls are perfectly elastic
• The total volume of the molecules is negligible compared with the volume occupied by the gas
• There are no intermolecular forces except during collisions

Answer: All molecules move with the same speed

The model assumes random motion with a distribution of speeds, not a single common speed. The other three statements are genuine assumptions of the kinetic model.

An ideal gas at a pressure of 1.0 × 10⁵ Pa and a temperature of 300 K occupies a volume of 2.49 × 10⁻³ m³. The number of moles of gas is closest to (R = 8.31 J K⁻¹ mol⁻¹):

• 0.010
• 0.10
• 1.0
• 10

Answer: 0.10

n = pV/(RT) = (1.0 × 10⁵ × 2.49 × 10⁻³)/(8.31 × 300) = 249/2493 ≈ 0.10 mol.

A sealed rigid container holds an ideal gas at 27 °C and 150 kPa. The gas is heated to 127 °C. What is the new pressure?

• 200 kPa
• 706 kPa
• 113 kPa
• 150 kPa

Answer: 200 kPa

At constant volume p/T is constant with T in kelvin: p₂ = 150 × (400/300) = 200 kPa. The value 706 kPa results from wrongly using the Celsius ratio 127/27.

Calculate the internal energy of 2.0 mol of an ideal monatomic gas at a temperature of 300 K. (R = 8.31 J K⁻¹ mol⁻¹)

• 2.5 kJ
• 5.0 kJ
• 15 kJ
• 7.5 kJ

Answer: 7.5 kJ

U = (3/2)nRT = 1.5 × 2.0 × 8.31 × 300 ≈ 7.5 × 10³ J = 7.5 kJ. Forgetting the factor 3/2 gives 5.0 kJ.