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MCAT Gas Laws Practice Questions: Ideal Gas, Dalton, Graham and Henry

By StudyStore · Published 2026-10-20 · 10 original questions · about 15 min

Ten free MCAT-style gas law practice questions on the ideal and combined gas laws, partial pressures, effusion, gas solubility, real gas behavior and kinetic theory, with full explanations.

Quick concept primer

An ideal gas obeys PV = nRT, with R = 0.0821 L·atm/(mol·K) or 8.314 J/(mol·K). Holding the amount of gas fixed gives the combined gas law, P1V1/T1 = P2V2/T2, which contains Boyle's, Charles's and Gay-Lussac's laws as special cases. Temperature must always be in kelvins. At STP (273 K and 1 atm), one mole of an ideal gas fills about 22.4 L. In a mixture, each gas contributes a partial pressure equal to its mole fraction times the total pressure (Dalton's law).

Kinetic molecular theory explains these laws: gas particles have negligible volume, do not attract one another, collide elastically, and have an average kinetic energy proportional to absolute temperature. Lighter particles therefore move faster at the same temperature, and effusion rate scales as 1/√M (Graham's law). Henry's law says the concentration of a dissolved gas is proportional to its partial pressure above the liquid. Real gases deviate from ideal behavior at high pressure, where particle volume matters, and at low temperature, where attractions matter.

Traps: convert Celsius to kelvins before using any ratio, and remember that equal temperature means equal average kinetic energy, not equal speed.

10 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

A flexible weather balloon holds 15 L of helium at 280 K and 0.90 atm on the ground. It rises to an altitude where the pressure is 0.30 atm and the temperature is 210 K. Assuming no helium escapes, about what is its new volume?

  1. 3.8 L
  2. 34 L
  3. 45 L
  4. 60 L
Show answer and explanation

Answer: B. 34 L

The combined gas law gives V2 = V1 x (P1/P2) x (T2/T1) = 15 x (0.90/0.30) x (210/280) = 15 x 3 x 0.75, about 34 L. The pressure drop triples the volume while the cooling shrinks it by a quarter. The 45 L choice ignores the temperature change, and 60 L inverts the temperature ratio.

Question 2

A rigid 4.1 L cylinder holds a gas at 300 K and 6.0 atm. Using R = 0.0821 L·atm/(mol·K), about how many moles of gas does it contain?

  1. 0.10 mol
  2. 0.50 mol
  3. 1.0 mol
  4. 10 mol
Show answer and explanation

Answer: C. 1.0 mol

Rearranging the ideal gas law gives n = PV/RT = (6.0 x 4.1)/(0.0821 x 300) = 24.6/24.6, which is about 1.0 mol. The 0.10 and 10 mol choices come from slipping a factor of ten in R, and 0.50 mol would correspond to half the stated pressure.

Question 3

A gas cylinder contains 0.50 mol of nitrogen, 0.30 mol of oxygen and 0.20 mol of carbon dioxide at a total pressure of 800 torr. What is the partial pressure of oxygen?

  1. 240 torr
  2. 160 torr
  3. 400 torr
  4. 800 torr
Show answer and explanation

Answer: A. 240 torr

The total moles are 0.50 + 0.30 + 0.20 = 1.00, so the mole fraction of oxygen is 0.30. Dalton's law gives PO2 = 0.30 x 800 = 240 torr. The 160 torr choice is the partial pressure of carbon dioxide, 400 torr is nitrogen's, and 800 torr is the total pressure of the mixture.

Question 4

Under identical conditions, an unknown gas effuses through a pinhole at exactly half the rate of methane (molar mass 16 g/mol). What is the molar mass of the unknown gas?

  1. 4 g/mol
  2. 8 g/mol
  3. 32 g/mol
  4. 64 g/mol
Show answer and explanation

Answer: D. 64 g/mol

Graham's law says rate1/rate2 = √(M2/M1). Setting rateunknown/ratemethane = 1/2 gives √(16/M) = 1/2, so 16/M = 1/4 and M = 64 g/mol. The 32 g/mol choice forgets the square root, and the smaller choices would make the unknown gas faster than methane, not slower.

Question 5

At 37 °C, a gas reaches a dissolved concentration of 0.30 mM in water when its partial pressure above the water is 0.20 atm. If its partial pressure is raised to 0.50 atm at the same temperature, what concentration will dissolve at equilibrium?

  1. 0.12 mM
  2. 0.30 mM
  3. 0.75 mM
  4. 1.5 mM
Show answer and explanation

Answer: C. 0.75 mM

Henry's law makes dissolved concentration proportional to partial pressure, C = kP. Raising the partial pressure by a factor of 0.50/0.20 = 2.5 raises the concentration to 0.30 x 2.5 = 0.75 mM. The 0.12 mM choice divides by 2.5 instead. This proportionality is why divers who breathe compressed air at depth carry more dissolved nitrogen in their blood.

Question 6

A chemist measures PV/(nRT) for a sample of a polar gas at a low temperature and a moderate pressure and obtains 0.90 instead of 1.00. What best explains this result?

  1. Attractions between gas particles reduce the force of their collisions with the container walls
  2. The finite volume of the gas particles makes the gas occupy more space than predicted
  3. The gas particles move faster than kinetic molecular theory predicts
  4. Collisions between gas particles create new particles
Show answer and explanation

Answer: A. Attractions between gas particles reduce the force of their collisions with the container walls

A value below 1 means the measured pressure is lower than ideal. At low temperature, intermolecular attractions pull particles toward each other just before they strike the wall, so wall collisions are softer and pressure drops. Finite particle volume (B) pushes PV/(nRT) above 1, mainly at high pressure. Choices C and D describe nothing that real gases do.

Question 7

Each of the following is an assumption of the kinetic molecular theory of ideal gases EXCEPT:

  1. Gas particles occupy negligible volume compared with their container
  2. Collisions between gas particles are perfectly elastic
  3. The average kinetic energy of the particles is proportional to absolute temperature
  4. All particles in a sample move at the same speed at a given temperature
Show answer and explanation

Answer: D. All particles in a sample move at the same speed at a given temperature

Kinetic theory assumes a distribution of speeds (the Maxwell-Boltzmann distribution), not one shared speed; only the average kinetic energy is fixed by temperature. Negligible particle volume, elastic collisions and kinetic energy proportional to kelvin temperature are all core assumptions of the model, so choices A, B and C are not the exception.

Question 8

A container holds a mixture of helium (4 g/mol) and argon (40 g/mol) at 300 K. Which statement is correct?

  1. Argon particles have 10 times the average kinetic energy of helium particles
  2. Both gases have the same average kinetic energy, and helium's rms speed is about 3.2 times argon's
  3. Both gases have the same rms speed because they are at the same temperature
  4. Helium particles have the higher average kinetic energy because they move faster
Show answer and explanation

Answer: B. Both gases have the same average kinetic energy, and helium's rms speed is about 3.2 times argon's

Average kinetic energy depends only on temperature, so both gases share it. Because ½mv2 is equal, the rms speed scales as 1/√M, and √(40/4) = √10, about 3.2. Choices A and D tie kinetic energy to mass or speed alone, and C confuses equal energy with equal speed.

Question 9

What mass of carbon dioxide (44 g/mol) occupies 5.6 L at STP, treating it as an ideal gas?

  1. 5.5 g
  2. 11 g
  3. 22 g
  4. 44 g
Show answer and explanation

Answer: B. 11 g

At STP one mole of ideal gas occupies about 22.4 L, so 5.6 L is 5.6/22.4 = 0.25 mol. The mass is 0.25 x 44 = 11 g. The 44 g choice is one full mole, 22 g is half a mole, and 5.5 g is an eighth of a mole, each from a misread of the volume ratio.

Question 10

A gas is sealed in a rigid steel container, and its pressure is recorded at three temperatures.

Temperature (°C)Pressure (kPa)
27120
77140
127?

What pressure is expected at 127 °C?

  1. 90 kPa
  2. 140 kPa
  3. 160 kPa
  4. 564 kPa
Show answer and explanation

Answer: C. 160 kPa

In a rigid container, pressure is proportional to kelvin temperature. Converting gives 300 K, 350 K and 400 K, and 120 x (400/300) = 160 kPa, which matches the trend set by the 77 °C point (120 x 350/300 = 140). The 564 kPa choice uses Celsius temperatures directly, and 90 kPa inverts the ratio.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.

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