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Thermal Physics · The Gas Laws

Gas Properties

Trap a gas in a container and change its pressure, volume, and temperature one at a time. Discover Boyle's law, Charles's law, and Gay-Lussac's law, then combine them into the ideal gas law PV = nRT — the single equation that links all the properties of a gas to the motion of its molecules.

Theory — Pressure, Volume, Temperature, and the Gas Laws

What Gas Pressure Is

A gas is made of fast-moving molecules that collide constantly with the walls of their container. Each collision gives the wall a tiny push; the sum of all these pushes, per unit area, is the pressure. More collisions, or harder collisions, mean higher pressure. Pressure is measured in pascals (Pa) or kilopascals (kPa); atmospheric pressure is about 101 kPa.

Pressure from Molecular Collisions Pressure = force of wall collisions / area
Faster molecules (higher T) or more crowding (smaller V) → more pressure

Boyle's Law — Pressure vs Volume

At constant temperature, squeezing a gas into a smaller volume packs the molecules closer, so they hit the walls more often and the pressure rises. Pressure and volume are inversely proportional.

Boyle's Law (constant T, n) P · V = constant  →  P₁V₁ = P₂V₂
Halve the volume → double the pressure

Charles's Law — Volume vs Temperature

At constant pressure, heating a gas makes its molecules move faster; to keep the pressure the same, the gas must expand. Volume is directly proportional to the absolute (Kelvin) temperature.

Charles's Law (constant P, n) V / T = constant  →  V₁/T₁ = V₂/T₂  (T in kelvin)
Double the temperature (K) → double the volume

Gay-Lussac's Law — Pressure vs Temperature

At constant volume, heating a gas makes the molecules strike the walls faster and more often, so the pressure rises. Pressure is directly proportional to the absolute temperature.

Gay-Lussac's Law (constant V, n) P / T = constant  →  P₁/T₁ = P₂/T₂  (T in kelvin)
Double the temperature (K) → double the pressure

The Ideal Gas Law

The three laws are special cases of one master equation that ties pressure, volume, the amount of gas (number of moles n), and the absolute temperature together.

Ideal Gas Law P · V = n · R · T

R = 8.314 J/(mol·K) (universal gas constant)
T must be in kelvin: T(K) = T(°C) + 273
Hold any two of P, V, T fixed and the law reduces to Boyle, Charles, or Gay-Lussac

Boyle

Constant T. P and V inversely related. PV = constant. Squeeze → pressure up.

Charles

Constant P. V and T directly related. V/T = constant. Heat → expands.

Gay-Lussac

Constant V. P and T directly related. P/T = constant. Heat → pressure up.

LawHeld constantRelationshipEquation
Boyle'sT, nP ∝ 1/V (inverse)P₁V₁ = P₂V₂
Charles'sP, nV ∝ T (direct)V₁/T₁ = V₂/T₂
Gay-Lussac'sV, nP ∝ T (direct)P₁/T₁ = P₂/T₂
Ideal gaslinks all fourPV = nRT

Instructions — Running the Virtual Experiment

The Gas Chamber tab lets you change conditions and watch the molecules; the Boyle's Law tab provides the quantitative test. Record every reading in your lab notebook.

Experiment 1 — Exploring the Three Laws (Gas Chamber tab)
1
Open Simulation → Gas Chamber. Set the law selector to Boyle (constant T). Reduce the volume with the slider and watch the pressure climb as the molecules crowd together and hit the walls more often. Confirm P×V stays roughly constant.
2
Switch to Charles (constant P). Raise the temperature and watch the piston rise as the gas expands. Confirm V/T stays constant (use kelvin).
3
Switch to Gay-Lussac (constant V). Raise the temperature with the volume locked and watch the pressure climb. Confirm P/T stays constant.
Experiment 2 — Verifying Boyle's Law (Boyle's Law tab)
1
Open Boyle's Law. The temperature and amount of gas are fixed (T = 300 K, n = 1 mol). Click each volume button (10–50 L) and then Record reading to log the pressure.
2
Click Plot & check. The pressure-versus-volume points form a curve (a hyperbola), and the product P×V should be constant for every reading — confirming Boyle's law and equalling nRT.

Simulation — Gas Chamber & the Gas Laws

Gas Properties Virtual LabChange conditions one at a time and watch the gas respond
gas molecules
Piston height = volume · molecule speed = temperature. The slider held constant by the chosen law is greyed out.

Controls

Readout
Pressure P— kPa
Volume V— L
Temperature T— K
P·V
Each point is one (volume, pressure) reading at fixed T = 300 K, n = 1 mol.

Fixed: T = 300 K, n = 1 mol

Set volume

Current reading (10 L)
Pressure— kPa
Volume (L)Pressure (kPa)P·V (kPa·L)
No readings yet — pick a volume and click "Record reading".

Team Questions

Question 1. A gas occupies 2.0 L at a pressure of 100 kPa. At constant temperature, it is compressed until the pressure is 250 kPa. What is the new volume? Use P₁V₁ = P₂V₂. (Type just the number in L)
Question 2. A gas occupies 3.0 L at 300 K. At constant pressure, it is heated to 450 K. What is the new volume? Use V₁/T₁ = V₂/T₂. (Type just the number in L)
Question 3. A gas has a pressure of 200 kPa at 300 K in a rigid (constant-volume) container. It is heated to 600 K. What is the new pressure? Use P₁/T₁ = P₂/T₂. (Type just the number in kPa)
Question 4. Which law describes the inverse relationship between the pressure and volume of a gas at constant temperature? (Name it)
Question 5. In the ideal gas law PV = nRT, the temperature must be expressed in which unit — Celsius or Kelvin? (One word)
Question 6. Using PV = nRT, find the volume occupied by 1.0 mol of an ideal gas at 273 K and 101.3 kPa (standard conditions). Use R = 8.314 J/(mol·K). (Type just the number in litres — e.g. 22.4)
Question 7 — Challenge. A gas at 27 °C is heated in a rigid container until its pressure doubles. What is the new temperature in °C? (Hint: convert to kelvin first.) (Type just the number in °C)

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.

Gas Properties: The Gas Laws and the Ideal Gas Equation

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To investigate the relationships among the pressure, volume, and temperature of a gas — Boyle's law, Charles's law, and Gay-Lussac's law — and to verify Boyle's law quantitatively by showing that the product of pressure and volume is constant at fixed temperature. The individual laws are shown to be special cases of the ideal gas law PV = nRT.

Theory

Gas pressure arises from molecular collisions with the container walls. At constant temperature, pressure and volume are inversely related (Boyle); at constant pressure, volume is proportional to absolute temperature (Charles); at constant volume, pressure is proportional to absolute temperature (Gay-Lussac). All three follow from the ideal gas law.

Boyle: P₁V₁ = P₂V₂ (const T)
Charles: V₁/T₁ = V₂/T₂ (const P)
Gay-Lussac: P₁/T₁ = P₂/T₂ (const V)
Ideal gas: PV = nRT, R = 8.314 J/(mol·K), T in K

Calculations — Sample: 1.0 mol at 300 K (Boyle's law data)

nRT (constant): (1.0)(8.314)(300) = 2494 J = 2494 kPa·L (since 1 J = 1 kPa·L)

At V = 20 L: P = nRT/V = 2494/20 = 124.7 kPa; P·V = 124.7 × 20 = 2494 kPa·L ✓

At V = 40 L: P = 2494/40 = 62.4 kPa; P·V = 62.4 × 40 = 2494 kPa·L ✓ (halving from 20→40 L? doubling volume halves pressure)

Standard volume check: 1 mol at 273 K and 101.3 kPa → V = nRT/P = (1)(8.314)(273)/101.3 = 22.4 L (the molar volume at STP)

Results Table — Boyle's Law (T = 300 K, n = 1 mol)

Volume (L)Pressure (kPa)P·V (kPa·L)
10249.42494
20124.72494
3083.12494
4062.42494
5049.92494

Charles trial (const P): doubling T from 300 K to 600 K doubled V. Gay-Lussac trial (const V): doubling T from 300 K to 600 K doubled P. Both confirm direct proportionality to absolute temperature.

Discussion

The pressure-versus-volume data formed a hyperbola, and the product P·V was constant at 2494 kPa·L for every volume tested, confirming Boyle's law. This constant equals nRT for the fixed temperature and amount of gas, as the ideal gas law predicts. Halving the volume doubled the pressure and vice versa, the signature of an inverse relationship.

The Charles and Gay-Lussac trials showed direct proportionality to the absolute temperature: at constant pressure the volume doubled when the kelvin temperature doubled, and at constant volume the pressure doubled when the kelvin temperature doubled. These results are consistent with the molecular picture — heating speeds up the molecules, increasing either the volume (to keep pressure constant) or the pressure (if the volume is fixed). All three laws were reproduced by holding two of the variables in PV = nRT constant, confirming that the ideal gas law unifies them.

Conclusion

The experiment verified Boyle's law (PV constant at fixed T) and demonstrated Charles's and Gay-Lussac's laws (V ∝ T and P ∝ T at fixed P and V respectively). The constant value of PV matched nRT, and a standard-conditions calculation recovered the molar volume of 22.4 L, confirming the ideal gas law PV = nRT as the unifying relationship among the gas properties.

Practice Questions

Show all work and include units. Remember to use kelvin for temperature.

Question 1
A 5.0 L sample of gas at 120 kPa is compressed at constant temperature to 2.0 L. What is the new pressure?
Hint: Boyle's law, P₁V₁ = P₂V₂.
Question 2
A balloon holds 2.5 L of air at 27 °C. It is warmed at constant pressure to 87 °C. What is its new volume? (Convert to kelvin first.)
Hint: Charles's law, V₁/T₁ = V₂/T₂. 27 °C = 300 K, 87 °C = 360 K.
Question 3
A rigid tank of gas reads 150 kPa at 290 K. It is cooled to 232 K. What is the new pressure?
Hint: Gay-Lussac's law, P₁/T₁ = P₂/T₂ (constant volume).
Question 4
How many moles of gas are in a 10.0 L container at 250 kPa and 300 K? Use PV = nRT with R = 8.314 J/(mol·K). (Watch units: 250 kPa = 250,000 Pa, 10.0 L = 0.0100 m³.)
Hint: n = PV/(RT). Keep SI units, then n comes out in moles.
Question 5
Explain, in terms of molecular collisions, why the pressure of a gas in a sealed rigid container increases when the gas is heated.
Hint: higher T → faster molecules → more frequent and harder wall collisions → higher pressure (Gay-Lussac).
Question 6 — Challenge
A gas occupies 4.0 L at 300 K and 100 kPa. It is heated to 450 K and simultaneously compressed to 2.0 L. What is the final pressure? Use the combined form P₁V₁/T₁ = P₂V₂/T₂.
Hint: solve P₂ = P₁V₁T₂/(T₁V₂) = (100)(4.0)(450)/[(300)(2.0)] = 300 kPa.