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Lesson 2 of 6

Kinetic Theory · Lesson 2 of 6

Gas Laws, Mixtures and Applications

“Use molecular and molar gas equations, interpret real-gas graphs and partial pressures, and solve practical gas-state problems.”

Learning Objectives

• Use PV = νRT = NkBT with absolute pressure and kelvin temperature. • Derive Boyle’s law, Charles’ law, the density form and the textbook STP molar volume. • Interpret ideal and real-gas plots, including molar versus sample-specific vertical axes. • Use partial pressures to compare molecule numbers and mass densities in a gas mixture. • Account correctly for gauge pressure, gas withdrawal, hydrostatic pressure and molecular counts.

One equation in molecular and molar language

A gas expands to fill its vessel and presses on the walls. Experiments on dilute gases relate its pressure P, container volume V, absolute temperature T and amount. The ideal-gas model summarises those measurements in a simple equation. The particle model will explain the equation in the next lesson; first learn exactly what it predicts and how to use it.

Ideal-gas equationLaTeX
ν is mole number, N is molecular count and n is number density. R ≈ 8.314 J mol⁻¹ K⁻¹, NA ≈ 6.022×10²³ mol⁻¹ and kB = R/NA ≈ 1.38×10⁻²³ J K⁻¹.

For one fixed gas sample, PV/T is constant and equals νR = NkB. It depends on the number of particles, not independently on volume. If the sample expands and its temperature also changes, PV changes with T while PV/T retains this value as long as the ideal-gas assumptions and fixed amount remain applicable. By contrast, the normalised ratio PV/(νT) equals the universal gas constant R for every ideal gas. Writing PV =KT is an equivalent abbreviation with K =νR =NkB, in J K⁻¹; K is fixed when the gas amount is fixed.

Always use absolute pressure, not gauge pressure, in a gas equation. A gauge reads pressure above the atmosphere, so Pabsolute = Pgauge + Patmosphere. Use kelvin: T = temperature in °C + 273.15, often rounded to +273 in textbook estimates. A litre is 10⁻³ m³ and a cubic centimetre is 10⁻⁶ m³. These conversions are part of the physics, not optional bookkeeping.

Gas law in terms of mass densityLaTeX
Divide the first equation by V and use ρ = msample/V. M is molar mass in kg mol⁻¹, so the result has pressure units Pa.

When do real gases resemble the ideal model?

An ideal gas obeys the ideal equation exactly by definition. Real molecules have finite size and interact, so no real gas is ideal at every density and temperature. The approximation is usually best at low density and temperatures well above liquefaction. Lowering pressure at a fixed temperature dilutes the gas. Raising temperature at fixed pressure also makes it less dense and reduces the relative importance of attractive binding. High temperature alone does not remove excluded-volume effects if the gas remains extremely dense.

Molar ratio PV/(νT): the ideal line is RPV/(νT) (J mol⁻¹ K⁻¹)P (atm)R0400800Higher T: shallower dipLower T: deeper dip
Real-gas departures from the ideal molar ratio— Schematic curves preserve the source’s qualitative low-pressure convergence and stronger attraction-related depression at lower temperatures. They are not digitised measurements or a quantitative equation of state.

The vertical coordinate is PV/(νT), in J mol⁻¹ K⁻¹, and the horizontal coordinate is pressure. An ideal gas lies on the horizontal line R. For a real gas, attractions can reduce pressure below the ideal prediction, producing a dip. At sufficiently high density, finite-size and repulsive effects can raise the ratio. The hotter curve in the illustrated regime has a shallower dip. All curves tend toward R in the dilute, low-pressure limit. Do not infer a universal temperature ordering from every imaginable real-gas curve; use the plotted trends and conditions.

Boyle’s and Charles’ laws are constrained versions of the equation

Fix ν and T. Then PV = νRT is constant, so P ∝ 1/V: Boyle’s law. Halving volume doubles pressure on an ideal isotherm. On a P–V graph these curves are decreasing hyperbolas, not straight lines, and the hotter isotherm lies above the colder one for the same amount. The slope is not constant; pressure changes more sharply at small volume.

Fix ν and P instead. Then V/T = νR/P is constant, giving Charles’ law: volume is proportional to absolute temperature. On a V-against-T graph the ideal prediction is a straight line through the kelvin origin. If the axes are interchanged, as in a T-against-V comparison, the line is still straight but its slope is P/(νR), so higher pressure gives a steeper line. A Celsius-temperature graph would not pass through its zero-temperature origin.

Ideal limiting curves used in gas-law comparisonsPVBoyle: fixed ν and THotter isotherms lie higherTVCharles comparison: T plotted against VHigher pressure gives the steeper ideal lineBoth panels use schematic scales and the ideal equations, not experimental data.
Interpreting the Boyle and Charles comparisons— The ideal families are drawn exactly from the stated proportionalities in dimensionless scales. The source compares solid experimental steam/CO₂ curves with dotted ideal lines; the lesson explains their departures rather than inventing experimental data.

The steam P–V comparison uses pressure units of 22 atm and volume units of 0.09 L. The CO₂ T–V comparison uses temperature units of 300 K and volume units of 0.13 L. The first shows theoretical inverse-volume curves versus real steam measurements; the second compares straight ideal T–V lines with real CO₂ curves. At lower temperatures or higher pressures real-gas departures can be substantial. At larger volumes and suitable high temperatures, the comparisons approach the ideal predictions. Axis normalisation changes the numerical scale, not the law being tested.

Fixed-amount comparison between two statesLaTeX
This relation does not apply unchanged when gas is withdrawn or added. In that case calculate the mole number at each state separately.
Why one mole occupies about 22.4 litres at textbook STP

Problem
Use T = 273.15 K, P = 1 atm = 1.01325×10⁵ Pa and R = 8.314 J mol⁻¹ K⁻¹ to find the ideal volume of one mole.

  1. 1.For ν = 1 mol, V = RT/P.
  2. 2.V = (8.314)(273.15)/(1.01325×10⁵) ≈ 2.2413×10⁻² m³.
  3. 3.This is 22.413 L, conventionally rounded to 22.4 L.
  4. 4.The same ideal molar volume applies to different gases because R is universal. It does not imply equal mass per mole. Here STP specifically means 0 °C and 1 atm, as in the problem; a different standard pressure gives a different molar volume.

Each gas contributes its own partial pressure

Suppose several ideal gases share one volume V at one equilibrium temperature T without reacting. Every species occupies the full container; it does not own a private portion of the volume. Species i would exert Pi = νiRT/V if it alone were present at that same V and T. This is its partial pressure. Summing the molecular contributions gives Dalton’s law: total pressure is the sum of partial pressures.

Ideal-gas mixture relationsLaTeX
The common T and V are essential. Pressure fraction equals mole fraction, not generally mass fraction.
Neon and oxygen with partial-pressure ratio 3:2

Problem
A non-reacting ideal mixture contains neon and oxygen at the same T and V. Their partial pressures satisfy PNe/PO₂ = 3/2. Find the ratios of particle count and mass density. Use molar masses 20.2 g mol⁻¹ for Ne and 32.0 g mol⁻¹ for O₂.

  1. 1.For each species PiV = νiRT. Since V and T are common, νNe/νO₂ = PNe/PO₂ = 3/2.
  2. 2.Particle counts are Ni = νiNA, hence NNe/NO₂ = 3/2 as well. Each neon particle is an atom, and each oxygen particle an O₂ molecule.
  3. 3.Mass density of species i is ρi = νiMi/V. Thus ρNe/ρO₂ = (3/2)(20.2/32.0) = 0.946875 ≈ 0.947.
  4. 4.There are more neon particles, yet their total mass per container volume is slightly smaller because a neon atom is lighter than an O₂ molecule.

Read the intercept: per mole and per sample are different

A fixed oxygen sample: PV/T tends to νRPV/T (J K⁻¹)PIdeal sample: 0.260 J K⁻¹T₁: shallower dipT₂: deeper dip; T₁ > T₂Pressure scale schematic; this ordinate is per sample, not per mole.
The oxygen-sample PV/T graph— This is a qualitative reconstruction of the exercise’s curve ordering. The dotted line is the ideal-sample prediction νR, not R. Both real curves approach this value at zero pressure.
Interpret every part of the oxygen graph

Problem
For a 1.00×10⁻³ kg oxygen sample, interpret the dotted line and temperature ordering, find the low-pressure PV/T intercept, and compare it with 1.00×10⁻³ kg hydrogen. Use MO₂ = 32.0×10⁻³ kg mol⁻¹, MH₂ = 2.02×10⁻³ kg mol⁻¹ and R = 8.31 J mol⁻¹ K⁻¹.

  1. 1.(a) The dotted horizontal line means ideal behaviour for this fixed sample: PV/T = νR independent of pressure.
  2. 2.(b) T₁ is the shallower-dip curve and T₂ the deeper-dip curve. In this plotted regime, the smaller departure corresponds to the higher temperature, so T₁ > T₂.
  3. 3.(c) Oxygen has ν = 0.00100/0.0320 = 0.03125 mol. The intercept is νR = 0.03125×8.31 = 0.2596875 J K⁻¹ ≈ 0.260 J K⁻¹.
  4. 4.(d) The same mass of hydrogen has far more moles, so its PV/T intercept is not the same; it would be about (0.00100/0.00202)×8.31 = 4.11 J K⁻¹.
  5. 5.To reproduce the oxygen intercept, hydrogen must have the same mole number 0.03125 mol, requiring mass (0.03125)(0.00202) = 6.3125×10⁻⁵ kg ≈ 6.3×10⁻⁵ kg.
  6. 6.The vertical unit is J K⁻¹ for this whole sample, whereas the earlier normalised molar graph has J mol⁻¹ K⁻¹ and ideal value R.

A cylinder losing gas: compare mole numbers, not a fixed-amount law

Oxygen withdrawn from a rigid cylinder

Problem
A 30 L oxygen cylinder initially has gauge pressure 15 atm at 27 °C. After gas is withdrawn, the gauge pressure is 11 atm and the temperature 17 °C. Estimate removed mass using atmospheric pressure 1 atm, R = 8.31 J mol⁻¹ K⁻¹, MO₂ = 0.032 kg mol⁻¹, 1 atm = 1.013×10⁵ Pa, and rounded temperatures 300 K and 290 K.

  1. 1.Convert gauge to absolute pressure: P₁ = 16 atm and P₂ = 12 atm. The rigid container volume is 30×10⁻³ = 0.030 m³.
  2. 2.Calculate ν₁ = P₁V/(RT₁) and ν₂ = P₂V/(RT₂). The gas amount changes, so P₁V/T₁ = P₂V/T₂ would be an incorrect assumption.
  3. 3.Removed mass is M(ν₁−ν₂) = (0.032)(0.030)(1.013×10⁵)/8.31 × (16/300 − 12/290).
  4. 4.This gives approximately 0.140 kg, or about 0.14 kg to the precision of the data.
  5. 5.Both cooling and loss of molecules contribute to the pressure drop. Using gauge pressures directly would give a different and physically incorrect result.

A rising bubble changes both pressure and temperature

An air bubble rising through a lake

Problem
An air bubble has volume 1.0 cm³ at a depth of 40 m, where the water temperature is 12 °C. Find its volume at the surface at 35 °C. Take water density 1000 kg m⁻³, g = 9.8 m s⁻² and atmospheric pressure 1.013×10⁵ Pa. Neglect gas dissolution, evaporation and surface-tension pressure, so the bubble’s molecule number stays fixed.

  1. 1.At depth the absolute pressure is P₁ = Patm + ρgh = 1.013×10⁵ + (1000)(9.8)(40) = 4.933×10⁵ Pa.
  2. 2.At the surface P₂ = 1.013×10⁵ Pa. Use T₁ ≈ 285 K and T₂ ≈ 308 K.
  3. 3.The fixed-amount relation gives V₂ = V₁(P₁/P₂)(T₂/T₁).
  4. 4.V₂ = (1.0×10⁻⁶)(4.933/1.013)(308/285) ≈ 5.26×10⁻⁶ m³ ≈ 5.3 cm³.
  5. 5.The bubble expands mainly because external pressure falls; warming adds a smaller expansion. Ignoring atmospheric pressure at the lake bottom or using Celsius ratios would give the wrong result.
Counting all air molecules in a room

Problem
Estimate the total number of gas molecules in a 25.0 m³ room at 27 °C and 1 atm. Use T ≈ 300 K, P = 1.013×10⁵ Pa and kB = 1.38×10⁻²³ J K⁻¹; treat the mixture as ideal.

  1. 1.Use PV = NkBT directly for the total particle count, including all gaseous species.
  2. 2.N = PV/(kBT) = (1.013×10⁵)(25.0)/[(1.38×10⁻²³)(300)] ≈ 6.12×10²⁶ molecules.
  3. 3.The estimate is about 6.1×10²⁶. The corresponding number density is N/V ≈ 2.45×10²⁵ m⁻³.
  4. 4.No separate oxygen and nitrogen percentages are needed for the total ideal-gas count; they would only be needed to divide this total among species.

Quick Quiz

Quiz

Quick check

The ideal-gas equation uses which pressure and temperature?

Explanation

Pressure must be referenced to vacuum, and temperature ratios must use an absolute scale.

Quick check

One fixed ideal-gas sample doubles its volume at fixed temperature. Its pressure becomes

Explanation

Boyle’s law gives P₂/P₁ = V₁/V₂ = 1/2.

Quick check

In an ideal mixture at common T and V, a component’s pressure fraction equals its

Explanation

Pi/P = νi/νtotal. Equal mole and mass fractions occur only in special cases.

Quick check

For 0.001 kg oxygen with M = 0.032 kg mol⁻¹ and R = 8.31 J mol⁻¹ K⁻¹, the ideal PV/T intercept is approximately

Explanation

νR = (0.001/0.032)×8.31 = 0.2597 J K⁻¹. This graph is per sample, not per mole.

Quick check

Why can a fixed-amount combined gas law not be used directly across oxygen withdrawal from a cylinder?

Explanation

Calculate ν = PV/(RT) separately before and after removal because the gas amount is not fixed.

Practice Problems

Practice Problems
  1. Derive PV = νRT, P = nkBT and P = ρRT/M from one another using the particle-count relations. Define every symbol and give the SI unit of each density.
  2. An ideal gas at absolute pressure 1.2×10⁵ Pa occupies 4.0 L at 300 K. Its amount stays fixed while pressure becomes 1.0×10⁵ Pa and temperature 350 K. Find its final volume.
  3. A mixture has 0.20 mol helium and 0.30 mol nitrogen in a 10.0 L vessel at 300 K. Find each partial pressure, total pressure and mole fractions using R = 8.314 J mol⁻¹ K⁻¹.
  4. Explain how a PV/(νT)-against-P plot differs from a PV/T-against-P plot for a specified gas mass. Find the ideal intercept of the latter for 0.0020 kg oxygen of molar mass 0.032 kg mol⁻¹, using R = 8.314 J mol⁻¹ K⁻¹.
  5. A rigid 20 L oxygen cylinder cools from 300 K to 290 K while its gauge pressure falls from 10 atm to 8 atm after gas is removed. Estimate removed mass using Patm = 1 atm = 1.013×10⁵ Pa, M = 0.032 kg mol⁻¹ and R = 8.314 J mol⁻¹ K⁻¹.
  6. A 2.0 cm³ air bubble at 20 m water depth and 280 K rises to the surface at 300 K. Estimate its final volume, assuming fixed molecule number, water density 1000 kg m⁻³, g = 9.8 m s⁻² and Patm = 1.013×10⁵ Pa. Neglect surface tension and dissolution.
  7. Find the total ideal-gas molecule count in a 10.0 m³ room at 300 K and 1.00×10⁵ Pa, using kB = 1.38×10⁻²³ J K⁻¹. Explain why the species percentages are unnecessary for this total.
  8. At fixed gas amount, sketch ideal Boyle P–V curves at two temperatures and ideal Charles T–V lines at two pressures. Label the ordering and explain why real gases can depart from those lines.

Key Takeaways

Key Takeaways

• PV = νRT = NkBT uses kelvin temperature and absolute pressure. • The constant PV/T depends on sample amount; PV/(νT) equals R in the ideal limit. • Boyle’s and Charles’ laws follow by holding the appropriate variables fixed. • Ideal mixture pressure is additive, and pressure fractions equal mole fractions. • Check fixed amount, gauge-pressure conversion and external pressure before comparing gas states.

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Molecules, Moles and Molecular Sizes

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Pressure, Temperature and Molecular Speeds