States of Matter · TOPIC LESSON
Translate between bulk gas measurements and the molecular motion behind pressure, speed distributions and effusion.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
A gas calculation starts with the boundary of the system. Is the vessel sealed, does a valve connect two compartments, or does gas escape? Is the container rigid, or can a piston move? Those details decide which of amount n, volume V, pressure P and temperature T are fixed. A familiar formula is useful only after those constraints are established.
The ideal-gas model treats particle volume as negligible compared with container volume and ignores intermolecular interaction energy between collisions. It is a limiting model, often useful for sufficiently dilute gases away from condensation. “Ideal” does not mean motionless or collision-free: impacts with a wall are essential to explaining pressure.
The equation PV=nRT describes an equilibrium state. It does not, by itself, specify how heat or work changes during a path between states. Two processes can share an initial and final ideal-gas state but exchange different amounts of heat and work. Conversely, during a rapidly changing nonuniform situation, assigning one temperature and pressure to the entire gas may not be justified.
Use absolute pressure, measured from vacuum. A gauge pressure measures the excess over local atmospheric pressure, so P_absolute=P_gauge+P_atmosphere. Pressure ratios formed directly from gauge readings usually fail. The subtraction of a vapour partial pressure, discussed later, is a different operation from converting gauge pressure to absolute pressure.
Temperature ratios require kelvin: T/K=t/°C+273.15. Doubling 20 °C to 40 °C is not doubling absolute temperature. Keep enough digits during conversion to avoid needless rounding error; if a problem explicitly uses the classroom approximation 273, apply it consistently rather than mixing conventions.
PV = nRT = Nk_B T
N counts molecules, n counts moles, and R=N_A k_B. P is absolute and T is in kelvin.
02 / UNDERSTAND & APPLY
With n and T fixed, PV is constant: reducing the available volume increases pressure. With n and P fixed, V/T is constant. With n and V fixed, P/T is constant. At fixed P and T, volume is proportional to the amount of gas. These are restrictions of the same state equation, not separate competing descriptions.
For a fixed amount between two equilibrium states, P₁V₁/T₁=P₂V₂/T₂. If a reaction or leak changes the number of gas molecules, include n explicitly. The more general ratio is P₂/P₁=(n₂/n₁)(T₂/T₁)(V₁/V₂). It immediately reveals whether two effects reinforce or oppose each other.
The gas constant is one physical constant with different numerical values in different unit systems. Use R=8.314 J mol⁻¹ K⁻¹ with pressure in pascals and volume in cubic metres. Since 1 kPa·L=1 J, the same numerical value works with kPa and litres. With bar and litres, use R=0.08314 L bar mol⁻¹ K⁻¹. With atmospheres and litres, R is approximately 0.082057 L atm mol⁻¹ K⁻¹.
Unit conversion is part of the physical solution. One litre is 10⁻³ m³; one bar is 10⁵ Pa; one atmosphere is 101325 Pa. In a speed formula using R in joules, molar mass must be in kg mol⁻¹, not g mol⁻¹. A hidden factor of one thousand in mass becomes a factor of about 31.6 after taking a square root.
“Standard conditions” are not a licence to insert 22.4 L mol⁻¹ everywhere. The ideal molar volume is RT/P and changes with the stated temperature and pressure. At 273.15 K and 1 atm it is about 22.414 L mol⁻¹; at 273.15 K and 1 bar it is about 22.711 L mol⁻¹. Read the convention supplied in the question.
ORIGINAL WORKED EXAMPLE
A sealed rigid ideal-gas vessel has gauge pressure 150 kPa when the atmosphere is 100 kPa and T=300 K. Find the gauge pressure at 360 K, assuming the atmosphere stays at 100 kPa.
Answer 200 kPa gauge.
03 / UNDERSTAND & APPLY
Write the amount as n=m/M, where m is sample mass and M its molar mass. Substituting into PV=nRT and dividing by volume gives P=ρRT/M. Thus ρ=PM/(RT) and M=ρRT/P. At fixed pressure and temperature, a heavier molar mass means a greater density, even though all ideal gases have the same number of molecules per unit volume.
An observed density can identify a pure gas only when conditions and candidate identities are sufficiently constrained. For a mixture, the same calculation returns the mole-fraction-weighted average molar mass, not necessarily the mass of any constituent molecule. A calculated value between two constituent molar masses can therefore be entirely reasonable.
For two ideal gases compared at the same P and T, ρ₁/ρ₂=M₁/M₂. If the pressures or temperatures differ, include their ratios. Relative vapour density is meaningful only with its reference gas and comparison conditions specified. With molecular hydrogen assigned molar mass 2 under a stated approximation, vapour density relative to hydrogen equals M/2; this is not a universal formula independent of the reference.
Check whether the weighed mass is the gas alone or includes the vessel. A tare subtraction is a separate mass balance. Also check for gas collected over a liquid, adsorption, leakage or association: the measured pressure and mass may not correspond to the assumed pure ideal gas. Such effects should be handled as given information, not silently ignored.
ORIGINAL WORKED EXAMPLE
An ideal gas has density 1.60 g L⁻¹ at 300 K and 100 kPa. Estimate its molar mass.
Answer 39.9 g mol⁻¹.
04 / UNDERSTAND & APPLY
For an ideal mixture, each component contributes pᵢ=nᵢRT/V, and P_total=Σpᵢ. Dividing the component expression by the total gives pᵢ=yᵢP_total, where yᵢ is the mole fraction among the gas-phase species. A partial pressure is the pressure that component would exert alone at the same T and V; it is not a statement that its molecules occupy only one corner of the vessel.
Mass fractions and mole fractions differ because particles have different molar masses. Convert masses to amounts before assigning partial pressures. The average mixture molar mass is ΣyᵢMᵢ, while the mass fraction of component i is yᵢMᵢ divided by that average. Both weighting schemes are correct for their own questions.
Gas collected over water generally contains water vapour. If the gas is saturated at the stated temperature, subtract the water saturation vapour pressure from the measured absolute total pressure to obtain the dry-gas partial pressure. Use that dry pressure with the collection volume and temperature to find the non-water gas amount.
The saturation subtraction assumes enough liquid water is present and gas–liquid equilibrium has been reached. An unsaturated sample has a smaller water partial pressure. A water-soluble or reactive collected gas introduces an additional material balance. If water levels differ across an apparatus, a hydrostatic pressure correction may also be needed before the vapour correction.
In a reactive gas mixture, first use reaction stoichiometry or equilibrium to determine each final gas-phase amount. Dalton’s law then converts those amounts to pressures. It cannot decide reaction extent on its own. A liquid or solid product does not contribute ideal-gas moles to the final gas-pressure sum.
ORIGINAL WORKED EXAMPLE
A saturated gas sample occupies 2.00 L at 300 K and total absolute pressure 100 kPa. The supplied water vapour pressure is 3.50 kPa. Find the dry gas amount, assuming negligible dissolution.
Answer 0.0774 mol dry gas.
05 / UNDERSTAND & APPLY
When two rigid vessels containing nonreacting ideal gases are connected and settle to a common temperature, total amount is conserved. Compute each initial amount separately, add them, and use the final total accessible volume. If both vessels start at the same temperature and finish at that temperature, P_f=(P₁V₁+P₂V₂)/(V₁+V₂). The familiar volume-weighted average is therefore a consequence of stated conditions.
If initial temperatures differ, use n_total=P₁V₁/(RT₁)+P₂V₂/(RT₂) and the actual final temperature. An insulated mixing problem may require an energy balance to find that temperature; the ideal gas law alone is insufficient. Include connecting-tube volume if it is not negligible. A valve opening does not justify averaging pressures without these checks.
For gas reactions at a common temperature and pressure, ideal gas volumes are proportional to amounts, so balanced stoichiometric mole ratios can become volume ratios. They do not apply to volumes measured at different temperatures or pressures without conversion. For a sealed fixed-volume reactor returned to its initial temperature, P_final/P_initial equals the ratio of final to initial gas moles.
Condensation can change this mole count substantially. If a combustion problem says water is removed or condensed completely with its residual vapour pressure neglected, omit it from dry gas moles. If it instead retains water as vapour at high temperature, include it. Explicitly state the interpretation rather than treating all balanced equation species as gases.
ORIGINAL WORKED EXAMPLE
Two nonreacting ideal gases at 300 K occupy vessels of 2.00 L at 150 kPa and 3.00 L at 50.0 kPa. They are connected with negligible tube volume and return to 300 K. Find final pressure.
Answer 90.0 kPa.
ORIGINAL WORKED EXAMPLE
Initially a rigid vessel contains 2.00 mol H₂ and 1.50 mol O₂. They react completely as far as stoichiometry allows. The sample returns to its initial temperature and all product water is removed, with residual vapour neglected. Find P_final/P_initial.
Answer P_final/P_initial=1/7.
06 / UNDERSTAND & APPLY
Consider one molecule of mass m in a cube of side L. Its velocity component normal to one wall is vₓ. An elastic reflection reverses that component, so the magnitude of momentum transferred to the wall in one collision is 2m|vₓ|. Between successive collisions with the same wall, it travels a round trip of 2L in the x direction, taking time 2L/|vₓ|.
The average force contribution is momentum per collision divided by the interval: m vₓ²/L. Summing over N molecules and dividing by the wall area L² gives P=(m/V)Σvₓ². For an equilibrium isotropic distribution, the mean squared components are equal, so ⟨vₓ²⟩=⟨v²⟩/3. This symmetry is the source of the factor one third.
The result is PV=(1/3)Nm⟨v²⟩=(2/3)E_trans,total. Comparing with Nk_BT gives the mean translational kinetic energy per molecule as (3/2)k_BT. Temperature fixes this mean energy, not the speed of every molecule. At the same temperature, light and heavy gases have equal mean translational energies but different characteristic speeds.
The derivation concerns translation of molecular centres of mass. A diatomic molecule can also rotate and, at sufficiently high temperature, vibrate; those internal modes affect heat capacity without changing the ideal relation between translational motion and pressure. Do not equate the total internal energy of every ideal gas to (3/2)nRT. That expression for total internal energy applies to the simple monatomic model when other excitations are negligible.
The vector average velocity is zero for a stationary isotropic gas, because motions in opposite directions cancel. The average speed is positive, and the mean squared speed is positive. A container may also move as a whole; bulk flow velocity is distinct from random thermal motion relative to that bulk motion.
P = ⅓ρ⟨v²⟩; E_trans/N = ³⁄₂ k_B T
ρ is gas mass density. The energy statement is translational, not a universal total-energy formula.
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07 / UNDERSTAND & APPLY
The most probable speed is the location of the maximum of the speed probability density. The mean speed is the average of individual speed magnitudes. The root-mean-square speed is the square root of the mean of squared speeds; it connects directly to kinetic energy. For a Maxwell distribution they obey v_mp < v_mean < v_rms.
With molar mass M in kg mol⁻¹, the formulas are v_mp=√(2RT/M), v_mean=√(8RT/(πM)) and v_rms=√(3RT/M). Their coefficients differ because a skewed distribution cannot be summarized by one universal “average”. In particular, squaring the mean speed does not give the mean squared speed.
At fixed species, every characteristic speed scales as √T. To double such a speed, quadruple absolute temperature. At fixed temperature, each speed scales as 1/√M. Equality of rms speeds for two gases requires T₁/M₁=T₂/M₂; equality of temperature instead implies equality of mean translational energies.
For a gas mixture, each component at thermal equilibrium has the same temperature but its own mass-dependent speed distribution. A mole-fraction-weighted average of component mean speeds describes one chosen particle-sampling statistic; inserting the mixture average molar mass into a pure-species speed formula does not generally produce that same statistic.
ORIGINAL WORKED EXAMPLE
Estimate the three characteristic speeds of N₂ at 300 K using M=0.0280 kg mol⁻¹ and R=8.314 J mol⁻¹ K⁻¹.
Answer Approximately 422, 476 and 517 m s⁻¹, respectively.
08 / UNDERSTAND & APPLY
A normalized Maxwell speed density f(v) has area one over all nonnegative speeds. A narrow interval from v to v+dv contains a fraction approximately f(v)dv, not f(v) alone. The vertical axis therefore has reciprocal-speed units. The peak height is not the total fraction of molecules at one exact speed, which is zero in the continuous model.
Raising temperature for the same gas shifts the most probable speed rightward and broadens the distribution. Because the total area remains one, the peak becomes lower. More particles exceed any specified positive speed threshold at higher temperature. A taller low-temperature peak does not imply a greater total number of molecules if both curves are normalized.
The three-dimensional distribution includes many more possible velocity directions at larger speed; this produces the v² factor in the speed density. The exponential factor suppresses very large speeds. Their competition gives a finite nonzero peak rather than a maximum at zero speed. A one-component velocity distribution, which includes positive and negative vₓ, has a different shape and interpretation.
For a simple dilute hard-sphere model, mean free path is λ=1/(√2 πd² n_number), where d is collision diameter and n_number=N/V. At fixed T, increasing pressure increases number density and shortens λ. Collision frequency scales roughly as characteristic speed divided by mean free path. These are model results with an assumed diameter, not universal constants independent of gas identity.
Collisions exchange energy among molecules, sustaining a distribution rather than forcing every molecule to the same speed. The macroscopic equilibrium temperature can be steady even though individual molecular speeds change from collision to collision. A molecular-speed diagram is a probability summary, not a trajectory simulation.
f(v)=4π(M/(2πRT))^(3/2) v² exp[−Mv²/(2RT)]
Maxwell speed probability density for equilibrium classical ideal gas; M in kg mol⁻¹ and v in m s⁻¹ with SI R.
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09 / UNDERSTAND & APPLY
Effusion is escape through an opening sufficiently small relative to the mean free path that molecule–molecule collisions in the opening are negligible. At equal temperature and upstream pressure, the molecular flux through the same opening is proportional to the mean molecular speed, hence inversely proportional to the square root of molar mass.
For two gases under matching conditions, r₁/r₂=√(M₂/M₁), where the rates compare numbers of molecules or amounts per unit time. Mass flow rates are not the same statistic: multiplying molecular flux by molecular mass changes the dependence. For equal collected amounts at effectively constant conditions, the corresponding time ratio is inverse to the rate ratio.
Diffusion is intermixing driven by concentration gradients through many collisions. Textbook Graham comparisons can be useful approximations in appropriately matched setups, but bulk diffusion in a background gas is not generally determined by molar masses alone. Collision sizes, mixture composition and transport conditions matter. Do not apply the pinhole effusion law to an arbitrary wide nozzle with pressure-driven flow.
An effusing mixture becomes compositionally fractionated because the lighter component leaves faster at a given partial pressure. Its escaping composition need not equal the vessel mole fractions, and those fractions evolve during appreciable depletion. A ratio calculated from the initial partial pressures describes the initial effusate, not every later sample.
ORIGINAL WORKED EXAMPLE
Equal small amounts of an unknown gas and methane take 60.0 s and 40.0 s respectively to effuse under matching effectively constant T and P through the same molecular-flow aperture. Take methane molar mass as 16.0 g mol⁻¹.
Answer 36.0 g mol⁻¹.
10 / UNDERSTAND & APPLY
A reliable final audit asks four things: which amount is conserved, whether pressure is absolute, whether all temperatures are in kelvin, and whether the units of R match the quantities inserted. Then compare the direction of change with the physical constraint. A smaller volume at unchanged n and T cannot produce a smaller ideal-gas pressure.
For kinetic questions, identify the statistic: mean vector velocity, mean speed, rms speed, translational energy or molecular flux. For mixtures and reactions, keep species amounts visible until the last step. The following original problems use the ideal-gas model and explicitly stated approximations; they are not attributed to a past examination.
ORIGINAL PRACTICE · 1
It becomes four times as large.
PV is constant, so P₂/P₁=V₁/V₂=8/2=4.
ORIGINAL PRACTICE · 2
180 kPa.
P₂=P₁T₂/T₁=120×450/300; n and V are fixed.
ORIGINAL PRACTICE · 3
12.47 L.
V=nRT/P=0.500×8.314×300/100.
ORIGINAL PRACTICE · 4
100 kPa for each gas.
Both amounts are 1.00 mol, so both mole fractions are one half despite the different masses.
ORIGINAL PRACTICE · 5
22.0 g mol⁻¹ and 1/22≈0.0455.
Average M=0.25×4+0.75×28=22. Helium mass contribution per mole of mixture is 1 g, so its mass fraction is 1/22.
ORIGINAL PRACTICE · 6
1800 K.
The speed ratio is √(T₂/T₁), hence T₂=9×200. This uses the stated ideal molecular model.
ORIGINAL PRACTICE · 7
v_He/v_O2=√8≈2.83.
Characteristic speeds scale as inverse square root of molar mass; mean translational energies remain equal.
ORIGINAL PRACTICE · 8
4.99 kJ for the one-mole sample (molar translational energy 4.99 kJ mol⁻¹).
The molar translational energy is 1.5RT=4988.4 J mol⁻¹. This does not automatically equal total internal energy for a polyatomic gas.
ORIGINAL PRACTICE · 9
It halves.
Number density doubles by P=n_number k_BT, while λ is inversely proportional to number density at fixed collision diameter.
ORIGINAL PRACTICE · 10
√5:1≈2.24:1.
Flux is proportional to partial pressure divided by √M at common temperature. Equal partial pressures leave the inverse-square-root mass factor.
ORIGINAL PRACTICE · 11
1.25.
Remaining A₂=0.75 mol, formed A=0.50 mol, total=1.25 mol. Pressure follows total gas amount at fixed T and V.
ORIGINAL PRACTICE · 12
No.
Opposite velocity vectors cancel in the vector average, but squared speeds add positively. Kinetic energy depends on ⟨v²⟩, not the square of the mean velocity vector.
KEEP THE BIG IDEAS
READ WITH CONTEXT
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.
Educational reference inspected in HTML. Teaching narrative, numerical choices and solved problems are original.
Educational reference inspected in HTML. Teaching narrative, numerical choices and solved problems are original.
Educational reference inspected in HTML. Teaching narrative, numerical choices and solved problems are original.
Educational reference inspected in HTML. Teaching narrative, numerical choices and solved problems are original.
Author-attributed educational HTML inspected; independently derived pressure and speed results. The displayed positive-exponent typo in an intermediate source equation is not adopted.
University course HTML inspected for collision cross section and dilute hard-sphere mean free path. Original solved examples.
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