Behaviour of Perfect Gases and Kinetic Theory of Gases | ISC Class 11 Physics Notes

This note covers the perfect gas equation, gas laws, Avogadro’s number, molecular assumptions, gas pressure, the kinetic interpretation of temperature, root mean square speed, compression work, degrees of freedom, equipartition of energy, heat capacities and mean free path.

What is a perfect gas and how is its state described?

A perfect gas, also called an ideal gas, is a theoretical gas that obeys the ideal gas equation exactly at all pressures and temperatures. No real gas is truly ideal. Real gases satisfy the equation approximately, more so at low pressures and high temperatures.

Pressure, written p, is force perpendicular to a surface per unit area. Volume, V, is the space occupied by the gas. Absolute temperature, T, is temperature on the kelvin scale. These bulk quantities describe a gas in equilibrium, when its macroscopic properties remain steady.

Let n be the amount of gas in moles and R the universal gas constant. A mole is an amount of substance containing Avogadro’s number of specified particles. The equation of state, meaning the relation between equilibrium state variables, is:

pV = nRT, with R = 8.314 J mol⁻¹ K⁻¹. Here J denotes joule, the unit of energy; mol denotes mole; and K denotes kelvin. Temperature ratios in gas equations require absolute temperatures.

Which units should be used?

The International System of Units, abbreviated SI, keeps numerical substitutions consistent. Use the same system for pressure, volume, temperature and the gas constant.

How do the gas laws specify their conditions?

Boyle’s law states that pressure varies inversely with volume for a fixed mass of gas at constant temperature: pV = constant. Charles’ law states that volume is proportional to absolute temperature for a fixed amount of gas at constant pressure: V/T = constant.

For a fixed amount of ideal gas at constant volume, p/T is constant. More generally, p₁V₁/T₁ = p₂V₂/T₂, where subscripts 1 and 2 identify the initial and final equilibrium states. This comparison does not apply unchanged if gas enters or leaves the sample.

Note: The ideal gas equation is an experimental result. It cannot be derived from the mechanical assumptions of kinetic theory alone. Comparing the experimental equation with the kinetic pressure equation supplies the interpretation of temperature.

How do moles, molecules and density connect the two descriptions?

Avogadro’s number, Nₐ, is the number of particles in one mole, approximately 6.02 × 10²³. Used as a conversion constant, Nₐ has unit mol⁻¹. Let N denote the total number of molecules in the sample. N here is a particle count, distinct from the unit symbol for newton.

N = nNₐ. Thus, a mole count and a molecule count describe the same sample on different scales. Avogadro’s hypothesis states that equal volumes of gases at the same temperature and pressure contain the same number of molecules, within the ideal gas description.

Let M be the molar mass, meaning mass per mole, in kg mol⁻¹, and m the mass of one molecule, in kilograms, kg. Then m = M/Nₐ. For a sample of total mass Mₛ, the amount is n = Mₛ/M.

What is number density?

Number density, η = N/V, is the number of molecules per unit volume, measured in m⁻³. Mass density, ρ = Mₛ/V, is mass per unit volume, measured in kg m⁻³. For identical molecules, ρ = ηm.

Boltzmann’s constant, kᵦ = R/Nₐ, connects molecular energy and temperature. Its approximate value is 1.38 × 10⁻²³ J K⁻¹. The equivalent equations pV = NkᵦT, p = ηkᵦT and p = ρRT/M connect particle counts or density with pressure.

Worked example 1. Estimate the volume of one water molecule. Take liquid water density as 1000 kg m⁻³, molar mass as 0.018 kg mol⁻¹ and Nₐ ≈ 6 × 10²³ mol⁻¹. Approximate molecular density by bulk liquid density.

Formula: m = M/Nₐ; Vₘ = m/ρ, where Vₘ is the estimated volume of one molecule.

Substitute: m = 0.018/(6 × 10²³) = 3 × 10⁻²⁶ kg; Vₘ = (3 × 10⁻²⁶)/1000.

Answer: Vₘ ≈ 3 × 10⁻²⁹ m³. This is a rough molecular volume, not the much larger average volume available to each molecule in a gas.

At standard temperature and pressure, abbreviated STP here as 273 K and one atmosphere, one mole of ideal gas occupies approximately 22.4 litres. The litre, L, is a volume unit with 1 L = 10⁻³ m³. The same molar volume does not imply equal molar masses.

Worked example 2. Find the molar volume of an ideal gas at 273 K and 1.01 × 10⁵ Pa. Take n = 1 mol and R = 8.31 J mol⁻¹ K⁻¹.

Formula: V = nRT/p.

Substitute: V = (1 × 8.31 × 273)/(1.01 × 10⁵).

Answer: V ≈ 2.25 × 10⁻² m³ with these rounded constants. The conventional molar volume at STP is approximately 22.4 L; the small difference reflects the rounded pressure used here.

Worked example 3. Estimate the number of air molecules in a room of volume 25.0 m³ at 300 K and pressure 1.01 × 10⁵ Pa. Treat air as ideal and take kᵦ = 1.38 × 10⁻²³ J K⁻¹.

Formula: N = pV/(kᵦT).

Substitute: N = (1.01 × 10⁵ × 25.0)/(1.38 × 10⁻²³ × 300).

Answer: Approximately 6.10 × 10²⁶ molecules occupy the 25.0 m³ room. The count includes all molecular constituents represented by the ideal gas model.

What assumptions make the kinetic theory of an ideal gas possible?

Kinetic theory explains macroscopic gas behaviour through the motion of a very large number of molecules. A gas that looks stationary contains molecules in incessant random motion. Equilibrium is dynamic: molecules continue to collide and change velocities while average properties remain constant.

1. Large molecular population: the gas contains a very large number of particles, so pressure and energy can be described by statistical averages.
2. Negligible molecular size: molecular dimensions are negligible compared with the distances between molecules in the ideal gas approximation.
3. Negligible interaction between collisions: intermolecular forces are neglected while molecules are apart. They move freely in straight lines according to Newton’s first law.
4. Elastic collisions: molecules collide with one another and with container walls without loss of total kinetic energy. Momentum is conserved for the complete colliding system.
5. Random directions: an equilibrium gas has no preferred direction of molecular motion. This property is called isotropy.
6. Brief collisions: the time spent in a collision is negligible compared with the time between collisions in the dilute gas treatment.

Why does a real gas depart from the model?

Real molecules have finite size and interact. At low pressures or high temperatures they are far apart and molecular interactions are negligible. At high temperatures much above liquefaction or solidification, and low pressures, the simple gas equation is a useful approximation.

An elastic collision can change an individual molecule’s speed. Conservation applies to the total kinetic energy of the collision, not separately to each participant. A stationary massive wall reverses the velocity component normal to the wall while leaving the parallel components unchanged.

The idealisation does not mean that collisions disappear. Collisions with walls provide the mechanism for pressure, and collisions between molecules continually redistribute molecular velocities. The pressure calculation relies on a steady distribution rather than identical speeds for all molecules.

How do molecular collisions produce the pressure equation?

Pressure arises from momentum transfer to a surface. Momentum is mass multiplied by velocity. A molecular impact changes momentum, and Newton’s second law relates the rate of this change to force. Newton’s third law gives the corresponding force exerted on the wall.

Consider a cube with side length L and wall area A = L², so its volume is V = L³. Let cₓ, cᵧ and c𝓏 be the velocity components along three mutually perpendicular axes x, y and z. The molecular speed c satisfies c² = cₓ² + cᵧ² + c𝓏².

A cube has x, y and z axes. An incoming molecular path and an outgoing path meet the right wall. The velocity labels show that the x component reverses while the other two components remain unchanged.

See Fig. 12.4 in your NCERT textbook

Derivation: Pressure from molecular impacts

Let Δt be a short time interval, where Δ indicates a small change. Let η𝗀 be the number density of a group with the same magnitude of normal velocity component. Angle brackets, as in ⟨c²⟩, denote an average over molecules.

1. Choose the positive x direction towards the wall. A molecule approaching with component cₓ rebounds with component −cₓ. Its momentum change is −2mcₓ, so the wall receives momentum 2mcₓ.
2. Only molecules within a layer of thickness cₓΔt can reach the wall during Δt. Its volume is AcₓΔt. On average, half of the relevant group moves towards the wall, giving η𝗀AcₓΔt/2 impacts.
3. Multiply the number of impacts by momentum transferred per impact. The momentum received by the wall is η𝗀Amcₓ²Δt. Dividing by Δt gives force; dividing by A gives the group pressure η𝗀mcₓ².
4. Adding all velocity groups gives total pressure p = ηm⟨cₓ²⟩. Isotropy gives ⟨cₓ²⟩ = ⟨cᵧ²⟩ = ⟨c𝓏²⟩ = ⟨c²⟩/3.
5. Use ρ = ηm to replace number density times molecular mass by mass density. The result relates measurable pressure to the mean square molecular speed.

p = ⅓ρ⟨c²⟩, or equivalently pV = ⅓Nm⟨c²⟩.

The mean square speed is the average of the squared molecular speeds. It is not generally the square of the average speed. The factor one third comes from the three equivalent spatial directions, not from a loss of energy during impact.

The final expression does not depend on wall area or the chosen time interval. A cubic vessel simplifies the argument, but the pressure relation is not restricted to cubes. Pressure also exists within the gas, not merely at its enclosing walls.

How does kinetic energy give temperature a molecular meaning?

Translational kinetic energy is energy associated with a molecule’s motion from place to place. Let Eₜᵣ be the total translational kinetic energy of the gas, measured in joules. Then Eₜᵣ = ½Nm⟨c²⟩, and the pressure result becomes pV = ⅔Eₜᵣ.

Derivation: The kinetic interpretation of temperature

1. Kinetic theory supplies pV = ⅓Nm⟨c²⟩ without introducing temperature into its mechanical assumptions.
2. The experimentally established ideal gas equation supplies pV = NkᵦT. Combining the two descriptions assumes that they apply to the same ideal gas state.
3. Equate their right sides and cancel the molecule count N. This gives ⅓m⟨c²⟩ = kᵦT.
4. Multiply by three halves to identify ½m⟨c²⟩, the average translational kinetic energy of one molecule.

⟨εₜᵣ⟩ = ½m⟨c²⟩ = ³⁄₂kᵦT, where ⟨εₜᵣ⟩ denotes average translational kinetic energy per molecule.

For one mole, multiply by Nₐ to obtain Eₜᵣ = ³⁄₂RT. For n moles, Eₜᵣ = ³⁄₂nRT. Absolute temperature therefore measures average translational kinetic energy per molecule, rather than the total energy of an unspecified amount of gas.

What is equal when different gases have the same temperature?

At a common temperature, molecules of different ideal gases have equal average translational kinetic energies. Their masses and speeds can differ. A heavier molecule has a smaller characteristic speed, because its greater mass compensates for that speed in the kinetic energy expression.

Internal energy, U, is the total microscopic energy of the gas. Translation accounts for all of U in the monatomic ideal gas model. Molecules with active rotational or vibrational motion have additional contributions, so Eₜᵣ must not automatically be substituted for U.

Worked example 4. Estimate the average thermal energy of a helium atom at 300 K. Treat helium as a monatomic ideal gas and use kᵦ = 1.38 × 10⁻²³ J K⁻¹.

Formula: ⟨εₜᵣ⟩ = ³⁄₂kᵦT.

Substitute: ⟨εₜᵣ⟩ = 1.5 × 1.38 × 10⁻²³ × 300.

Answer: The average energy is 6.21 × 10⁻²¹ J per atom. It is an average over many atoms, not a claim that every atom has this energy at an instant.

What determines the root mean square speed of gas molecules?

The root mean square speed, cᵣₘₛ, is obtained by squaring molecular speeds, taking their mean, and then taking the square root. Thus cᵣₘₛ = √⟨c²⟩ = √(3p/ρ) = √(3kᵦT/m) = √(3RT/M). Its SI unit is metre per second, m s⁻¹, where s denotes second.

For a given gas, cᵣₘₛ is proportional to √T. For gases at the same temperature, it is inversely proportional to √M. Molar mass must be expressed in kg mol⁻¹ when R is used in J mol⁻¹ K⁻¹.

Does increasing pressure necessarily increase molecular speed?

At constant temperature, changing the pressure of a given ideal gas does not change its rms speed. Its density changes in the same proportion as pressure, leaving p/ρ unchanged. If density is kept constant instead, cᵣₘₛ varies as √p because temperature then changes.

The phrase “pressure increases” alone therefore does not specify the speed change. State what remains constant before using a proportionality. Random molecular motion also differs from the bulk motion of a gas: a stationary sample can have a large rms speed.

Worked example 5. Compare air and hydrogen at 273 K. Treat each as ideal, with air represented by effective molar mass 29.0 × 10⁻³ kg mol⁻¹ and hydrogen by 2.02 × 10⁻³ kg mol⁻¹. Use R = 8.31 J mol⁻¹ K⁻¹.

Formula: c(air) = √(3RT/Mₐ); c(hydrogen) = √(3RT/Mₕ), where the names in parentheses identify the gas and subscripts a and h identify its molar mass.

Substitute: c(air) = √(3 × 8.31 × 273/0.0290); c(hydrogen) = √(3 × 8.31 × 273/0.00202).

Answer: c(air) ≈ 484 m s⁻¹ and c(hydrogen) ≈ 1.84 × 10³ m s⁻¹. Hydrogen’s rms speed is approximately 3.79 times the effective air value at this temperature.

For comparison, the measured speed of sound in air at STP is about 331 m s⁻¹. The effective air rms speed is of the same order, but the two speeds describe different things: random molecular motion and the propagation of a sound disturbance.

Worked example 6. Estimate the rms speed of nitrogen at 300 K. Take the mass of one nitrogen molecule as 4.65 × 10⁻²⁶ kg and kᵦ = 1.38 × 10⁻²³ J K⁻¹.

Formula: cᵣₘₛ = √(3kᵦT/m).

Substitute: cᵣₘₛ = √[(3 × 1.38 × 10⁻²³ × 300)/(4.65 × 10⁻²⁶)].

Answer: cᵣₘₛ ≈ 517 m s⁻¹ with these rounded inputs. The result represents the square root of the mean square speed, not the speed of every nitrogen molecule.

Worked example 7. At what temperature does argon have the same rms speed as helium at 253 K? Use molar masses Mₐᵣ = 39.9 g mol⁻¹ and Mₕₑ = 4.0 g mol⁻¹; g denotes gram. Treat both gases as ideal.

Formula: Tₐᵣ/Mₐᵣ = Tₕₑ/Mₕₑ, where the subscripts identify the gases.

Substitute: Tₐᵣ = 253 × 39.9/4.0.

Answer: Tₐᵣ ≈ 2500 K, to two significant figures. Consistent molar mass units cancel in this ratio; the heavier argon needs a higher temperature to have the same rms speed.

How is work done when a gas is compressed?

A piston is a movable boundary that changes the gas volume. Work transfers energy when a force displaces that boundary. Let Wᵦᵧ denote work done by the gas and Wₒₙ work done on it; both are measured in joules.

For constant opposing pressure p, Wᵦᵧ = p(V₂ − V₁), where V₁ and V₂ are the initial and final volumes. Expansion gives positive work by the gas. Compression gives negative work by the gas and positive work on it: Wₒₙ = −Wᵦᵧ.

What if pressure changes during compression?

In a quasi-static process, changes are slow enough for the gas to pass through equilibrium states. The pressure difference between gas and surroundings is infinitesimal in the ideal limit. Adding the small work contributions gives Wᵦᵧ = ∫p dV, with integration from V₁ to V₂.

Here dV is an infinitesimal volume change and ∫ means summation by integration. For an isothermal process, meaning constant temperature, substitution of p = nRT/V gives Wᵦᵧ = nRT ln(V₂/V₁), where ln denotes the natural logarithm.

For isothermal compression, Wₒₙ = nRT ln(V₁/V₂). This is positive because V₁ exceeds V₂. The formula assumes a fixed amount of ideal gas undergoing quasi-static compression; it must not be applied to every possible compression without its conditions.

What is the molecular explanation?

Molecules striking an inward-moving massive piston can rebound faster, gaining kinetic energy. If energy is retained, compression raises temperature. During isothermal compression, energy must be transferred out as heat so that the average translational kinetic energy remains unchanged.

Heat is energy transferred because of a temperature difference. A compressed gas expanding against a piston transfers energy through work. The temperature response depends on heat exchange as well as work, so compression alone does not specify a unique temperature change.

What are degrees of freedom and the law of equipartition?

A degree of freedom is an independent way in which molecular motion can occur. A molecule moving along a line needs one position coordinate, in a plane two, and in space three. Motion of the whole molecule through space is called translation.

Each molecule free to move in space has three translational degrees of freedom. A monatomic gas such as argon is treated as having these three energy contributions. A diatomic molecule contains two atoms and can also rotate about its centre of mass.

How do rotation and vibration add energy?

A rigid diatomic molecule is modelled as two atoms at fixed separation, without vibration. It has two effective rotational degrees of freedom about independent axes perpendicular to the line joining the atoms. Together with translation, this gives five contributions to its energy.

Two joined circles represent the atoms. Two sketches show rotation about dotted axes labelled (1) and (2). Curved arrows indicate rotation, and each axis passes through the middle of the molecular pair.

See Fig. 12.6 in your NCERT textbook

Vibration is oscillation of the atoms relative to each other. A vibrational mode has both kinetic energy and potential energy, meaning energy associated with motion and with displacement from equilibrium. Each appears as a quadratic term, a term containing the square of a motion variable.

Definition: The law of equipartition of energy states that, in thermal equilibrium at absolute temperature T, each quadratic energy term has average energy ½kᵦT. Each translational or rotational degree contributes ½kᵦT; each active vibrational mode contributes kᵦT.

Thermal equilibrium means there is no net heat exchange between bodies in thermal contact. Equipartition concerns averages in this condition. It does not require every molecule to have equal instantaneous energy or every type of motion to have the same speed.

Let f count the active quadratic energy terms. Then the average total molecular energy is ½fkᵦT. One vibrational mode adds two to this energy count because its kinetic and potential terms each contribute ½kᵦT.

The rigid rotator approximation is valid for oxygen at moderate temperatures, but it is not always valid. When vibrational energy contributes, ignoring it underestimates the predicted heat capacity. The number of active energy contributions must therefore accompany a heat capacity prediction.

How does equipartition determine the heat capacities of gases?

Specific heat capacity is heat supplied per unit mass per unit temperature rise under specified conditions, measured in J kg⁻¹ K⁻¹. Molar heat capacity is the corresponding quantity per mole, measured in J mol⁻¹ K⁻¹. These are distinct quantities.

Let Cᵥ be molar heat capacity at constant volume and Cₚ that at constant pressure. Write cᵥ and cₚ for the corresponding capacities per unit mass. Then cᵥ = Cᵥ/M and cₚ = Cₚ/M, using molar mass M in kg mol⁻¹.

How are the general expressions obtained?

For a fixed number f of active quadratic energy terms, U = ½fnRT. At constant volume no boundary work occurs. Thus heat supplied increases internal energy, giving Cᵥ = ½fR.

At constant pressure the gas also expands and does work. For an ideal gas, Cₚ − Cᵥ = R, so Cₚ = ½(f + 2)R. The heat capacity ratio is γ = Cₚ/Cᵥ = (f + 2)/f, where γ is the dimensionless Greek letter gamma.

These heat capacities follow from the temperature dependence of energy, not from energy itself. For example, one mole of monatomic ideal gas has U = ³⁄₂RT, but its Cᵥ is ³⁄₂R. The extra temperature factor must not remain in a heat capacity.

A polyatomic molecule contains more than two atoms. For a non-linear polyatomic model, three translational and three rotational contributions give six quadratic terms before vibration. If q is the number of active vibrational modes, Cᵥ = (3 + q)R and Cₚ = (4 + q)R.

A linear molecule has its atoms along one line; a non-linear molecule does not. Linear polyatomic molecules have two effective rotational contributions, so their rigid-model count is five. The molecular arrangement, as well as the number of atoms, matters when counting rotational freedom.

The predicted capacities agree well with measured values for several gases. Discrepancies occur for several others. Usually, their experimental capacities are greater than predictions that ignore vibration, suggesting that including vibrational modes can improve the agreement.

Worked example 8. A fixed-volume cylinder holds helium in 44.8 L at STP. Take the molar volume as 22.4 L mol⁻¹ and R = 8.31 J mol⁻¹ K⁻¹. Find the heat needed for a 15.0 K temperature rise, treating helium as monatomic and ideal.

Formula: n = V/(molar volume); Q = nCᵥΔT, where Q is heat supplied and ΔT is temperature rise. For helium, Cᵥ = 3R/2.

Substitute: n = 44.8/22.4 = 2 mol; Q = 2 × (3/2) × 8.31 × 15.0.

Answer: Q = 374 J to three significant figures. Cᵥ is used because the cylinder volume is fixed; using Cₚ would include expansion work that is absent here.

What is mean free path and why does diffusion take time?

The mean free path, λ, is the average distance travelled by a molecule between successive collisions. It is a length, measured in metres. Individual paths have different lengths, so λ describes an average rather than a fixed distance between all collisions.

Diffusion is the spreading and mixing of molecules through random motion. Large molecular speeds do not mean a molecule crosses a room in one straight journey. Repeated collisions deflect it, and spreading through a gas therefore takes time.

How do molecular size and number density enter?

Model molecules as spheres of diameter d. A collision occurs when their centres come within distance d. A molecule moving a distance ⟨c⟩Δt sweeps a collision volume πd²⟨c⟩Δt, where ⟨c⟩ is average speed and π is the circle constant.

A sloping dashed cylinder surrounds a straight molecular path. Blue circular molecules lie along and beside it. The labels show a travel length given by average speed times the time interval, and the collision distance d.

See Fig. 12.7 in your NCERT textbook

If other molecules were stationary, this argument would give λ = 1/(πηd²). In reality they also move. Accounting for their relative motion gives the more accurate result λ = 1/(√2πηd²).

Thus, for a given molecular diameter, mean free path decreases when number density increases. For a fixed number density it decreases with the square of molecular diameter. Using η = p/(kᵦT) gives λ = kᵦT/(√2πd²p) in the ideal gas approximation.

At constant temperature and molecular diameter, λ varies inversely with pressure. At constant pressure and diameter, it increases with absolute temperature. These conditions distinguish the effect of temperature from changes in density caused by compression or expansion.

Let τ denote the average time between successive collisions. Then λ = ⟨c⟩τ. The corresponding collision rate is 1/τ, measured in s⁻¹. Average speed and rms speed are different averages and should not be silently interchanged.

Mean free path also differs from mean intermolecular separation, the typical distance between molecules at an instant. In a highly evacuated tube, number density is small and the mean free path can become as large as the tube’s length.

Glossary

• Perfect gas — A theoretical gas that obeys the ideal gas equation exactly at all pressures and temperatures.
• Equation of state — A relation connecting the macroscopic variables that describe an equilibrium state of a system.
• Avogadro’s number — The number of specified particles in one mole, approximately 6.02 × 10²³.
• Number density — The number of molecules present per unit volume of a gas sample.
• Elastic collision — A collision in which the total kinetic energy of the colliding system is conserved.
• Isotropy — The absence of a preferred direction in the molecular motion of an equilibrium gas.
• Rms speed — The square root of the mean of the squared speeds of molecules.
• Isothermal process — A thermodynamic process in which the temperature of the system remains constant throughout.
• Degree of freedom — An independent way molecular motion can occur, contributing to the description of its energy.
• Equipartition — Equal average energy of ½kᵦT per active quadratic energy term in thermal equilibrium.
• Molar heat capacity — Heat required per mole per unit temperature rise under specified thermodynamic conditions.
• Mean free path — The average distance travelled by a molecule between two successive collisions.

Common errors and misconceptions

• Misconception: The ideal gas equation follows from molecular mechanics alone. Correct: It is experimental and is combined with the kinetic pressure equation to interpret temperature.
• Misconception: Gas law temperature ratios can use Celsius readings. Correct: The gas equations use absolute temperature in kelvin.
• Misconception: Equal temperatures give equal molecular speeds. Correct: They give equal average translational kinetic energies; heavier molecules have smaller rms speeds.
• Misconception: Raising pressure necessarily raises rms speed. Correct: At constant temperature, pressure and density change together and rms speed remains unchanged.
• Misconception: A vibrational mode contributes just ½kᵦT. Correct: Its kinetic and potential terms together contribute kᵦT when the mode is active.
• Misconception: Every ideal gas has U = ³⁄₂nRT. Correct: This gives translational energy; active rotations and vibrations add to total internal energy.
• Misconception: Compression always increases temperature. Correct: Temperature also depends on heat transfer; isothermal compression keeps temperature constant by transferring heat out.
• Misconception: Mean free path equals molecular separation. Correct: One is distance travelled between collisions; the other describes separation between molecules at an instant.

Exam-style questions with model answers

Q1. Define an ideal gas and state the conditions under which real gases approach ideal behaviour. [2 marks]

1. An ideal gas is a theoretical gas that obeys pV = nRT exactly at all pressures and temperatures.
2. Real gases approach this behaviour at low pressures and high temperatures, well above liquefaction or solidification.

Q2. An ideal gas is compressed at constant temperature. Explain the effects on its density, mean translational kinetic energy and rms speed. [3 marks]

1. The same gas mass occupies a smaller volume, so its mass density increases. Its pressure increases in the same proportion, leaving the ratio of pressure to density constant.
2. The average translational kinetic energy per molecule remains unchanged because it depends only on the absolute temperature.
3. The rms speed remains unchanged because it depends on temperature and molecular mass, both of which are unchanged.

Q3. For an isotropic ideal gas of identical molecules undergoing elastic wall collisions, derive p = ⅓ρ⟨c²⟩ using momentum transfer. Define the symbols introduced. [6 marks]

1. Let m be molecular mass and cₓ the incoming velocity component normal to a wall. Elastic reflection changes it to −cₓ, so each collision transfers momentum 2mcₓ to the wall.
2. Let A be wall area, Δt a short time and η𝗀 the number density of a group with this component magnitude. The accessible layer has volume AcₓΔt.
3. On average half the group travels towards the wall, so the number of impacts is η𝗀AcₓΔt/2.
4. Multiplying by 2mcₓ and dividing by AΔt gives the group pressure η𝗀mcₓ². Adding groups gives p = ηm⟨cₓ²⟩, where η is total number density.
5. Isotropy gives ⟨cₓ²⟩ = ⟨c²⟩/3, where c is molecular speed and angle brackets denote a molecular average.
6. Since mass density ρ = ηm, substitution gives p = ⅓ρ⟨c²⟩. Thus gas pressure depends on density and mean square molecular speed.

Q4. Air and hydrogen are treated as ideal gases at 273 K. Their molar masses are 29.0 × 10⁻³ and 2.02 × 10⁻³ kg mol⁻¹ respectively. Using R = 8.31 J mol⁻¹ K⁻¹, calculate both rms speeds and compare them. [4 marks]

1. Use cᵣₘₛ = √(3RT/M), where M is molar mass. The supplied masses are already in the units required by the given gas constant.
2. For air, cᵣₘₛ = √(3 × 8.31 × 273/0.0290) ≈ 484 m s⁻¹.
3. For hydrogen, cᵣₘₛ = √(3 × 8.31 × 273/0.00202) ≈ 1.84 × 10³ m s⁻¹.
4. Hydrogen’s speed is about 3.79 times the air value. Its smaller molecular mass permits a higher speed at the same temperature.

Q5. A fixed-volume cylinder contains 44.8 L of monatomic ideal helium at STP. Its temperature is raised by 15.0 K. Given the STP molar volume 22.4 L mol⁻¹ and R = 8.31 J mol⁻¹ K⁻¹, calculate the heat supplied. [3 marks]

1. The amount of helium is its volume divided by molar volume: n = 44.8/22.4 = 2 mol. The matching litre units cancel in this calculation.
2. Helium is monatomic, so Cᵥ = 3R/2. Constant volume means the gas performs no expansion work and the heat increases its internal energy.
3. Therefore Q = nCᵥΔT = 2 × (3/2) × 8.31 × 15.0 = 374 J to three significant figures.

Q6. An ideal gas undergoes quasi-static isothermal compression from volume V₁ to a smaller volume V₂. It contains n moles at temperature T, and R is the gas constant. Obtain the work done on the gas and explain the heat transfer. [4 marks]

1. At each stage p = nRT/V. Quasi-static compression permits the work by the gas to be obtained by integrating p dV from V₁ to V₂.
2. Integration gives Wᵦᵧ = nRT ln(V₂/V₁), where ln is the natural logarithm. This is negative because V₂ is smaller than V₁.
3. Work done on the gas is Wₒₙ = −Wᵦᵧ = nRT ln(V₁/V₂), a positive quantity.
4. At constant temperature an ideal gas has unchanged internal energy. The work supplied is transferred out as heat.

Q7. State equipartition and use it to obtain Cᵥ, Cₚ and γ for a rigid diatomic ideal gas. Then explain how one active vibrational mode changes these values. Cᵥ and Cₚ are molar heat capacities; γ = Cₚ/Cᵥ; R is the gas constant. [5 marks]

1. Equipartition assigns an average energy ½kᵦT to each active quadratic energy term in thermal equilibrium. Here kᵦ is Boltzmann’s constant and T is absolute temperature.
2. A rigid diatomic molecule has three translational and two effective rotational degrees of freedom. Its five contributions give energy 5kᵦT/2 per molecule and 5RT/2 per mole.
3. At constant volume, heat increases internal energy without expansion work. Differentiating the molar energy with respect to temperature gives Cᵥ = 5R/2.
4. For an ideal gas, Cₚ − Cᵥ = R. Therefore Cₚ = 7R/2 and γ = 7/5 for the rigid diatomic model.
5. One active vibration adds both a kinetic and a potential term. Consequently Cᵥ becomes 7R/2, Cₚ becomes 9R/2, and γ becomes 9/7.

Q8. Define mean free path and explain how it varies with number density and molecular diameter. Why does a high molecular speed not imply immediate diffusion across a room? [3 marks]

1. Mean free path λ is the average distance between successive molecular collisions. For spherical molecules of diameter d and number density η, λ = 1/(√2πηd²), with π the circle constant.
2. At fixed diameter it varies inversely with number density. At fixed number density it varies inversely with the square of diameter.
3. Collisions repeatedly change the direction of molecular motion. Molecules therefore do not travel straight across the room unhindered, even though their speeds are large.

Key takeaways

• The experimental perfect gas equation connects pressure, volume, temperature and amount; real gases approach it at low pressures and high temperatures.
• Kinetic pressure comes from momentum transferred in molecular collisions and equals one third of density times mean square speed.
• Absolute temperature measures average translational kinetic energy per molecule; equal temperatures do not require equal molecular speeds.
• Rms speed increases with the square root of absolute temperature and decreases with the square root of molar mass.
• Compression does positive work on a gas; its temperature response also depends on the heat transferred during the process.
• Equipartition assigns half a Boltzmann constant times temperature to each active quadratic energy term in thermal equilibrium.
• Count translation, rotation and active vibration before predicting heat capacities; distinguish molar capacity from capacity per unit mass.
• Mean free path is an average collision-free travel distance, and repeated deflections explain why diffusion takes time despite high molecular speeds.

Test yourself

Which part of the temperature argument comes from experiment?

The ideal gas equation pV = nRT is experimental. It is combined with the mechanical pressure relation to interpret temperature.

Why is there a factor of one third in the pressure equation?

Isotropy makes the mean square velocity components equal along three perpendicular directions, each contributing one third of the mean square speed.

Do molecules stop colliding when a gas reaches equilibrium?

No. Molecular motion and collisions continue; equilibrium means that the average macroscopic properties remain steady.

What stays equal for two ideal gases at the same temperature?

The average translational kinetic energy per molecule is equal. Rms speeds differ if their molecular masses differ.

Why does a vibrational mode count twice in equipartition?

It contains both a quadratic kinetic energy term and a quadratic potential energy term, each contributing ½kᵦT when active.

Why is Cₚ greater than Cᵥ for an ideal gas?

Heating at constant pressure supplies expansion work as well as increasing internal energy; at constant volume there is no expansion work.

What is the sign of work done by a gas during compression?

Work done by the gas is negative because its volume decreases. Work done on the gas is positive.

At fixed temperature and molecular diameter, how does pressure affect mean free path?

Mean free path varies inversely with pressure because increasing pressure increases the molecular number density.