Temperature and Ideal Gases

01

Temperature on an absolute scale

Temperature determines the direction of net heating when systems are put in thermal contact. Gas equations use an absolute temperature scale, so a Celsius reading must be converted before substitution.

A temperature does not depend on the thermometer's substance

A thermodynamic temperature scale has an absolute zero and is independent of the property of any particular substance. Mercury expansion, electrical resistance and a gas's pressure can all change with temperature. Calibrated thermometers use such properties to indicate the same physical quantity; the choice of material does not define a different thermodynamic temperature.

The SI unit is the kelvin, K, written without a degree symbol. Here T denotes thermodynamic temperature and θ denotes the Celsius reading. Other texts may use T for a Celsius reading too, so read the named quantity and its unit.

Convert a reading, then compare it

This equation relates the numerical readings on the two scales. Absolute zero is 0 K = -273.15°C. A reading of 26.85°C corresponds to 300.00 K.

A temperature interval of 1 K is the same size as an interval of 1°C. The 273.15 offset cancels when two readings are subtracted, but it does not cancel when their ratio is taken.

Interpret a thermometer reading

For a contact thermometer, establish good thermal contact and wait for a stable reading before using it as the object's temperature. Select a suitable range, record the scale or display resolution and check calibration. The surroundings' temperature is not automatically the temperature of an object that is still warming or cooling.

In the classical ideal-gas model, mean translational kinetic energy tends to zero as T tends to zero. This does not establish that every possible microscopic motion in every real material stops. Likewise, extending an ideal-gas graph to 0 K does not show that a real gas stays gaseous and ideal all the way there.

02

Particles, moles and mass

Specify what you are counting. For a molecular gas, the gas particles are molecules; counting every atom inside those molecules gives a different number.

Connect an amount to a particle count

N is the number of specified particles and is dimensionless. n is the amount of substance, measured in mol. One mole contains the Avogadro number of specified entities. Using the supplied rounded Avogadro constant:

The unit mol cancels with mol^-1, leaving a count. Before a precise calculation, roughly half a mole already suggests order 10²³ particles. A result of only a few hundred particles would be inconsistent with that amount.

Distinguish mass quantities

Relative atomic mass A_r compares the average mass of an atom with one twelfth of the mass of a carbon-12 atom. Relative molecular mass M_r compares the mass of a molecule with that same reference. Both are ratios and have no unit.

Molar mass M is mass divided by amount, in kg/mol. It is different from the mass m of one particle, in kg. A sample's mass is the mass of all its particles together.

Move between particle and mole equations

The Boltzmann constant k relates energy per particle to temperature. The molar gas constant R uses amount in moles instead. With supplied rounded values:

Substitute N = nN_A:

Thus pV = NkT and pV = nRT describe the same particle count. Use either N with k or n with R; pairing N with R counts the mole conversion twice. Small numerical differences from separately rounded constants are rounding effects.

03

Use the ideal-gas equation

An ideal-gas state is described by absolute pressure, gas volume, particle number and thermodynamic temperature. Convert the units and name the fixed quantities before comparing two states.

Use one consistent set of quantities

Use p in Pa, V in m³ and T in K. N is the number of gas particles; n is the amount in mol, with N = nN_A and Nk = nR. For a molecular gas, N counts molecules.

The units agree: Pa m³ = N m = J, where N here is the newton unit. On the particle side, a dimensionless count multiplied by J/K and K also gives J. This is a relation between equilibrium state quantities; pV is not automatically the energy transferred by heating during a process.

Estimate the pressure scale

For roughly half a mole near room temperature, use T of order 300 K, R of order 8 J/(mol K) and V of order 10^-2 m³. Then nRT/V is of order 10⁵ Pa. These rough supplied scales help catch a volume conversion error before precise substitution.

Use absolute pressure and number density

Gauge pressure is a pressure difference relative to the ambient pressure. A sensor reading 20 kPa above a supplied ambient pressure of 100 kPa gives an absolute gas pressure of 120 kPa. Zero gauge pressure means equality with ambient pressure, not zero absolute pressure or no particles.

The standard atmosphere is a defined pressure unit. The actual local atmospheric pressure need not equal 1 atm.

Number density is particle count per unit volume, N/V, in m^-3. For the 0.500 mol sample above:

Here n means amount in mol. In another context, n may denote number density instead. Check the quantity and unit rather than assuming the letter always has the same meaning.

Derive controlled comparisons

For a fixed number of particles, pV/T = Nk is constant. Hence p₁V₁/T₁ = p₂V₂/T₂. Holding one further quantity fixed gives:

For that sample, warming from 300 to 360 K at constant pressure gives V₂ = 12.0(360/300) = 14.4 litres. If volume is held fixed instead, p₂ = 103875(360/300) = 124650 Pa, or about 1.25 × 10⁵ Pa. These are two different changes from the same starting state.

A leaking container does not keep N fixed. A graph against Celsius temperature does not pass through its zero-temperature origin. State the controls and scale whenever claiming proportionality.

Investigate pressure at fixed volume

A sealed rigid vessel in a controlled water bath can be connected to an absolute-pressure sensor. Vary the bath temperature in steps and record settled gas pressure and temperature, aiming to keep the amount and total gas volume fixed.

Choose suitable instrument ranges and record pressure and thermometer resolutions. Check the sensor's zero or pressure reference, check for leaks and use a temperature range that gives clearly resolvable pressure changes. Allow time for the gas to settle after each change.

The gas in the connecting tube is part of the system too. Include its volume and arrange for the gas to be approximately at one temperature; a large tube volume left at room temperature complicates the comparison. The bath thermometer does not prove that the gas has already reached that temperature.

Read a separate generated data example

The following values describe a different model sample, whose pressure is 100.000 kPa at 300 K. They do not describe the 103.875 kPa starting state above. They are generated from p/T = 100000/300 Pa/K and rounded to 0.001 kPa; the displayed digits do not claim an instrument's resolution.

The unrounded generating model has p/T = 333.333... Pa/K. Its kelvin graph passes through zero. On a Celsius horizontal axis, the same model instead has p = 91.05 kPa at 0°C and extrapolates to zero pressure at -273.15°C.

Only the supplied range is represented by the data. A line fitted there does not establish that a real gas stays ideal down to 0 K: changes such as condensation invalidate that extension.

Explain a practical departure causally

• Reading too early during warming: gas still below the recorded bath temperature has lower pressure than the equilibrium prediction at that bath temperature. Repeating the same rushed procedure does not remove this bias.
• Leakage: the particle number changes, so successive points no longer test the same fixed-N relationship.
• Uncorrected gauge readings: a constant atmospheric offset changes the intercept and p/T. It need not change the slope of a pressure-temperature line.
• Vessel expansion: volume may change slightly as temperature rises, weakening the assumed fixed-volume control.

For real readings, retain instrument precision and uncertainties and compare the measured gradient and intercept with the stated model before claiming agreement. An area under a pressure-temperature graph is not heat supplied: its units are not joules.

04

Pressure from particle collisions

A wall experiences pressure because gas particles repeatedly transfer momentum to it. Find the impulse from one collision, divide by the interval between repeated impacts, then sum and average over the gas.

State the ideal-gas assumptions

Consider N particles of the same mass m in a stationary container at equilibrium. The kinetic model assumes:

• Very many particles obey Newton's laws and move continually and randomly. Motion is isotropic: there is no preferred direction and no bulk drift.
• The particles' own volume is negligible compared with the gas volume.
• Forces between particles are negligible except during collisions. Between collisions, each particle travels at constant velocity.
• Collisions between particles and with stationary walls are elastic. Collision durations are negligible compared with travel times.

Particles need not all have the same speed. Elastic collisions conserve the relevant total kinetic energy, but can change individual velocities. These assumptions describe a model; they are not equally accurate for every real gas state.

Find the wall impulse

Let the box length normal to a selected pair of walls be L, and let each wall's area be A, so V = AL. Take +x towards the right wall. For the incoming particle, let c_x > 0 be the magnitude of its normal velocity component. The outgoing x-component is -c_x.

A smooth stationary wall reverses the normal component while leaving tangential velocity components unchanged. Using momentum change = final momentum - initial momentum:

The wall pushes the particle inward; the particle pushes the wall outward. These are equal and opposite forces on different bodies. The wall impulse is positive in the chosen x direction.

Use the same-wall return time

In the simple bouncing-particle construction, the particle crosses to the opposite wall and returns before striking the original wall again. Using the positive normal speed c_x:

L/c_x accounts for only one crossing. The brief collision duration gives the much larger force during contact, not the force averaged over the intervening flight time.

Sum the normal contributions

The N particles share mass m but have different velocity components. In the following equations, mean(c_x²) means the average of the squared x-components over the particles:

Pressure is the time-averaged normal force per unit area:

Use three-dimensional isotropy

Isotropic motion gives equal mean-square components along any three perpendicular axes:

For each particle, the square of its speed is the sum of the squared components. Taking the mean gives:

The factor of one third applies to the mean-square components, not to every particle's individual components. A random gas can have zero mean velocity while mean(c²) is positive, so it still exerts pressure.

The simple return-time construction follows uninterrupted flights between walls. In a gas with elastic interparticle collisions, velocities are redistributed between particles; averaging the equilibrium ensemble retains the same isotropic pressure result. Do not claim that every actual particle always follows one uninterrupted round trip.

05

Temperature and particle energy

Absolute temperature determines the mean translational kinetic energy per ideal-gas particle. Equal temperature gives equal mean energy, even when different particle masses lead to different characteristic speeds.

Connect the two pressure equations

For N particles of mass m, the ideal-gas relation is pV = NkT. The collision model gives pV = Nm mean(c²)/3. Equate them and cancel N:

This is the average kinetic energy associated with translation of each gas particle. Particles have a range of velocities; the equation does not assign the same energy to every particle. It also does not by itself give every contribution to the total internal energy of a molecular gas or another phase.

Distinguish mean speed from rms speed

Root-mean-square speed means square the speeds, find their mean, then take the square root:

It is generally different from the mean speed and from the magnitude of the mean velocity. A zero mean velocity can result from opposite directions cancelling; speeds and their squares are non-negative.

Estimate microscopic scales

With supplied k of order 10^-23 J/K and T of order 300 K, mean translational energy is of order a few 10^-21 J per particle. For a supplied light-particle mass of order 10^-26 kg, √(3kT/m) is of order 10³ m/s. These are rough scale checks, not precise measurements.

For the same gas, doubling the absolute temperature doubles mean translational energy but multiplies rms speed by √2, not two. A Celsius reading doubling does not establish either of these changes.

06

Revision summary

Name the particles, use absolute pressure and temperature, and distinguish an individual collision from a time or particle average.

Temperature, amount and counting

The thermodynamic scale is independent of the property of a particular substance. Use T/K = θ/°C + 273.15. Temperature differences have equal numerical size in K and °C; ratios in gas and particle-energy equations require kelvin.

N counts the specified particles. For 0.500 mol of O₂, it is 3.01 × 10²³ molecules, not the twice-as-large atom count. Relative atomic and molecular masses are dimensionless comparisons with one twelfth of a carbon-12 atom's mass; molar mass M has unit kg/mol.

Gas states and controlled changes

Use Pa, m³ and K. Add ambient pressure to gauge pressure. Recall 1 atm = 101325 Pa, 1 litre = 10^-3 m³ and 1 cm³ = 10^-6 m³. The actual atmosphere may differ from the standard unit.

• Fixed N and T: p ∝ 1/V; a reciprocal pressure-volume curve.
• Fixed N and V: p ∝ T; a straight pressure-kelvin line.
• Fixed N and p: V ∝ T; a straight volume-kelvin line.

The 0.500 mol, 12.0 litre, 300 K model gives p ≈ 1.04 × 10⁵ Pa and N/V ≈ 2.51 × 10²⁵ m^-3. Keep it separate from the generated graph sample whose 300 K pressure is 100.000 kPa.

For a fixed-volume investigation, control leaks and total gas volume, wait for a common settled gas temperature, and use absolute pressure. A graph's gradient has meaning only with the correct axes and units. A pressure-temperature area is not energy transferred, and extrapolation is not evidence that a real gas remains ideal.

Rebuild the pressure derivation

The model assumes many Newtonian particles in random isotropic motion, negligible particle volume, negligible forces between collisions, and brief elastic collisions. For N particles of the same mass m, use a positive incoming normal speed c_x:

The round-trip interval is not collision duration. Mean-square isotropy is an ensemble statement, not equal components for every particle. Zero mean velocity does not imply zero mean-square speed or zero pressure. Revisit the wall impulse and averaging steps if the factor of one third is unclear.

Connect temperature to an average energy

Equal T gives equal mean translational energy per particle. Four times the particle mass gives half the rms speed at that T. Doubling T gives twice the mean energy and √2 times the rms speed for the same gas. Mean speed, rms speed and mean velocity are different averages.

Quantities and units

The same letter can denote different quantities in different models: c here is microscopic speed, whereas c can mean specific heat capacity in thermal calculations. Use the definition and units to identify the quantity.