An ideal gas links amount, volume, and absolute temperature. Its internal energy still needs a molecular model.
By the end: predict pressure changes and explain why pressure, particle speed, and internal energy are different quantities.
Read the ideal-gas relationship
Gas molecules transfer momentum to container walls in collisions. More particles per volume create more collision opportunities, and greater mean translational kinetic energy increases momentum transfer. The ideal-gas model neglects molecular volume and interactions except during collisions. Its equation relates absolute pressure p, volume V, amount n in moles, and absolute temperature T. One mole contains about 6.022 × 10²³ entities; R is about 8.314 J per mole per kelvin.
For the same amount of gas, pV divided by T stays constant. Use absolute pressure, not a gauge reading measured relative to atmospheric pressure. The model starts at 100 kPa, 10 L, and 300 K. Since one kilopascal times one litre equals one joule, its initial pV is 1,000 J. These conditions specify about 0.4009 mol.
Separate collision strength from particle concentration
Pressure measures force per area, not energy per particle. Imagine a molecule approaching a flat wall perpendicular to its surface. An elastic reflection reverses that component of momentum while leaving its kinetic energy unchanged. The wall receives momentum during the collision. Repeated impacts produce an average force, even when the wall does not move and no boundary work occurs. More particles can increase the collision rate at unchanged temperature. Faster particles can increase both the momentum delivered per encounter and the number of encounters per second. These are different ways to raise pressure.
Reflection transfers momentum without moving the wall
Schematic before-and-after views of one elastic, normal collision with a stationary wall. Arrow lengths match because speed is unchanged. The molecule reverses its momentum; repeated wall impulses produce average pressure without boundary work.
For a fixed number of molecules in equilibrium, lowering the volume raises the number per unit volume. At unchanged temperature their mean translational energy stays the same, but more momentum reaches a unit wall area per second. This is why pressure can rise without a temperature increase. In a cubic container, changing volume also changes distances between opposite walls, so a single-particle travel argument needs care. The macroscopic result combines all directions of motion and gives pressure proportional to number density times absolute temperature, rather than pressure proportional to speed alone.
Check the sample amount using compatible units. Ten litres is 0.010 m³ and 100 kPa is 100,000 Pa. Their product is 1,000 Pa m³, equal to 1,000 J. Dividing by 300 K gives nR equal to 10/3 J/K. Dividing again by R, approximately 8.314 J per mole per kelvin, gives about 0.4009 mol. Multiplying that amount by Avogadro's constant gives about 2.41 times ten to the power twenty-three particles. The twelve visible markers encode this fixed amount schematically, not an actual microscopic census.
Absolute pressure has its zero at no pressure. A gauge instead reports a difference from its surrounding pressure. In a separate arithmetic example, let the surroundings be at 100 kPa and a gauge read 50 kPa. The gas pressure is then 150 kPa absolute. If fixed-volume heating doubles its kelvin temperature, ideal-gas pressure becomes 300 kPa absolute. With the same surroundings, the gauge would read 200 kPa. Doubling the original gauge reading would give 100 kPa and the wrong absolute endpoint. Pressure ratios in the gas law require the absolute values.
The gauge origin moves the numerical ratio
Linear stacked lengths from zero absolute pressure, 1.7 drawing units per kPa. Gold is the fixed 100 kPa surroundings; the remaining length is gauge pressure. Doubling kelvin temperature at fixed volume doubles absolute pressure from 150 to 300 kPa.
Temperature needs the same attention to its zero. A change from 300 K to 450 K multiplies ideal-gas pressure by 1.5 at fixed volume and amount. Those endpoints are 26.85 °C and 176.85 °C, whose ratio does not have that physical meaning. Conversely, doubling sample amount while keeping volume and temperature fixed also doubles pressure, but it doubles total translational energy by adding particles rather than increasing each particle's mean energy. The gas comparison holds amount fixed, so a pressure change there cannot be explained by secretly adding molecules.
Separate motion, density, and energy
At fixed volume, doubling kelvin temperature doubles ideal-gas pressure. At fixed temperature, doubling volume halves pressure without changing average translational energy per molecule. Typical molecular speeds scale with the square root of temperature for a fixed species, so doubling temperature multiplies a root-mean-square speed by about 1.414, not two. Molecules have a distribution of speeds, not one common speed.
For a classical monatomic ideal gas, three translational motion components give an internal energy of three halves of nRT, or three halves of pV. The chosen sample therefore has U = 1,500 J at 300 K in that model’s energy reference. Doubling volume at fixed temperature leaves U unchanged; doubling temperature doubles U. The twelve drawn markers represent a fixed amount of gas, not literally twelve molecules. At fixed volume, this sample needs 5 J per kelvin because heat only changes U. At fixed pressure, it also does expansion work, requiring an extra nR per kelvin. Its total constant-pressure capacity is therefore about 8.333 J/K. The heat-capacity ratio, gamma, is 5/3 for this classical monatomic model.
Derive the sample's two heat capacities
For a classical monatomic ideal gas, translation has three independent directional contributions. Each contributes an average of one half kB T per particle in equilibrium. Adding them gives three halves kB T, and multiplying by particle number gives three halves nRT for the sample. This model omits a constant energy reference, so U equal to 1,500 J at 300 K is not an absolute inventory of every possible atomic energy. Its useful prediction here is that changing temperature by 60 K changes the sample's internal energy by 300 J, regardless of volume.
At fixed volume, with only heat and pressure-volume work possible, the stationary boundary does no work. The first law then gives Q equal to ΔU. For this sample, three halves nR is 5 J/K, so warming from 300 K to 360 K requires 300 J. This is the whole-sample constant-volume capacity. Dividing by approximately 0.4009 mol gives a molar capacity of about 12.47 J per mole per kelvin. A mass-specific capacity would require dividing by sample mass instead, which needs the molecular species or its molar mass.
Three directional increments add to the sample energy change
Each equal rectangle represents 100 J of the sample’s energy increase for a 60 K rise. These are directional energy contributions, not three separate groups of particles. The monatomic translational model gives 300 J in total and whole-sample CV = 5 J/K.
Now consider the same 60 K rise at constant external pressure of 100 kPa in a mechanically reversible, quasistatic limit. The volume grows from 10 L to 12 L. The gas does 100 kPa times 2 L, or 200 J, of positive work on its surroundings. Internal energy still rises by 300 J because its temperature change is the same. Therefore Q must be 500 J. The whole-sample constant-pressure capacity is 500 J divided by 60 K, or 25/3 J/K. Extra heat supplies work rather than an extra internal-energy rise.
The difference between the two sample capacities is nR, or 10/3 J/K. The corresponding molar difference is R. Keeping these quantities separate prevents a factor-of-amount error when applying a textbook's molar formula to the sample. At constant pressure, pΔV equals nRΔT, explaining the difference directly. This argument assumes a closed sample and pressure-volume work only. A quasistatic gas remains close to internal equilibrium; calling a process reversible additionally excludes friction and requires vanishing driving differences for its exchanges. Slow piston motion alone would not establish all those conditions.
A moving boundary adds a work requirement
Piston sketches use the same cross-section and 10 drawing units of gas height per litre. The constant-pressure gas grows from 10 to 12 L; the outlined band marks its added 2 L. Only pressure-volume work is included. A reversible route requires more than simply moving slowly.
The two 360 K endpoints have the same U but different volumes and pressures. The fixed-volume endpoint has 10 L and 120 kPa; the constant-pressure endpoint has 12 L and 100 kPa. Neither heat capacity is a stored amount of heat, and neither implies that temperature tells you every state variable. For this ideal gas U depends only on temperature at fixed amount, whereas a full equilibrium state still requires volume or pressure. Comparing the two energy accounts explains why identical temperature changes need not demand identical heat inputs when the boundary can move.
Work through the mechanism
Equation of state
p is absolute pressure in Pa, V is volume in m³, n is amount in mol, R ≈ 8.314 J/(mol K), and T is in K. One kPa times one litre also equals one joule.
Internal energy of a classical monatomic ideal gas
This translational model is not the general energy formula for every gas or every temperature. An arbitrary constant energy reference is omitted.
Heat capacities of the whole monatomic sample
Here C values are whole-sample capacities in J/K, not molar capacities. CV is three halves nR because fixed-volume heat changes U. At fixed pressure, expansion adds nR of work per kelvin, giving Cp = five halves nR. The ratio Cp/CV is gamma = 5/3.
Change density and thermal motion separately
Selected pressure 200 kPa absolute; internal energy 3,000 J.
Both panels represent the same amount of classical monatomic ideal gas. Container widths encode volume at fixed cross-section. Twelve schematic markers stand for that fixed amount, not literal molecular counts. The short line below each box encodes RMS-speed ratio; it is not a trajectory or a distribution of actual speeds. Presets replace both inputs; each slider changes only its named input. There is no playback.
State
Pressure
Monatomic U
10 L, 300 K
100 kPa
1,500 J
10 L, 600 K
200 kPa
3,000 J
20 L, 300 K
50 kPa
1,500 J
Same amount of classical monatomic ideal gas in all rows; pressures are absolute.
Test two changes that cancel in pressure
Start the gas comparison with Original state so both panels show 300 K and 10 L. Predict the effects of choosing Double temperature, fixed volume. The selected box keeps its width because volume is unchanged. Pressure rises to 200 kPa and U rises to 3,000 J, but the RMS-speed indicator becomes only about 1.414 times its reference length. Squaring that speed ratio gives two, matching the energy ratio. A doubled pressure is therefore not evidence for doubled speed. Temperature controls a mean squared speed for a specified molecular species.
Keep temperature at 600 K and change only Selected volume to 20 L. Now both temperature and volume are twice their reference values, so their factors cancel in p proportional to T/V. The selected pressure returns to 100 kPa. Internal energy remains 3,000 J and the speed ratio remains about 1.414 because temperature has not changed during this second selection. The markers spread through twice the volume at the same amount. This comparison gives equal pressures with unequal temperatures, densities, and internal energies. Pressure alone cannot identify any of those quantities.
Equal pressure can hide unequal temperature and energy
The gas boxes use equal cross-sections and a common volume scale. Each contains the same twelve schematic markers representing the same sample amount. The hotter state has half the number density and twice the mean translational energy per particle, leaving pressure unchanged.
Next choose Double volume, fixed temperature. That preset replaces both inputs, giving 20 L and 300 K rather than preserving the previous 600 K. Pressure is now 50 kPa and U is 1,500 J. Move only the temperature slider to 450 K and predict a pressure of 75 kPa from 50 times 450/300. The internal energy should be 2,250 J and RMS speed about 1.225 times the reference. Check all three readouts. The speed uses the square root of 1.5, while both pressure at fixed volume and U use 1.5 directly.
For a lower-temperature comparison, select 150 K and 10 L, then change volume to 30 L without changing temperature. The pressure falls from 50 kPa to about 16.67 kPa. Internal energy stays at 750 J and the speed ratio stays about 0.707. The amount does not shrink with the pressure, and the lower density does not by itself cool this ideal gas's endpoint. These selections compare equilibrium states. The interface does not calculate a physical route that keeps temperature constant while the volume changes, or the heat and work required along such a route.
Finally compare the original state with 600 K and 20 L again. Equal p means equal average force per unit area on the walls, but the hotter sample's lower number density offsets its greater mean translational energy. In the kinetic relation, pressure is two thirds of translational energy per volume. The reference has 1,500 J in 10 L, or 150 J/L; the selected state has 3,000 J in 20 L, also 150 J/L. Multiplying by two thirds gives 100 J/L, numerically 100 kPa. This reconciles equal pressure with unequal total energy.
Verify each state against the fixed sample amount
State T, V
p
U
RMS / reference
300 K, 10 L
100 kPa
1,500 J
1
600 K, 10 L
200 kPa
3,000 J
√2 ≈ 1.414
600 K, 20 L
100 kPa
3,000 J
√2 ≈ 1.414
450 K, 20 L
75 kPa
2,250 J
√1.5 ≈ 1.225
150 K, 30 L
16.67 kPa
750 J
√0.5 ≈ 0.707
All rows have nR = 10/3 J/K and use the classical monatomic model. Pressures are absolute. RMS speeds are ratios for one fixed species, not measured velocities. Rows compare endpoints and do not prescribe heat or work.
Know when the molecular model changes
Equipartition assigns an average energy of one half kB T to each independent quadratic contribution to a classical particle’s energy. The Boltzmann constant kB is about 1.381 × 10⁻²³ J/K and relates individual-particle energies to temperature. Three translational contributions give three halves kB T per particle. Rotations and vibrations can add energy storage in molecules; they do not all become active at every temperature.
Quantum energy spacings can suppress modes at low temperature, so constant classical heat capacities cannot be extended to absolute zero. Dense gases, strong interactions, and conditions near condensation also violate ideal-gas assumptions. A fit to pV = nRT does not establish that a gas is monatomic or that its heat capacity is three halves nR.
Choose which microscopic model supports an inference
Speed also depends on particle mass. At the same temperature, two classical ideal-gas species have equal mean translational kinetic energy per particle, not equal RMS speed. If one species has four times the particle mass, its RMS speed is half as large because mass times mean squared speed must match. A hypothetical comparison can express this with relative masses one and four and relative RMS speeds two and one. The products of mass and squared speed are both four. Temperature equality can coexist with very different individual speed distributions.
Equal temperature fixes mean energy, not speed
Circle areas encode relative particle masses 1 and 4; arrow lengths encode relative RMS speeds 2 and 1. Directions are schematic. Equal m times mean squared speed gives equal mean translational kinetic energy at the same temperature.
For molecular gases, energy can also enter rotation and vibration. An active rotational contribution changes the relation between total internal energy and temperature even when pV still closely follows nRT. The pressure relation primarily describes translational motion and density; it does not count every internal molecular energy-storage mechanism. Thus a measured-looking agreement with the ideal-gas equation would not justify using the monatomic capacity automatically. In the present comparison, the choice of three translational contributions is a separate assumption, not something proved by its pressure readout or by the twelve schematic markers.
Quantum mechanics makes the allowed rotational and vibrational energies discrete. If a gap is large compared with kB T, thermally accessing the next state is unlikely, so that mode contributes little to capacity in that range. Increasing temperature can make additional states appreciably occupied and change capacity. This is a gradual population effect, not a universal temperature at which every molecule suddenly starts rotating. At still lower temperatures, quantum statistics and condensation can also invalidate a dilute classical gas description. Extending the constant 5 J/K sample capacity down to absolute zero would ignore these changes.
Microscopic collision diagrams explain pressure without predicting an actual molecular history. A real gas has positions and momenta for all its particles, and an equilibrium ensemble describes many possible detailed states. Twelve stationary dots and a speed indicator are not those states, their probabilities, or trajectories. Likewise, the later left/right counting example groups detailed possibilities into coarse assignments; its multiplicity is not a complete gas phase-space count. Inferring a probability requires specified weights as well as a set of alternatives. The present pressure calculation needs neither a literal twelve-particle gas nor an unshown random-motion simulation.
A state description also does not fix transfers. For the sample at 300 K, endpoints at 10 L and 20 L have equal ideal monatomic U, so ΔU is zero and the first law requires net Q equal to W. Reversible isothermal expansion can have both positive; insulated free expansion can have both zero. Boundary work depends on opposing external pressure during expansion, not merely on gas pressures listed at the endpoints. Specifying the route is therefore necessary before interpreting a pressure-volume area as work. The gas controls supply endpoint comparisons, leaving that route question to the next lesson.
The same endpoints allow different transfers
Schematic state-and-route diagram, not a pressure path or time scale. Both routes connect the same fixed amount of ideal gas at 300 K from 10 to 20 L. The isothermal work uses nRT ln 2; the actual free expansion has zero opposing pressure and zero work.
Exercises
Change pressure without changing temperature
The model sample expands from 10 L to 30 L while ending at 300 K. Find its final pressure and internal energy. Can you infer the heat transfer from endpoints alone?
Hint
Use the fixed pV/T ratio for pressure and U = 3pV/2 for energy.
Worked answer
Quantity
Final value
p
100 × 10 / 30 ≈ 33.33 kPa
pV
1,000 J
U
1,500 J
Constraint on transfers
Q = W, but neither is determined alone
A state equation does not specify a process.
The final pressure is about 33.33 kPa and U remains 1,500 J. Thus ΔU is zero. This does not determine Q or W separately. The first law only requires their net values to be equal for this change. The next lesson compares routes with these same kinds of endpoints.
Avoid a speed shortcut
Temperature rises from 300 K to 1,200 K for the same classical ideal-gas species. How do mean translational energy and root-mean-square speed change?
Hint
Energy is proportional to T; squared speed is proportional to T.
Worked answer
Quantity
Multiplier
Absolute temperature
4
Mean translational energy
4
Root-mean-square speed
2
A statistical speed measure differs from each particle’s instantaneous speed.
Mean translational energy becomes four times as large. Root-mean-square speed becomes twice as large, because the square root of four is two. This says nothing about a particular molecule keeping a fixed speed ratio through its collisions.
Sources and further reading
These optional references support the stated laws and model assumptions. The numerical examples and diagrams are authored teaching models, not experimental measurements or engineering specifications. Selected relevant passages were reviewed, not entire textbooks or a comprehensive research literature.