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Temperature and Ideal Gases overview

Topic 3 of 5

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

pV = NkT = nRT

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 = nNA 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.

Worked gas state

0.500 mol in 12.0 litres

The supplied gas has Celsius temperature 26.85°C. Convert both temperature and volume:

T = 26.85 + 273.15 = 300.00 K
V = 12.0 × 10-3 = 0.0120 m3
p = nRT/V
= [0.500(8.31)(300)]/0.0120
= 103875 Pa ≈ 1.04 × 105 Pa

Using N = 3.01 × 1023 and k = 1.38 × 10-23 J/K instead gives 103845 Pa, the same three-significant-figure result. The small difference arises from the supplied constants being rounded separately.

For volume conversions, 1 litre = 10-3 m3 and 1 cm3 = 10-6 m3. Leaving 12.0 litres as 12.0 m3 makes the pressure a thousand times too small.

Optional check An ideal gas has amount 0.500 mol, volume 12.0 litres and Celsius temperature 26.85 degrees C. With R = 8.31 J/(mol K), what is its absolute pressure?
An ideal gas has amount 0.500 mol, volume 12.0 litres and Celsius temperature 26.85 degrees C. With R = 8.31 J/(mol K), what is its absolute pressure?

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.

pabsolute = pgauge + pambient
1 atm = 101325 Pa
1.20 atm = 121590 Pa ≈ 1.22 × 105 Pa

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:

N/V = (3.01 × 1023)/0.0120
≈ 2.51 × 1025 m-3
p = (N/V)kT

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:

Ideal-gas comparisons with particle number fixed
Also fixedResultGraph
Temperature TpV constant; p ∝ 1/Vp against V is a reciprocal curve
Volume Vp/T constant; p ∝ Tp against T is a line through the kelvin origin
Pressure pV/T constant; V ∝ TV against T is a line through the kelvin origin

Name the fixed quantities before comparing states

These two ideal-model comparisons use the 0.500 mol sample, with the same fixed particle number N. They are separate controlled changes from its 300 K, 12.0 litre, 103.875 kPa state.

Fixed N and T = 300 K: reciprocal pressure-volume curve

At fixed temperature, doubling volume halves absolute pressureThe horizontal axis is gas volume in litres, from zero to 24; the vertical axis is absolute pressure in kilopascals. The ideal curve is plotted over volumes six to 24 litres, with pV equal to 1246.5 joules. Point A is six litres and 207.75 kilopascals. B is twelve litres and 103.875 kilopascals. C is 24 litres and 51.9375 kilopascals. Both N and temperature 300 kelvin stay fixed. These are calculated states, not experimental readings. No finite pressure is assigned to zero volume, and the curve does not reach the origin.06121824050100150200ABCAbsolute pressure / kPaVolume V / litre

A: 6.0 litres, 207.75 kPa. B: 12.0 litres, 103.875 kPa. C: 24.0 litres, 51.9375 kPa. Here 1 kPa litre = 1 J, so all three products pV equal 1246.5 J.

Fixed N and p = 103875 Pa: volume follows kelvin temperature

At constant absolute pressure, 300 to 360 kelvin gives twelve to 14.4 litresThe horizontal axis is thermodynamic temperature from zero to 360 kelvin; the vertical axis is volume in litres. The model relation is V in litres equals 0.0400 times T in kelvin. Point A is 300 kelvin and twelve litres. Point B is 360 kelvin and 14.4 litres. The part between A and B is solid; the lower-temperature model extrapolation to the origin is dashed. N and absolute pressure 103875 pascals stay fixed. This ideal-model extrapolation does not establish that a real gas stays gaseous or ideal down to zero kelvin.0100200300360051015ABVolume / litreTemperature T / K

A: 300 K, 12.0 litres. B: 360 K, 14.4 litres. The dashed part is an ideal-model extrapolation; a real gas need not remain a gas or remain ideal at much lower temperatures.

Both comparisons use the original 0.500 mol model sample. One fixes temperature at 300 K; the other fixes pressure at 103875 Pa. The curves describe ideal states, not experimental observations.

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.

Generated fixed-volume gas pressures, rounded to 0.001 kPa
T / Kp / kPa
300100.000
320106.667
340113.333
360120.000

Measure one gas system after its temperature settles

The graphs below use a separate fixed-N, fixed-V model sample: its pressure is 100.000 kPa at 300 K. It is not the 103.875 kPa sample in the preceding comparison.

A sealed rigid vessel in a controlled bath

The bath thermometer and connected absolute-pressure sensor measure different quantitiesA sealed rigid gas vessel is immersed in a controlled water bath. A thermometer bulb is in the bath outside the gas vessel. A hollow tube joins the vessel's open port to a sealed absolute-pressure sensor above the bath. The vessel wall is closed everywhere except that connected port; the vessel, tube and sensor cavity together contain the fixed gas sample. The tube's section above the bath is shown explicitly, because its volume and temperature can matter. The bath temperature is not assumed to be the instantaneous gas temperature; allow equilibrium and account for a significant connecting volume at a different temperature. This is a schematic apparatus, with no implied product range, resolution or literal gas volume.AbsolutepressuresensorBaththermometerGas inconnecting tubeRigid vesselSealed gasControlled water bath

A bath reading is usable as the gas temperature only after suitable thermal equilibrium. Keep N and the effective gas volume fixed, measure absolute pressure, and account for leaks and connecting-tube volume.

Absolute pressure against kelvin temperature

The generating pressure-temperature line has zero intercept in kelvinBoth gas-temperature plots use the same physical temperature range, zero to 360 kelvin, and the same pressure scale, from zero to 120 kilopascals and slightly above. Four supplied generated points are 300, 320, 340 and 360 kelvin at 100.000, 106.667, 113.333 and 120.000 kilopascals. The solid part and pale band identify this supplied temperature range. The dashed line below it is an extrapolation of the ideal generating model, not measured evidence at low temperature. The horizontal coordinate is T in kelvin, so the generating line reaches the origin and has slope one third of a kilopascal per kelvin. N and V are fixed for this separate model sample. Rounding of supplied points is much too small to resolve on these axes. Real-gas behaviour is not established by the low-temperature extrapolation.010020030036004080120Absolute pressure / kPaTemperature T / K

Solid line and filled points: the supplied 300-360 K range. Dashed line: ideal-model extrapolation. The generating ratio is p/T = 333.333... Pa/K.

The same model against Celsius temperature

The same ideal pressure line has a shifted temperature origin in CelsiusBoth gas-temperature plots use the same physical temperature range, zero to 360 kelvin, and the same pressure scale, from zero to 120 kilopascals and slightly above. Four supplied generated points are 300, 320, 340 and 360 kelvin at 100.000, 106.667, 113.333 and 120.000 kilopascals. The solid part and pale band identify this supplied temperature range. The dashed line below it is an extrapolation of the ideal generating model, not measured evidence at low temperature. The horizontal coordinate is theta in degrees Celsius, from minus 273.15 to positive 86.85. The brown vertical reference marks zero degrees Celsius; there the model pressure is 91.05 kilopascals. Its zero-pressure extrapolation lies at minus 273.15 degrees Celsius, the left edge, rather than at Celsius zero. This graph has the same physical span and line gradient as the kelvin plot, with a shifted temperature coordinate. N and V are fixed for this separate model sample. Rounding of supplied points is much too small to resolve on these axes. Real-gas behaviour is not established by the low-temperature extrapolation.-273.15-100086.8504080120At 0 °C:91.05 kPaAbsolute pressure / kPaCelsius temperature θ / °C

Solid line and filled points: the supplied 300-360 K range. Dashed line: ideal-model extrapolation. The brown reference is 0 °C, not zero kelvin. The model's zero-pressure intercept is -273.15 °C.

These pressures were generated from an ideal relation and rounded to 0.001 kPa. Those digits do not specify an instrument's resolution. A real gas can condense or depart from the ideal model before an extrapolated zero is approached.

The apparatus separates the gas system from its surrounding bath. The two graphs use the separate 100.000 kPa at 300 K model sample, with the same pressure scale. Dashed regions extend the ideal model beyond the supplied 300-360 K range.

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.

Optional spreadsheet application

Find a pressure-temperature gradient

Use the four generated pairs above. The Motion spreadsheet method explains manual entry, copying formulas and numeric scatter graphs.

  1. Put temperature / K in A1 and pressure / kPa in B1. Enter the four pairs in A2:B5 as numbers.
  2. Use C1 for absolute pressure / Pa and D1 for pressure / temperature / Pa K^-1. Enter =B2*1000 in C2 and =C2/A2 in D2. Fill both formulas through row 5 only.
  3. Create a numeric XY scatter graph with A2:A5 horizontally and C2:C5 vertically. Label the axes T / K and p / Pa. Fit a straight line with a free intercept and display its equation; do not force it through zero before examining the result.
  4. Interpret the gradient using p = (N/V)kT. Divide its value in Pa/K by k = 1.38 × 10-23 J/K to estimate N/V. Change displayed precision without rounding away stored values.
Compare the fitted model

The ratios in column D are close to 333.333 Pa/K. The rounded data give a free-intercept fit with gradient 333.330 Pa/K and intercept +1.10 Pa.

N/V = gradient/k
= 333.330/(1.38 × 10-23)
≈ 2.42 × 1025 m-3

The unrounded generating model has zero intercept and gradient 333.333... Pa/K. These small fitted departures arise from rounding, not experimental scatter. They do not provide new experimental evidence for the model.

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.