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Chemistry · Ch 6 — Gaseous State

Ideal gas equation

6.3

Ideal gas equation

The complete physical state of a gas needs four variables: pressure P, volume V, temperature T and amount n. The three gas laws already met each isolate one pairwise relationship between these variables:

Boyle's law: V∝1PV\propto\dfrac{1}{P} (n, T constant)

Charles's law: V∝TV\propto T (n, P constant)

Avogadro's law: V∝nV\propto n (T, P constant)

Combining all three into a single proportionality, V∝nTPV\propto\dfrac{nT}{P}, and introducing a proportionality constant R (the universal gas constant) turns this into an equality:

V=nRTP⟹PV=nRT(6.11)V=\frac{nRT}{P}\qquad\Longrightarrow\qquad PV=nRT\qquad(6.11)

This is the ideal gas equation. Because it links all four state variables together, it is also called the equation of state of an ideal gas.

Evaluating R. Since pressure can be expressed in several different units (Table 6.1), R itself takes a different numerical value depending which pressure (and volume) unit is used, so it is worth deriving it in more than one system.

Using the "STP" reference point of 1 atm, 22.414 dm3^3, for 1 mole at 273.15 K:

R=PVnT=1 atm×22.414 dm31 mol×273.15 K=0.0821 dm3 atm mol−1K−1R=\frac{PV}{nT}=\frac{1\ \text{atm}\times22.414\ \text{dm}^3}{1\ \text{mol}\times273.15\ \text{K}}=0.0821\ \text{dm}^3\,\text{atm}\,\text{mol}^{-1}\text{K}^{-1}

Using the modern "standard conditions" reference point of 1 bar (10510^5 Pa), 22.71 ×10−3\times10^{-3} m3^3, for 1 mole at 273.15 K:

R=105 Pa×22.71×10−3 m31 mol×273.15 K=8.314 Pa m3K−1mol−1R=\frac{10^5\ \text{Pa}\times22.71\times10^{-3}\ \text{m}^3}{1\ \text{mol}\times273.15\ \text{K}}=8.314\ \text{Pa}\,\text{m}^3\text{K}^{-1}\text{mol}^{-1}

which, converted, is the same constant expressed in several equivalent forms:

R=8.314×10−5 bar m3K−1mol−1=8.314×10−2 bar dm3K−1mol−1=8.314×10−2 bar L K−1mol−1=8.314 J K−1mol−1R=8.314\times10^{-5}\ \text{bar}\,\text{m}^3\text{K}^{-1}\text{mol}^{-1}=8.314\times10^{-2}\ \text{bar}\,\text{dm}^3\text{K}^{-1}\text{mol}^{-1}=8.314\times10^{-2}\ \text{bar}\,\text{L}\,\text{K}^{-1}\text{mol}^{-1}=8.314\ \text{J}\,\text{K}^{-1}\text{mol}^{-1} …

Misc 6.3-worked-sf6Worked example: pressure of sulphur hexafluoride from the ideal gas equation

Worked out. For n=2n=2 mol of SF6_6 in a V=6 dm3V=6\ \text{dm}^3 steel vessel at 70 ∘C70\ ^\circ\text{C} (T=343T=343 K), using R=0.0821 dm3 atm K−1mol−1R=0.0821\ \text{dm}^3\,\text{atm}\,\text{K}^{-1}\text{mol}^{-1}: P=nRTV=2×0.0821×3436≈9.39P=\dfrac{nRT}{V}=\dfrac{2\times0.0821\times343}{6}\approx9.39 atm. …