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Concept 01 / 07States of Matter

Gas Laws: Boyle, Charles, Gay-Lussac and Avogadro

The empirical gas laws relating pressure, volume, temperature and amount, the absolute temperature scale from Charles' law, graphical forms (isotherms, isobars, isochores), the combined gas law and typical applications such as bubbles rising in a lake.

Figure 1
The gaseous state is the simplest state of matter: molecules are far apart, intermolecular forces are weak, and gases fill any container. Four measurable quantities describe a gas: pressure PP, volume VV, temperature TT and amount nn. The gas laws, found experimentally, relate them two at a time. Boyle's law (1662). At constant TT and nn, P∝1VP \propto \frac{1}{V}, that is PV=PV = constant and P1V1=P2V2P_{1}V_{1} = P_{2}V_{2}. A plot of PP against VV is a rectangular hyperbola (an isotherm); higher temperatures give isotherms farther from the axes. Plots of PP against 1V\frac{1}{V} are straight lines through the origin, and PVPV against PP is a horizontal line. Since density d=mVd = \frac{m}{V}, at constant TT density is proportional to pressure: air is denser at sea level than on a mountain. Charles' law (1787). At constant PP and nn, V∝TV \propto T where TT is absolute temperature: V1T1=V2T2\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}. In Celsius, Vt=V0(1+t273.15)V_{t} = V_{0}\left(1 + \frac{t}{273.15}\right): volume changes by 1273.15\frac{1}{273.15} of its 0∘C0^{\circ}\mathrm{C} value per degree. Every isobar, extrapolated, reaches zero volume at −273.15∘C-273.15^{\circ}\mathrm{C}. This defines absolute zero and the Kelvin scale T=t+273.15T = t + 273.15. (Real gases liquefy long before this.) Gay-Lussac's law. At constant VV and nn, P∝TP \propto T: P1T1=P2T2\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}. A pressure cooker or a tyre heated in summer shows it. The graphs of PP against TT are isochores. Avogadro's law (1811). At constant TT and PP, V∝nV \propto n: equal volumes of all gases contain equal numbers of molecules. The molar volume at STP (273.15 K273.15\,\mathrm{K}, 1 bar1\,\mathrm{bar}) is 22.7 L22.7\,\mathrm{L}; at 273.15 K273.15\,\mathrm{K}, 1 atm1\,\mathrm{atm} it is 22.4 L22.4\,\mathrm{L}. Combined gas law. For a fixed amount, P1V1T1=P2V2T2\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}. Always use kelvin; pressure and volume units only need to match on both sides. Units. 1 atm=101.325 kPa=760 mm Hg=760 torr=1.01325 bar1\,\mathrm{atm} = 101.325\,\mathrm{kPa} = 760\,\mathrm{mm\,Hg} = 760\,\mathrm{torr} = 1.01325\,\mathrm{bar}; 1 bar=105 Pa1\,\mathrm{bar} = 10^{5}\,\mathrm{Pa}. Key Takeaways: - Boyle PV=PV = const (TT, nn fixed); Charles VT=\frac{V}{T} = const (PP, nn fixed); Gay-Lussac PT=\frac{P}{T} = const (VV, nn fixed); Avogadro Vn=\frac{V}{n} = const (TT, PP fixed). - Temperatures in gas laws must be in kelvin. - Isotherm: hyperbola in PP-VV; isobar: line through 0 K0\,\mathrm{K} in VV-TT. - Molar volume: 22.4 L22.4\,\mathrm{L} at 0∘C0^{\circ}\mathrm{C}, 1 atm1\,\mathrm{atm}; 22.7 L22.7\,\mathrm{L} at 0∘C0^{\circ}\mathrm{C}, 1 bar1\,\mathrm{bar}. - Pitfall: doubling Celsius temperature does not double volume; convert to kelvin first. Brain-Teaser 1: An air bubble at the bottom of a lake, where the pressure is 3 atm3\,\mathrm{atm} and the temperature 7∘C7^{\circ}\mathrm{C}, rises to the surface, where the pressure is 1 atm1\,\mathrm{atm} and the temperature 27∘C27^{\circ}\mathrm{C}. By what factor does its volume change? Solution: V2V1=P1T2P2T1=3×3001×280=3.21\frac{V_{2}}{V_{1}} = \frac{P_{1}T_{2}}{P_{2}T_{1}} = \frac{3 \times 300}{1 \times 280} = 3.21. The volume grows about 3.23.2 times: mostly from the drop in pressure, a little from the warming. Brain-Teaser 2: An open flask contains air at 27∘C27^{\circ}\mathrm{C}. To what temperature must it be heated so that one-third of the air escapes? Solution: Pressure and volume are fixed (open flask), so nT=nT = constant. If one-third escapes, n2=23n1n_{2} = \frac{2}{3}n_{1} and T2=32T1=32×300=450 K=177∘CT_{2} = \frac{3}{2}T_{1} = \frac{3}{2} \times 300 = 450\,\mathrm{K} = 177^{\circ}\mathrm{C}. Brain-Teaser 3: A gas occupies 500 mL500\,\mathrm{mL} at 1 bar1\,\mathrm{bar}. To what pressure must it be compressed at constant temperature to occupy 200 mL200\,\mathrm{mL}, and what happens to its density? Solution: P2=P1V1V2=1×500200=2.5 barP_{2} = \frac{P_{1}V_{1}}{V_{2}} = \frac{1 \times 500}{200} = 2.5\,\mathrm{bar}. Mass is unchanged and volume falls by the factor 2.52.5, so density rises 2.52.5 times, in proportion to pressure.