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Concept 02 / 07States of Matter

Ideal Gas Equation and Dalton's Law of Partial Pressures

Combining the gas laws into PV = nRT, values and units of R, density and molar mass from the gas equation, Dalton's law of partial pressures, mole fractions, and correcting for aqueous tension when a gas is collected over water.

Figure 1
Ideal gas equation. Combining Boyle, Charles and Avogadro, V∝nTPV \propto \frac{nT}{P}, so PV=nRTPV = nRT. An ideal gas obeys this exactly at all conditions. Real gases approach it at low pressure and high temperature. The gas constant RR has the same value for all gases: R=8.314 J mol−1 K−1=0.0821 L atm mol−1 K−1=0.08314 L bar mol−1 K−1=1.987 cal mol−1 K−1R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}} = 0.0821\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}} = 0.08314\,\mathrm{L\,bar\,mol^{-1}\,K^{-1}} = 1.987\,\mathrm{cal\,mol^{-1}\,K^{-1}}. Choose RR to match the units of PP and VV: litres and atmospheres go with 0.08210.0821; SI units (Pa\mathrm{Pa}, m3\mathrm{m^{3}}) go with 8.3148.314. Density and molar mass. Writing n=mMn = \frac{m}{M} gives PV=mMRTPV = \frac{m}{M}RT, so d=mV=PMRTd = \frac{m}{V} = \frac{PM}{RT} and M=dRTPM = \frac{dRT}{P}. At fixed TT and PP, gas density is proportional to molar mass. The vapour density (relative to hydrogen) is VD=M2\text{VD} = \frac{M}{2}. Dalton's law of partial pressures (1801). In a mixture of non-reacting gases, the total pressure equals the sum of the partial pressures, each being the pressure that gas would exert alone in the whole volume: Ptotal=p1+p2+p3+…P_{\text{total}} = p_{1} + p_{2} + p_{3} + \ldots Since pi=niRTVp_{i} = \frac{n_{i}RT}{V}, dividing by Ptotal=ntotalRTVP_{\text{total}} = \frac{n_{\text{total}}RT}{V} gives pi=xiPtotalp_{i} = x_{i}P_{\text{total}}, where xi=nintotalx_{i} = \frac{n_{i}}{n_{\text{total}}} is the mole fraction. For gases, mole fraction also equals volume fraction. Gas collected over water. Such a gas is saturated with water vapour, so the pressure of the dry gas is pdry gas=Ptotal−aqueous tensionp_{\text{dry gas}} = P_{\text{total}} - \text{aqueous tension}, where aqueous tension is the vapour pressure of water at that temperature (about 24 mm Hg24\,\mathrm{mm\,Hg} at 25∘C25^{\circ}\mathrm{C}). The average molar mass of a mixture is Mavg=∑xiMiM_{\text{avg}} = \sum x_{i}M_{i}; for air, about 29 g mol−129\,\mathrm{g\,mol^{-1}}. Gases lighter than 2929 (H2\mathrm{H_{2}}, He\mathrm{He}, NH3\mathrm{NH_{3}}, CH4\mathrm{CH_{4}}) rise in air. Key Takeaways: - PV=nRTPV = nRT; d=PMRTd = \frac{PM}{RT}; M=dRTPM = \frac{dRT}{P}. - R=0.0821 L atm mol−1 K−1=8.314 J mol−1 K−1R = 0.0821\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}} = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}. - pi=xiPp_{i} = x_{i}P; mole fraction == volume fraction for gases. - Dry gas pressure == total pressure minus aqueous tension. - Pitfall: Dalton's law applies only to non-reacting gases; NH3\mathrm{NH_{3}} and HCl\mathrm{HCl} react. Brain-Teaser 1: 8 g8\,\mathrm{g} of O2\mathrm{O_{2}} and 4 g4\,\mathrm{g} of H2\mathrm{H_{2}} are placed in a 1 L1\,\mathrm{L} flask at 27∘C27^{\circ}\mathrm{C}. Find the total pressure and the partial pressure of hydrogen. Solution: n(O2)=832=0.25n(\mathrm{O_{2}}) = \frac{8}{32} = 0.25, n(H2)=42=2n(\mathrm{H_{2}}) = \frac{4}{2} = 2, total 2.25 mol2.25\,\mathrm{mol}. P=2.25×0.0821×3001=55.4 atmP = \frac{2.25 \times 0.0821 \times 300}{1} = 55.4\,\mathrm{atm}. x(H2)=22.25=0.889x(\mathrm{H_{2}}) = \frac{2}{2.25} = 0.889, so p(H2)=0.889×55.4=49.3 atmp(\mathrm{H_{2}}) = 0.889 \times 55.4 = 49.3\,\mathrm{atm}. Though the masses differ only twofold, hydrogen supplies 89%89\% of the pressure, because pressure counts molecules, not mass. Brain-Teaser 2: A gas has density 1.25 g L−11.25\,\mathrm{g\,L^{-1}} at 0∘C0^{\circ}\mathrm{C} and 1 atm1\,\mathrm{atm}. Identify a possible gas. Solution: M=dRTP=1.25×0.0821×273=28.0 g mol−1M = \frac{dRT}{P} = 1.25 \times 0.0821 \times 273 = 28.0\,\mathrm{g\,mol^{-1}}. It could be N2\mathrm{N_{2}}, CO\mathrm{CO} or C2H4\mathrm{C_{2}H_{4}} (all 2828). Density alone cannot distinguish them. Brain-Teaser 3: Oxygen collected over water at 27∘C27^{\circ}\mathrm{C} has a total pressure of 760 mm Hg760\,\mathrm{mm\,Hg} in a 250 mL250\,\mathrm{mL} jar. Aqueous tension at 27∘C27^{\circ}\mathrm{C} is 27 mm Hg27\,\mathrm{mm\,Hg}. How many moles of O2\mathrm{O_{2}} were collected? Solution: p(O2)=760−27=733 mm Hg=733760=0.964 atmp(\mathrm{O_{2}}) = 760 - 27 = 733\,\mathrm{mm\,Hg} = \frac{733}{760} = 0.964\,\mathrm{atm}. n=PVRT=0.964×0.2500.0821×300=9.79×10−3 moln = \frac{PV}{RT} = \frac{0.964 \times 0.250}{0.0821 \times 300} = 9.79 \times 10^{-3}\,\mathrm{mol}. Ignoring the water vapour would overestimate this by about 3.7%3.7\%.

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