Chemistry 2nd Paper · Classes 11-12 · Chapter 1 · 1.3.6

The Ideal Gas Equation.

How many molecules are in a balloon, or in a single breath? Boyle, Charles and Avogadro each gave a separate relation. Put the three together and the four quantities P, V, n and T are tied into one equation: PV = nRT. From that one equation you can get moles, mass, density and molar mass.

1. Three laws, one equation

Boyle's law, with n and T fixed, gives V ∝ 1/P. Charles's law, with n and P fixed, gives V ∝ T. Avogadro's law, with P and T fixed, gives V ∝ n. Join the three and you get V ∝ nT/P, and putting in a constant makes PV = nRT. This relation is the ideal gas law, or the ideal gas equation. For one mole n = 1, so PV_m = RT, where V_m is the molar volume.

R is called the universal gas constant, or the molar gas constant. Why universal? At the same temperature and pressure, one mole of any gas has the same molar volume. So whether the gas is hydrogen or oxygen, R keeps the same value.

Where does the unit of R come from? R = PV/nT = (force/area) × V/(nT). Volume divided by area leaves one length, so R = (force × length)/(moles × kelvin). Force times length is work, so the unit of R is work (or energy) per mole per kelvin.

What R means physically: keep the pressure fixed and warm one mole of ideal gas by one kelvin. The work done by the expansion is exactly R.

Pressure = force/area, so the SI unit of pressure is N m⁻² = kg m⁻¹ s⁻². You will need this in the next part, when the units of P and V decide which R to use.

2. The value of R in each unit system

Take one mole and R comes out of PV = RT. In litre-atmosphere units V = 22.414 L, P = 1 atm and T = 273.15 K, so R = 1 × 22.414/(1 × 273.15) = 0.082058 L atm mol⁻¹ K⁻¹. In SI units put V = 22.414×10⁻³ m³, P = 101.325×10³ N m⁻² and T = 273.15 K, and R = 8.314 J mol⁻¹ K⁻¹, because 1 N m = 1 J.

In CGS units P = 76 × 13.6 × 981 dyne cm⁻², V = 22414 cm³ and T = 273.15 K give R = 8.32×10⁷ erg K⁻¹ mol⁻¹, because 1 dyne cm = 1 erg. Options may write it as 8.314×10⁷ or 8.316×10⁷ erg; it is the same value. To reach calories, divide the joule value by 4.184 (1 cal = 4.184 J): 8.32 ÷ 4.184 = 1.987 cal K⁻¹ mol⁻¹. In engineering units P = 14.7 × 144 lb ft⁻², V = 359 ft³ and T = 273 K give R = 2783.63 ft lb mol⁻¹ K⁻¹.

The five values at a glance: 0.082 L atm mol⁻¹ K⁻¹, 8.314 J mol⁻¹ K⁻¹ (SI), 8.32×10⁷ erg mol⁻¹ K⁻¹ (CGS), 1.987 cal mol⁻¹ K⁻¹ and 2783.63 ft lb mol⁻¹ K⁻¹. The calorie unit is used only for the food value of foodstuffs. The number changes, the meaning does not; now drop each value into its unit system below.

3. The Boltzmann constant

R counts per mole. The gas constant per molecule is the Boltzmann constant k, and k = R/N_A.

In L atm units k = 0.082/6.022×10²³ = 1.36×10⁻²⁵ L atm K⁻¹ molecule⁻¹. In SI units k = 8.314/6.022×10²³ = 1.38×10⁻²³ J K⁻¹ molecule⁻¹.

4. Molar mass and density

If the mass of gas is w and its molar mass is M, then n = w/M. Put that into PV = nRT and PV = wRT/M, that is M = wRT/PV.

Density is d = w/V. Turn the same equation round and you get d = PM/RT.

Two things follow from d = PM/RT. At constant temperature d ∝ P, so raising the pressure raises the density. At constant pressure and temperature d ∝ M, so a heavier gas is denser.

When a mass is given, find n = w/M first and then write the equation. For 44 g of CO₂ n = 1, so PV = RT. For 8 g of He n = 2, so PV = 2RT. For 4 g of H₂ n = 2, so PV = 2RT.

5. Working with the equation

Four quantities, one equation. Lock one, move the other three, and switch the units to see R.

Now do it by hand. Choose the unknown, give the other three, and put T in kelvin. The examples below follow the book's problems 1.25, 1.30 and 1.31.

Problem 1.21: gas at 27°C has its pressure raised from 0.8 to 1.18 atm, and its volume must double. T₂ = 1.18 × 2x × 300/(0.8 × x) = 885 K = 612°C. So the temperature must rise by 585°C.

Problem 1.22: take 1000 mL of gas at standard conditions to 27°C and 74/76 atm, and V₂ = 1 × 1 × 300/(273 × 74/76) = 1.1286 L.

Problem 1.23: 3×10⁻³ m³ of gas at STP has mass 4.5×10⁻³ kg. Then 4.0×10⁻⁴ m³ at 27°C and 101.99 kPa has mass 4.5×10⁻³ × 3.664×10⁻⁴/(3×10⁻³) = 5.496×10⁻⁴ kg.

Problem 1.24: a gas with a solid present has volume 0.12 dm³ at 20°C, and 71.5 cm³ at 47°C when the pressure is doubled. From (120 − x)/293 × P = 2P(71.5 − x)/320 the solid's volume is x = 13.154 cm³.

Problem 1.25: moles of 20 L of methane at 27°C and 77/76 atm are n = PV/RT = (77/76) × 20/(0.082 × 300) = 0.8237 mol. Problem 1.26: 1 mL of gas at 30°C and 0.987 atm has n = 0.987×10⁻³/(0.082 × 303) = 3.97×10⁻⁵ mol, that is 2.3907×10¹⁹ molecules.

Problem 1.27: 250 mL of methane at 27°C and 102 kPa has n = (102/101.325) × 0.25/(0.082 × 300) = 0.010230302 mol, so its mass = 0.1637 g.

Problem 1.28: if 0.842 g of gas occupies 400 mL at 20°C and 740 mm Hg, M = wRT/PV = 0.842 × 0.082 × 293 × 760/(740 × 0.4) = 51.942 g mol⁻¹. Problem 1.29: if 250 mL of gas at 30°C and 95 kPa has mass 0.2 g, M = 0.2 × 0.082 × 303 × 101.325/(95 × 0.25) = 21.2 g mol⁻¹ (as printed in the book).

Problem 1.30: the volume of 15 g of N₂ (M = 28) at 27°C and 0.526 atm is V = wRT/PM = 15 × 0.082 × 300/(0.526 × 28) = 25.05 L. Problem 1.31: the pressure of 12 g of CO₂ in a 5 L vessel at 30°C is P = wRT/VM = 12 × 0.082 × 303/(5 × 44) = 1.35524 atm.

Problem 1.32: the density of CO₂ at 100°C and 1.0526 atm is d = PM/RT = 1.0526 × 44/(0.082 × 373) = 1.514 g L⁻¹. Problem 1.33: a 10 L evacuated cylinder holds 2 g of H₂ and 2 g of D₂ at 27°C, so n = 1 + 0.5 = 1.5 mol and the total pressure is P = 1.5 × 0.082 × 300/10 = 3.69 atm.

Problem 1.34: an adult's lung capacity is 3.8 L, so at 1 atm and 37°C n = 1.0 × 3.8/(0.08206 × 310) = 0.15 mol. Problem 1.35: 1.0 L of ammonia at 25°C and 733.4 mm Hg has mass 0.672 g. At STP V₀ = 733.4 × 1.0 × 273/(298 × 760) = 0.884 L, so the density d = 0.672/0.884 = 0.760 g L⁻¹.

Problem 1.36: an unknown gas has density 5.380 g L⁻¹ at 15°C and 736 mm Hg. n = (736/760) × 1.0/(0.08206 × 288) = 0.041 mol and M = 5.380/0.041 = 131 g mol⁻¹, so the gas is probably xenon (atomic mass 131.3). In the same way, if 3.2 g of a gas occupies 2.24 L at STP, n = 0.1 mol and M = 3.2/0.1 = 32 g mol⁻¹, so the gas is O₂.

Counting molecules: 0.2 L of air at 27°C and 1.0 atm has n = PV/RT = 0.2/(0.0821 × 300) = 8.12×10⁻³ mol, that is 8.12×10⁻³ × 6.022×10²³ = 4.89×10²¹ molecules. Likewise 11.2 L of CO₂ at 1 atm and 25°C has n = 11.2/(0.0821 × 298) = 0.458 mol, which is 2.76×10²³ molecules.

For equal masses of gases at the same temperature and volume, the gas with the larger molar mass has fewer moles and so a lower pressure: P ∝ 1/M.

Vessels of the same volume, pressure and temperature hold the same number of molecules, so monatomic gases such as He, Ne, Ar and Kr also hold equal numbers of atoms. And since d ∝ M, the heaviest, Kr, has the highest density.

Volume of steam: 1 kg of water vapour at 100°C and 1 atm has n = 1000/18 = 55.6 mol, so V = nRT/P = 55.6 × 0.0821 × 373/1 = 1701 L. The other way round: O₂ at 15.5 atm in a 10 dm³ cylinder at 30°C has n = 15.5 × 10/(0.0821 × 303) = 6.23 mol, so its mass is 6.23 × 32 = 199.4 g.

6. Check yourself

The first question is on this topic's formulas.

In the second, watch the units and the value of R.

The last is a calculation.

The whole note on seven cards.

Source: Chemistry 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Environmental Chemistry”, section 1.3.6.