Use of the Ideal Gas Equation What volume does 1 mole of gas occupy at STP (Standard Temperature and Pressure)? STP corresponds to T = 273 K and P = 1 atm PV = nRT (1 atm) x V = (1 mole) x (0.08206 L atm mol-1 K-1) x (273.15 K) V = 22.414 L or 101325 Pa x V = (1 mol) x (8.3145 J K-1 mol-1) x (273.15 K) V = 0.022414 m3 = 22.414 L

Scuba Diving A scuba diver’s tank contains 0.3 kg O2 in a volume of 2.32 L. Estimate the gas pressure at 5 °C and the volume it would occupy at 30 °C at atmospheric pressure. 0.3 kg = 300 g ≡ (300 g) / (32 g mol-1) = 9.375 mole Hence at 5 °C (278K), P = (9.375 mol) x (0.08206 atm L K-1 mol-1) x (278 K) (2.32) L = 92.2 atm V = (9.375 mol) x (0.08206 atm L K-1 mol-1) x (308 K) 1 atm = 237 L

More Calculations So what is the number density of molecules at atmospheric pressure (the number of molecules per unit volume)? 1 mole occupies 22.414 L. Hence at STP there are 2.687 x 1025 molecules m-3 or 2.687 x 1019 cm-3 So what is the average distance between molecules? 2.687 x 1025 m-3 - hence each molecule has an effective volume available to it of (1 / 2.687 x 1025) = 3.72 x 10-26 m3. Take the cube root of both sides of this equation: (3.72 x 10-26)1/3 = 3.34 x 10-9 m = 3.3 nm

So Our Picture of a Gas Becomes …. e.g., air at STP Average distance between molecules ~3 nm ~0.15 nm N2 bond length

Density of a Gas Because PV = nRT and n = m/M (mass / rel mol mass) we can write: PV = mRT / M Or density = m/V = PM / RT e.g., calculate the density of H2 at 1.32 atm and -45 ºC density = (1.32 atm) x (2.02 g mol-1) (0.08206 atm L K-1 mol-1) x (228 K) = 0.143 g L-1 (care with units)

Dalton’s Law of Partial Pressures Very often we encounter mixtures of gases rather than pure gases (e.g., air contains N2, O2, Ar, CO2 etc.) The partial pressure of a gas in a mixture is defined to be the pressure it would exert if it were the only gas present. According to Dalton’s Law The total pressure of a mixture of gases is the sum of the partial pressures of the individual gases. So, e.g., for air: Pair = PN2 + PO2 + PCO2 + PAr + …..

Sample Problem Involving Partial Pressures If we have 2 moles of H2, 4 moles of O2, and 6 moles of He in a 5 liter vessel at 27 C, determine the partial pressure of each gas and the total pressure of the mixture. P(total) = 9.85+19.7+29.55 = 59.10 atm Solution: Convert the temperature to Kelvin: K = 27 + 273 = 300 K Use the ideal gas law for each gas: P(H2) = nRT / V = (2 moles H2) (0.0821 liter-atm / mol-K) (300 K) / 5 liters = 9.85 atm P(O2) = nRT / V = (4 moles O2) ( 0.0821 liter-atm / mol-K) ( 300 K) / 5 liters = 19.7 atm P(He) = nRT / V = ( 6 moles He) ( 0.0821 liter-atm / mol-K) ( 300 K) / 5 liters = 29.55 atm