A Level Chemistry Measuring moles of GASES by measuring their volume, pressure and temperature PV = nRT

 molecules move freely (no inter-molecular forces) and randomly FACTORS AFFECTING THE VOLUME OF A GAS Gas  molecules move freely (no inter-molecular forces) and randomly  gas mols will always DIFFUSE to fill whatever volume is available. Volume (V) of a gas depends on the : 1. number of moles (n) of gas 2. pressure (P) applied to the gas 3. temperature (T) of the gas Volume does NOT depend on the type of gas!

AVOGADRO’S LAW: V n V  n provided P and T fixed V = k1n

BOYLE’S LAW: P V V  1/P provided n and T fixed PV = k2

CHARLES’S LAW: V  T V = k3T provided n and P fixed 0oC 273K V T /K provided n and P fixed V  T and T measured in KELVIN V = k3T T (/K) = T(/OC) + 273

the IDEAL GAS EQUATION PV = nRT All 3 laws can be combined into one equation the IDEAL GAS EQUATION PV = nRT where "R" is called the UNIVERSAL GAS CONSTANT NB : Must use appropriate S.I. units R = 8.31 J mol-1 K-1 VARIABLE V P T n S.I. UNIT m3 Nm-2 (or Pa) K Moles

Notes : 1. Nm-2 = Pascal = "Newtons per square metre" (1 kg spread over 1m2  approx. 10Pa) 2. 1 kPa = 1000 Pa 3. 1m3 = 1000 dm3 (1000 litres) = 1000000 cm3 4. Use PV = nRT to calculate n, given P,V and T  calculate Mr if also know mass, m (/g) Mr = m / n

IDEAL GAS EQUATION not perfect at all T and P for any gas OK within experimental error provided : a. P is not too high b. T is not too low c. Molecules are not too large. ie equation less reliable if gas is close to liquefying A truly "IDEAL" gas would OBEY the equation at ALL T and P. NO SUCH GAS EXISTS! He and H2 are the nearest because these show the weakest intermolecular forces

Q1. Calculate the number of moles present, if a Q1 Calculate the number of moles present, if a gas exerts 110 kPa pressure at 20oC in a volume of 20.0 dm3 P V n R T PV = nRT 110 kPa = 110000 Pa RT PV n = 20.0 dm3 = 0.020 m3 = 110000 x 0.020 8.31 x 293 ???? = 0.904 moles 8.31 J mol-1 K-1 20oC = 293 K

PV = nRT P V n R T 100 kPa = 100000 Pa RT PV n = 100 cm3 = 0.0001 m3 = Q2 A sample of gas weighs 0.272g. It has a volume of 100cm3 at a temperature of 16oC and pressure 100 kPa. Calculate the number of moles of gas in the sample and hence the mass of one mole of the gas. P V n R T PV = nRT 100 kPa = 100000 Pa RT PV n = 100 cm3 = 0.0001 m3 = 100000 x 0.0001 8.31 x 289 ???? = 4.16 x 10-3 moles 8.31 J mol-1 K-1 Mr = m/n = 0.272 4.16 x 10-3 16oC = 289 K = 65.4 g

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