Gases Chapter 9 What parameters do we use to describe gases? pressure: force/unit area 1 atm = 101 kPa; volume: liters (L) Temperature: K Amount of gas: moles A gas consists of small particles that move rapidly in straight lines until they collide; they have enough kinetic energy to overcome any attractive forces; gas molecules are very far apart; gas molecules have very small volumes compared to the volumes of the containers they occupy; have kinetic energies that increase with an increase in temperature; collisions of the gas cause pressure (force /unit area) What is meant by % volume? In principle, the actual molecular volume of a gas is so small in comparison to the volume it will occupy that we treat gases at mathematical points. What is air pressure due to? A column of air 1 m2 has a mass of 10,300 kg, producing a pressure of 101 kPa due to gravity (14.7 pounds/in2) 1 atm = 76 cm Hg; 101 kPa Atmospheric pressure is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth; Pressure is: about 1 atmosphere at sea level; depends on the altitude and the weather; is lower at high altitudes where the density of air is less; How do we measure pressure? A barometer measures the pressure exerted by the gases in the atmosphere indicates atmospheric pressure as the height in mm of the mercury column A water barometer would be 13.6 times taller than a mercury barometer because the density of Hg is 13.6 times as dense as water • What is the relationship between the variables used to describe gases? Boyle’s Law is concerned with the relationship of pressure and volume using a fixed amount of gas ( a fixed number of mols of gas) P*V = constant at constant temperature Initial Final Note that the initial product between pressure and volume is 4L*1atm = 4 L atm In the final diagram: 2L*2atm = 4 Latm Decreasing the volume to ½ results in a doubling of the pressure We find that the product of volume and pressure is equal to a constant PV = a constant Suppose we had a frictionless piston and a gas Temperature and Volume enclosed within the piston. What could we say about the pressure inside and outside of the piston if the piston was notionless? pressure P (inside) = P (outside) temperature (inside) = temperature (outside) Supppose we now heat the gas inside the piston What happens to the number of collisions a molecule makes against the wall if we heat a gas? the number of collisions increase causing an increase in the pressure inside What will happen to the volume of the gas if the pressure outside the piston does not change and the pressure inside increases? V/T(K) = constant Charles’ Law is concerned with the relationship of temperature and volume when dealing with a constant amount of gas (mols) V ∝ T when T is expressed in K. The K temperature scale is derived from the behavior of gases if V ∝ T then V = kT where k is a constant at constant pressure Suppose now that we double the Pressure and the amount of gas , n number of molecules in the same volume and at the same temperature. What will that do to mols is proportional to the number of collisions with the volume if T and P is walls? kept constant the pressure? What will it do to the volume if we allow the pressure to return to its orginal value? Avogadro’sIf we allow theLaw pressure is concerned to return to itwith theoriginal relationship value, what between will happen the to the volume? number of molecules or mols (n) and the volume of a gas under conditionsn is directly of proportionalconstant pressure to pressure and temperature Summary Ideal gas law: PV = nRT where R is a constant R = 0.0821 L.atm/K.mol Note that at constant n and T, PV = constant Boyle’s Law Note that at constant P and T, V/n = constant Avogadro’s Law Note that at constant P and n, V/T = constant Charles’s Law Standard conditions of pressure and temperature T = 0 °C (273 K) Pressure: 1 atm What volume does a mol of any ideal gas occupy at STP? PV = nRT V = 1mol(0.0821 L*atm/K*mol)(273 K)/(1 atm) V = 22.4 L This means that equal volumes of gases under identical conditions of temperature and pressure contain equal number of molecules What is the difference between an ideal gas and a real gas? The ideal gas equation was generated from the kinetic theory of gases making the following assumptions or approximations: 1. The molecules could be treated as points (ie molecular volume = 0) 2. There are no attractive interactions between molecules. 3. Gas particles move around at random 4. Collision of gas molecules with the wall are totally elastic 5. The kinetic energy of the gas particle is ∝ to temperature (K) In general, the ideal gas law works best at low pressures and high temperatures Real Gases: van der Waal’s equation (P + an2/V2)(V-nb) = n RT an2/V2 corrects for intermolecular attractions nb corrects for the real volume of molecules Dalton’s Law of partial pressures: Total atmospheric pressure = 1 atm; How much of the pressure is contributed by N2? Pressure is a consequence of molecules colliding with each other and the walls of the container *062 For air if If PTV = nTRT and nT =(nO2 + n N2 + ...) at constant T, PTV = (no2 + n N2 + ...)RT Since the actual volume of the molecules is small in comparison to the volume occupied by the gas, all molecule occupy the same volume V. The contribution to the total pressure is dependent on the number of collision of each gas with the wall and this is dependent on the number of molecule of each gas. Hence: P = (PN2 + PO2 + ...) PO2V = nO2RT ; PN2V = nN2RT ... Temperature: a measure of the average kinetic energy of molecules Distribution of molecular speeds as a function of temperature What are some of the consequences associated with the fact that molecules at the same temperature have different speeds? Diffusion: mixing of gases at Effusion: escape the same pressure through a small opening The size of the pinhole needs to be small for effusion *07 . .. .. Two molecules of different mass at the same temperature effusing through an opening If M > m, then : 1/2Mv2 = 1/2mV2 V > v . .. .. Two molecules of different mass at the same temperature effusing through an opening From the kinetic theory of gases speed of a molecule v = (3RT/M)1/2 For two gases at the same temperature 2 2 1/2mava = 1/2mbvb va = average speed of molecule a vb = average speed of molecule b 2 2 1/2 ma /mb = vb /va or vb /va = (ma /mb ) The rate at which molecule a hits the pinhole ∝ v if the comparisons are made at the same concentration and temperature. If you wait long enough ... .. .. .. Two molecules of different mass at the same temperature effusing through an opening at equilibrium Solving some problems involving gases 1. A sample of gas at 25 °C and 2 atm pressure in a 5 L vessel was found to have a mass of 18 g. What is its molecular weight? PV = n RT 2 atm*5 L = n*0.0821 (Latm/K mol)*298 K n = 10/(0.0821*298) mol; 0.4087 n = wt/ mw; 0.4087 = 18g/mw mw = 44 g/mol Suppose the gas at the right exerted a pressure of 15 cm as shown. Would the pressure of the gas be greater or less than 1 atm? How many atm of pressure is 15.2 cm the gas exerting? 1 atm = 76 cm Hg 76 - 15.2 = 60.8 Suppose the gas at the right exerted a pressure of 15 cm as shown. Would the pressure of the gas be greater or less than 1 atm? How many atm of pressure is 15.2 cm the gas exerting? 1 atm = 76 cm 76 - 15.2 = 60.8 Suppose we have a sample of equal amounts of H2 and D2 in a vessel and a small opening is introduced. What will be the initial rates of effusion? 0.5 0.5 vH2/vD2 = (mD2/mH2) = (4/2) = 1.42 Will the relative rate change with time? What is the density of natural gas (CH4) at STP? PV = nRT density is g/mL or g/L We know the molar volume of any gas is 22.4 L at STP How many g of methane in a mole? 16g/22.4 L = 0.714 g/l or 7.14*10-4 g/mL or PV =(wt/mw)*RT; mw*P/RT = (wt/V) The surface temperature of Venus is about 1050 K and the pressure is about 75 Earth atmospheres. Assuming these conditions represent a “Venusian STP, what is the standard molar volume of a gas on Venus? PV = nRT 75 atmV =1mol*0.0821(Latm/K mol)*1050 K; V = 1.15 L Natural gas is a mixture of a number of substances including methane (mol fraction, 0.94); ethane (mol fraction, 0.04); propane (mol fraction, 0.015). If the total pressure of the gases is 1.5 atm, calculate the actual pressure contributed by each of the gases described. mol fraction = mol A/(mol A + mol B + ....) PT = 1.5 = P CH4+ PC2H6 + ... Px V = nxRT nCH4/nC2H6 = PCH4/PC2H6 = 0.94/.04 CH4 = 0.94*1.5 C2H6 = 0.04*1.5 C3H8 = 0.015*1.5.