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Gas Laws Physical Characteristics of Gases Physical Characteristics Typical Units Volume, V liters (L) Pressure, P Kilopascals (kPa) or atmosphere (atm) Temperature, T Kelvin (K) Number of atoms or molecules, n mole (1 mol = 6.023x1023 atoms or molecules) 8-Boyle’s Law Pressure and volume are inversely related at constant temperature. PV = K As one goes up, the other goes down. P1V1 = P2V2 “Father of Modern Chemistry” Robert Boyle Chemist & Natural Philosopher Listmore, Ireland January 25, 1627 – December 30, 1690 Boyle’s Law: P1V1 = P2V2 9 & 10-Charles’ Law Volume of a gas varies directly with the absolute temperature at constant pressure. V = KT V1 / T1 = V2 / T2 Jacques-Alexandre Charles Mathematician, Physicist, Inventor Beaugency, France November 12, 1746 – April 7, 1823 Charles’ Law: V1/T1 = V2/T2 Charles’ Law: V1/T1 = V2/T2 10- Gay-Lussac Law At constant volume, pressure and absolute temperature are directly related. P=kT P1 / T1 = P2 / T2 Joseph-Louis Gay-Lussac Experimentalist Limoges, France December 6, 1778 – May 9, 1850 11-Avogadro’s Law: V1/n1=V2/n2 12- Avogadro’s Hypothesis • Equal volumes of gases at the same P and T all contain the same number of molecules. Ex: at STP conditions, the following is true: 2H2 2mol H2 44.8L H2 12.04 x 1023 + O2 --> 1mol O2 22.4L O2 6.02 x 1023 2H2O 2mol H2O 44.8L H2O 12.04 X 1023 That is, VOLUMES of gases at same T and P are in same ratios as the coefficients (like moles are!) 13- Combined Gas Law • Combination of ‘minor’ gas laws which • enables the solving of problems involving changes of P, V, T and n all together P1V1 = P2V2 n1T1 n2T2 If a variable is held constant, omit it from the formula If a variable is doubled with no initial value given, just substitute it for x and 2x or ‘10’ and ‘20’ Example • 2 mol of gas X has a volume of 1L at 25oC. If its pressure is tripled and temperature raised to 100oC, what would be its new volume? n1 = n2 = 2 mol (variable can be eliminated) V1 = 1L T1 = 25 oC + 273 = 298K V2 = ? T2 = 100oC + 273 = 373K V2 = P1V1T2 = (x)(1L)(373K) T1P2 (298K)(3x) P1 = x P2 = 3x = 0.417L 14- Molar Volume • Is the volume of ONE MOLE of any gas • • at a specific T and P This value changes with any change in P and T At STP molar volume = 22.4 L Ideal Gas vs Real Gas • Ideal Gases Hypothetical gases that ‘obey’ all gas laws under all conditions of P and T Don’t exist in real life Gas laws assume that gases behave as ideal gases under any T and P Assume that at any P and T ideal gases: Exhibit no attraction between particles ie: won’t liquefy or solidify Occupy no volume at zero kelvin • Real Gases In reality, all gases will eventually liquefy or solidify at very low temperatures or high pressures. At these conditions , particles attract each other Gas Laws are therefore not applicable under these conditions of T and P 15- Ideal Gas Law • Involves the collective relationship • between the P, V, n and T of ANY gas. PV = nRT P = Pressure of gas V = Volume of gas n = moles of gas T = Temperature (Kelvin) R = Universal gas constant 16 - Universal Gas Constant (R) • • • Value applied to make ideal gas law a valid math equation Value changes depending on pressure UNIT Calculated at STP conditions for one mole of any gas If P in atm, R calculated by: R = PV = 1 atm x 22.4L = 0.082 atm L nT 1mol x 273K mol K If P in kPa, R calculated by: R = PV = 101.3 kPa x 22.4L = 8.31 kPa L nT 1mol x 273K mol K Ideal Gas Law Example • n=? What mass of Ne gas must be put into an 880ml rigid vessel to obtain a gas with a P of 240kPa and temperature of 35oC? P = 240kPa V = 880ml (0.88L) T = 35oC + 273 = 308K n = PV = 240kPa x 0.88L = 0.083mol Ne gas RT 8.31 kPa L x 308K mol K Mass Ne = 0.083mol x 20g/1mol = 1.66g 17 - Partial Pressure of Gas • Note: equal VOLUME of gases (containing same mols of gas) at same T apply: THE SAME PRESSURE Dalton’s Law of Partial Pressure • • Ptotal= P1 + P2 + P3… Example #1 2L of gas A, 4L of gas B and 8L of gas C all at the same temperature are put in a container. The total pressure of the mixture is 235kPa. What is the partial pressure (PP) of each? 1) PPA = 2L x 235kPa = 33.6kPa 14L 2) PPB = 4L x 235kPa = 67.1kPa 14L 3) PPC = 8L x 235kPa = 134.3kPa 14L Ptotal = 235kPa Dalton’s Law Ex #2 • 7.84L of CO2 and 0.40 mol H2 at SATP are put into a 10.5L vessel. The temperature of the mixture drops to -12.5OC. What is the PP of each gas? (SATP same as STP BUT at 25OC) 1) nCO2 = PV = 101.3kPa x 7.84L = 0.32mol CO2 RT 8.31 x 298K 2)ntotal = 0.32 mol + 0.40 mol = 0.72 mol 3) Ptotal = nRT = 0.72 mol x 8.31 x 260.5 K = 148.4 kPa V 10.5L 4) PPCO2 = 0.32/0.72 x 148.4kPa = 65.96 kPa PPH2 = 0.4/0.72 x 148.4kPa = 82.44kPa STP vs SATP • SATP • Standard ambient temperature and pressure 25oC (298k) 101.3kPa STP Standard temperature and pressure 273K 101.3kPa 18- How to calculate the molar volume of a gas at any T and P • Ex: What is the molar volume of CO2 gas at 890kPa and 15oC? •P = 890 kPa •T= 288 K •n= 1mol •R = 8.31 kPa L mol K V= nRt P V= 1mol x 8.31 x 288 890 kPa V= 2.69 L *The volume of a gas is affected by any change in T and P but 19- Density of Gases •D=m • V If any 2 equivalent variables of a gas are known, they can be used as a conversion factor to solve a problem. Example • The density of a gas is 1.8g/L at 25OC and 101.3kPa. What is its molar mass? Volume for one mol: V=? n= 1 mol T= 298K P=101.3kPa V=nRT = 1 x 8.31 x 298 = 24.45L 101.3 Molar Mass = 24.45L x 1.8g = 44g 1L Example #2 • In the reaction N2 + 3H2 -> 2NH3 , if 4.7mol H2 are reacted, what volume of NH3 would be produced at a T of 45OC and P of 156 kPa? a) 4.7mol H2 x 2mol NH3 = 3.13 mol NH3 produced 3mol H2 b) VNH3 = nRT = 3.13 x 8.31 x 318K = 53.02L P 156 kPa Example #3 • For the equation in example 2, if 24.44L NH3 are produced at 1.43 atm and 550C, what mass of H2 must be reacted? V= 24.44L N = PV P = 1.43atm x 101.3kPa 1 atm RT = 144.9kPa x 22.4L = 144.9 kPa 8.31 x 328K T = 550C + 273 = 328K n= ? = 1.3mol 1.3mol NH3 x 3mol H2 = 1.95molH2 x 2g 2mol NH2 1mol = 3.9gH2 20- Relative Masses of Gases • Remember Avagadro’s Hypothesis!! • Ex #1 What is the mass of 15.68L of gas X at 350C and 116kPa, if its 1.8 times more dense than O2? a) 32g/mol x 1.8 = 57.6g/mol Gas X b) n= PV = 116kPa x 15.68L = 0.71mol Gas X RT 8.31 x 308K c) 0.71 mol x 57.6g/mol = 40.9g Example #2 • At SATP, 7.35 L of gas A weighs 17.5g. • • What is its molar mass? V= nRT = 1 x 8.31 x 298 = 24.5L P 101.3kPa Molar Mass = 24.5L x 17.5g = 58g/mol mol 7.35L 21- Graham’s Law of Diffusion • Diffusion Process whereby a substance spreads spontaneously in all directions from regions of high to low concentration At the same T gases diffuse/move at different rates The lighter the gas, the faster it diffuses Graham’s Law • The relative rate of diffusion of 2 gases is inversely proportional to the square root of their molar masses or densities. ______________ RateA = √molar mass B RateB √molar massA Example #1 Calculate the RR (Relative Rate) of diffusion of H2 and CO2 __________ √44g/mol = 6.63 = 4.7* √2g/mol 1.41 • There is no unit for the RR of diffusion • The 4.7 indicates that H2 diffuses at a rate 4.7 times faster than CO2