Question

6. A 120 L container contains 0,750 moles of gas at 23°C. What is the pressure of the gas? (6 pts.) 212 ?L-12 296 7. A gas mixture contains 225 mm Hg N, 312 mm Hg H, and 219 mm Hg 03 a) What is the total pressure of the gas mixture? (5 pts.) PN2X +PH2 + Poz-225+ 3127 'Hz & Poz=225+ 312 1219 mmHg = 756 mmity b) If the mixture is in a 110 liter container at 315 K, how many moles of each gas are present? (8 pts.) C l, NH 8. a) Which gas particles move fastest at 75°C, H Explain your reasoning. (5 pts.) b) Which gas will have the greatest density at 35°C and 745 mm Hg. Xe Cl; NOExplain your reasoning. (5 pts.)

Answer 1

6) Assuming the gas behaves as an ideal gas: PV = nRT P = nRT/V Given, n= 0.750mol; T = 23°C = 23+273 = 296K and R = 0.082057L.atm/kmol P = (0.750*0.082057*296)/12.0 = 1.52 atm Thus pressure of a gas would be 1.52 atm. 7) as per the Dalton's Law of Partial Pressure Ptotal = PN2 + PH2 + PO2 = 225 + 312 + 219 = 756 mmHg Total number of moles (Xtotal) = PV/RT = (756*11.0) / (62.364LmmHg/Kmol*315) = 0.4233 moles Let's say total mole fraction of each gas: XN2 = x1; xH2=x2; x02=x3 PN2 = x 1*Ptotal x1 = PN2/Ptotal = 225/756 = 0.2976 x1 = number of moles of N2 / Xtotal number of moles of N2 = x1*Xtotal = 0.2976*0.4233= 0.126 moles

PH2 = x2*Ptotal x2 = PH2/Ptotal = 312/756 = 0.4127 x2 = number of moles of H2 / Xtotal number of moles of H2 = x2*Xtotal = 0.4127*0.4233= 0.175 moles PO2 = x3*Ptotal x3 = PO2/Ptotal = 219/756 = 0.2897 x3 = number of moles of O2 / Xtotal number of moles of O2 = x3*Xtotal = 0.2897"0.4233= 0.123 moles 8) a) lighter the molar mass of the gas molecules the faster the gas molecules move. H2 has the lowest molar mass and hence H2 will move fastest. b) PV = nRT so. n*M P*M thus, PM TRT RT VRT If we keep, pressure and temperature constant then higher the molar mass, higher the density of that gas. Hence Xe has highest molar mass 131.3 g/mol hence Xe will have greatest density.