Topic 6 Gases and Colligative Properties Boyle noticed an inverse relationship between volume and pressure. Pressure x volume = constant PV = a V V P 1/P Charles found the volume of a gas, at constant pressure, increased linearly with temperature. Volume = constant x temperature V = bt V He Different gases extrapolated to zero volume at the same temperature. This is absolute zero at -273.15 C = 0 K. CH 4 H 2 Avagadro proposed that equal volumes of gases at the same temperature and pressure contained the same number of particles. Combining these equations gives Volume = constant x No of moles V = cn T Ideal Gas Equation: PV = nrt R is the ideal gas constant (8.314 J K 1 mol 1 or 0.082 L atm K 1 mol 1 ) 45
Example: The mass of 1.00 L of a gas at 2.00 atm and 25 C is 2.76 g. What is the molecular weight of the gas? PV = nrt and n = m/m (2.0 atm)(1.00 L) = n (0.082 L atm K -1 mol -1 )(273 + 25 K) n = (2.00 / 24.436) mol = 0.0818 mol n = mass / molar mass or molar mass = mass/ moles Molar mass = 2.76 g / 0.0818 mol = 33.7 g mol -1 Question: If a gas occupies a volume of 1.5 L at 105 kpa and 20 C, what volume will it occupy at standard temperature and pressure (0 C and 1 atm = 101.3 kpa)? Partial Pressure In a mixture of gases, the total pressure exerted is the sum of the partial pressures that each gas would exert if it were alone. P TOTAL = p A + p B + p C +. The partial pressure is related to the number of moles present, the mole fraction, x, Mole fraction of A = x A = (No of moles of A) / (Total No of moles present) Thus p A = x A P TOTAL and p B = x B P TOTAL Example: Air consists of approximately 20 % oxygen and 80 % nitrogen. What are the partial pressures of these gases at 10 atm? Mole fraction of O 2 = x O2 = (20 %) / (20 % + 80 %) = 0.20 Mole fraction of N 2 = x N2 = (80 %) / (80 % + 20 %) = 0.80 (check: mole fractions = 1) At 10 atmosphere: p O2 = 0.20 x 10 atm = 2 atm p N2 = 0.80 x 10 atm = 8 atm (check: partial pressure = total pressure) 46
Question: Exactly 500 ml of a gas contains 0.18 mol of PCl 3 and 0.050 mol of Cl 2 at 298 K. What is the total pressure and partial pressures of these gases? (Hint: i. Calculate total no of moles, ii. Use this and the data given, together with the Ideal Gas Equation, to find P TOTAL iii. Calculate mole fractions, iv. Calculate partial pressures). Example: 8.00 g of hydrogen and 16.00 g of oxygen are combined in a 20.0 L vessel at 200 C. If the gases react according to the equation: 2H 2 (g) + O 2 (g) 2H 2 O (g) Determine the partial pressure of each gas and the total pressure before and after reaction. Before reaction. Amount of H 2 = 8.00 g / 2.00 g mol -1 Amount of O 2 = 16.00 g/ 32.00 g mol -1 Total amount of gas = 4.00 mol = 0.500 mol = 4.50 mol Mole fraction of H 2 x(h 2 ) = mol of H 2 / Total mol = 4.00 / 4.50 = 0.889 Mole fraction of O 2 x(o 2 ) = mol of O 2 / Total mol = 0.500 / 4.50 = 0.111 Use PV = nrt to determine total pressure P (Total) = n (Total) RT/V = (4.50 mol)(8.314 J K -1 mol -1 )(473 K) / (20.0 L) = 885 J L -1 = 885 kpa (Note J = Nm, L = 10 +3 m 3, so JL -1 = 10-3 Nm -2 and Pa = Nm -2, so kpa = 10-3 Pa = 10-3 Nm -2 and JL -1 = kpa) p(h 2 ) = x(h 2 )P (Total) = (0.889)(885 kpa) = 786 kpa p(o 2 ) = x(o 2 )P (Total) = (0.111)(885 kpa) = 98 kpa check 786 kpa + 98 kpa = 885 kpa 47
After reaction: 2H 2 (g) + O 2 (g) 2H 2 O (g) Initial / mol 4.00 0.500 (oxygen is limiting) Final / mol 3.00-1.00 Total moles = 4.00 mol P (Total) = n (Total) RT/V = (4.00 mol)(8.314 J K -1 mol -1 )(473 K) / (20.0 L) = 786 J L -1 = 786 kpa p(h 2 ) = x(h 2 )P (Total) = (3.00/4.00)(786 kpa) = 590 kpa p(h 2 O) = x(h 2 O)P (Total) = (1.00/4.00)(786 kpa) = 197 kpa check 590 kpa + 197 kpa = 787 kpa (NB rounding error) Kinetic Molecular Theory of Gases The volume of the particle is negligible (zero) The particles are in constant motion No intermolecular forces between particles Average kinetic energy of the gas is directly proportional to the temperature (Kelvin) P = 2/3 { [ n N A ( 1 / 2 m u 2 ) ] / V } Relative No with a given speed 273 K 1273 K No of molecules N 2 H 2 O 2273 K He H 2 1000 2000 Speed (m /s) Molecular speed 48
Colligative Properties Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure The presence of a solute affects the physical properties of a solvent. Colligative properties all depend on the number, not the identity, of the solute particles in a solution. Use salt to melt ice on a footpath - causes the freezing point of the water to decrease Add antifreeze to the water in the radiator of a car to lower the freezing point of the water Osmotic Pressure The flow of solvent molecules into a solution through a semi-permeable membrane is called OSMOSIS. In osmosis a semi-permeable membrane prevents the transfer of solute particles. More solvent molecules enter the solution than leave it. As a result the solution volume increases and its concentration decreases. The solution exerts a backward pressure - the osmotic pressure () that eventually equalises the flow of solvent molecules in both directions The osmotic pressure is also defined as the applied pressure required to prevent this volume change. = crt is the osmotic pressure in atm, c is the total molarity of all solute particles (mol L -1 ) R is the gas law constant and T is the temperature in K Example: Contact lens rinses consist of 0.015 M NaCl solution to prevent any changes in volume of corneal cells. What is the osmotic pressure of this saline solution at 38 C? = crt = (2 x 0.015 mol L -1 )(0.082 L atm K -1 mol -1 )(273 + 38 K) = 0.77 atm (NB The saline solution contains 0.015 M Na + ions and 0.015 M Cl ions, so the total concentration of solute particles is 2 x 0.015 M = 0.030 M) 49
Question: Lysozyme is an enzyme that breaks down bacterial cell walls. A solution containing 0.150 g of this enzyme in 210 ml of solution has an osmotic pressure of 0.00125 atm at 25 C. What is the molar mass of lysozyme? = crt and c = mol/vol and mol = mass/molar mass Sodium ions and cells Of the four major biological cations (Na +, K +, Mg 2+, Ca 2+ ), Na + is essential for all animals to regulate fluid volume. Na + accounts for >90mol% of all cations outside a cell. A high concentration of Na + draws water out of the cell by osmosis; a low concentration leaves more inside. The primary role of Na + is to regulate the water volume and the primary role of the kidneys is to regulate [Na + ]. Dialysis - the blood is passed through a cellophane tube a semipermeable membrane - this is immersed in dialyzing solution (with the same concentration of ions and small molecules as blood without the waste products). The unwanted metabolites are transferred to the dialyzing solution. A similar phenomena to dialysis occurs at the walls of most plant and animal cells - however transfer of solvent and SMALL solute molecules and ions occurs 50
ISOTONIC SOLUTION - identical osmotic pressure to cell fluid HYPOTONIC SOLUTIONS lower osmotic pressure than cell fluid - cell swells & ruptures - transfer of fluid INTO CELL - LYSIS HYPERTONIC SOLUTIONS - higher osmotic pressure than cell fluid - cell shrivels - transfer of fluid OUT OF CELL CRENATION The word tonocity refers to the tone, or firmness of a biological cell. Question: Solutions for the intravenous injection must have the same osmotic pressure as blood. According to the label on the bottle each 100 ml of normal saline solution used for intravenous injection contains 900 mg of NaCl. What is the molarity of NaCl in normal saline solution? What is the osmotic pressure of normal saline solution at the average body temperature of 37 C? What concentration of glucose (C 6 H 12 O 6 ) has the same osmotic pressure as normal saline solution. 51