I. Ideal gases. A. Ideal gas law review. GASES (Chapter 5) 1. PV = nrt Ideal gases obey this equation under all conditions. It is a combination ofa. Boyle's Law: P 1/V at constant n and T b. Charles's Law: V T at constant n and P c. Avogadro's Law: equal numbers of moles of gases at the same T and P occupy the same volume or V n at constant T and P ). Note that according to this equation, 1 mole of an ideal gas at STP (Standard B. Uses. Temperature and Pressure, that is, 73 K and 1.00 atm or 760 Torr ) will occupy L atm (1.00 mol)(0.0806 )(73 K) mol K =.414 L 1.00 atm This is called the molar volume at STP. The symbol for the molar volume is V 1. PV = nrt = grt/m There are five variables in this expression. If any four are known, the fifth can be calculated.. This expression can be used a. in stoichiometric calculations SEE PREVIOUS UNIT b. in molar mass calculations from gas densities. grt M = = PV V g RT RT P = d. P M d at constant T and P. c. Example: A compound was analyzed and found to contain 53.3% C, 11.1% H, and 35.6% O by mass. In another experiment it was found that a 0.503g sample of the gaseous compound occupied a volume of 697 ml at 17 C and 00 Torr pressure. Calculate the molecular formula of the compound. Steps: a. Determine the empirical formula from the analysis data. b. Determine the Molar Mass c. Deduce the molecular formula. 1
a. Empirical formula. 53.3g mole of C = 1.0g/mol = 4.44!!!!".0 divide by.5 mole of H = 11.1g 1.0g/mol =11.111! divide!! by!.5 " 5.0 35.6g mole of O = 16.0g/mol =.5!!!!" 1.0 divide by.5 Empirical formula C H 5 O b. Molar Mass (MM) From ideal gas law: M = grt PV = (0.503)(6.4)(400) = 90 (00)(0.697) EFM = mass of C H 5 O = (1)+5(1)+1(16) = 45 molecular formula is C 4 H 10 O 3. Mixtures of gases-dalton's Law of Partial Pressures. a. The partial pressure of a component, i, of a mixture of gases is P i. P i = n i V RT RT RT P Tot =! P i =! n i = ntot i i V V b. Partial pressure and mole fraction ( X i ) P i P Tot n = i = X n i = Mole fraction of i Tot c. Example: Suppose that a mixture of 0.80 g of He and an unknown amount of O when placed in a 3.0 L container at 500 K exerted a pressure of 3.59 atm. 1) What are the partial pressures of the two gases? n He = 0.80g = 0.0 mol 4.0g/mol atm L (0.0 mol)(0.0806 mol P He = K)( 500K) 3.0 L = 3.59 atm.74 atm = 0.85 atm P O ) What is the mass of O present? =.74 atm n O = P O!n O = P O 0.85 atm n He = (0.0mol) = 0.06 mol n He P He P He.74 atm g O = (0.06mol)(3 g mol) = 1.99 g
4. Gas behavior and the ideal gas law can be understood using the Kinetic Molecular Theory. This is one of the most frequently used theories in Chemistry II. Kinetic Molecular Theory as Applied to Ideal Gases. A. Postulates for ideal gases. 1. Gases are composed of tiny particles ( molecules or atoms ) in constant, random, straight-line motion. On the average the particles are so far apart that the actual volumes of the particles themselves are negligible compared to the total gas volume. a. Although for real gases this assumption is not exactly correct, it is not an unreasonable assumption at moderate temperatures and pressures. For example, at STP the volume of He gas is about 99.91% empty space. b. At high pressures the assumption of point masses (zero volume for the particles) will not be valid and deviations will be expected. c. Random motion means that the gas molecules have equal probability of moving in any direction. The phenomenon of pressure is due to the bombardment of gas particles with the sides of the container. For a small contained volume of a gas, the pressure should be independent of where the pressure gauge is located ( that is, either the top or the bottom of the container ). Although this is true for small samples of a gas, gases are effected by gravity. Our gaseous atmosphere is held to the earth by gravitational attraction.. For ideal gases no forces of attraction exists between the gas particles. Because of the constant random motion, collisions will occur between the gas particles themselves and also with the sides of the container. The collisions are elastic collisions and will be totally governed by chance. a. In an elastic collision, the total kinetic energy of the two particles before collision is equal to the total kinetic energy after collision. There can be a transfer of kinetic energy between the two particles, but the totals before and after collision must be the same. b. In an elastic collision with the sides of the container, the speed of the gas particle before and after collision must be the same. 3. The average kinetic energy per mole of a gas (KE) is directly proportional to the temperature of the gas measured in Kelvin. It can be shown that KE = 3/RT a. For a collection of gas molecules, KE = 1 Mµ = 3 RT where M = the molar mass of the gas in kg / mol, µ = the average of the square of the speed in m / s and KE = the average kinetic energy per mole in Joules. b. At a temperature of 0 K, the average kinetic energy is zero. Since it is not possible to have negative kinetic energies, the temperature can never be lower than 0 K; this 3
is absolute zero. c. Because of collisions, an individual molecule is continually changing its kinetic energy, yet the average kinetic energy per mole changes only with temperature. This means that at a particular temperature there is a fixed distribution of kinetic energies and the only way to change that distribution is to change the temperature. B. Distribution of kinetic energies. (Maxwell-Boltzmann distribution) 1. The distribution can be seen by plotting the fraction of molecules having a particular kinetic energy vs. kinetic energy. The general shape of the curve is given below. 0.050 0.040 KE mp Fraction. 0.030 0.00 0.010 KE 0.000 0 5 10 15 0 5 30 35 KE, kj a. In the plot, the most probable value of the kinetic energy is KE mp. Note that the distribution is not symmetric about KE mp but is skewed towards the high kinetic energy end. b. The average kinetic energy, KE, is the value of the kinetic energy such that half the molecules will have kinetic energies greater than this value and half will have values less than it (the area under the curve to the left of KE is equal to the area to the right). c. Because of the shape of the distribution, KE is always greater than KE mp.. As the temperature changes, the distribution will change. When the temperature increases, the following changes take place. a. The distribution shifts towards much higher kinetic energies so that a larger fraction of molecules will have extremely high kinetic energies. b. The distribution will tend to flatten out. 4
c. The curves below show these changes for temperatures of 500 K and 1000 K. 0.050 0.040 500 K Fract. 1000 K Fract. Fraction 0.030 0.00 0.010 6.4 1.47 0.000. 0 5 10 15 0 5 30 35 KE, kj 3. The fraction of molecules having extremely high kinetic energies increases as the temperature increases. This causes the average kinetic energy per mole (KE) to increase such that KE T. Note that if the temperature in K were doubled, the kinetic energy of each particle would not double. The distribution would shift such that KE would be twice as large as before. C. Gas Dynamics. 1. Average Speed. KE = 1 Mµ = 3 RT! µ = RMS speed = 3RT M a. The letters RMS stands for Root-Mean-Square, that is, the square root of the average value of the square of a quantity. In cases where there are a number of molecules having different speeds there are several ways to express the "average" speed. 1) Suppose we have two molecules, one having a speed of 4m/s and the other having a speed of 6m/s. The mean speed µ = 4 +6 m/s = 5m/s. The square of this average µ = 5 m s. The average of the square of the speed = µ = 4 + 6 = 16+36 m = 6 s 5
The RMS speed = µ = 6 m s = 5.1 m/s ) Note that the two "average" speeds are slightly different. The RMS speed is the larger of the two. b. Examples. 1) Calculate the average kinetic energy per mole and the RMS speed of O at 500 C. KE = 3/RT = 3/(8.314 J/mol K)(773 K) = 9640 J/mol RMS speed = 3RT = M 3 8.314 Kg m /s 773 K mol K 3x10-3 Kg/mol = 6.05x10 5 m /s RMS speed = 7.76x10 m/s Note that the molar mass must be expressed in the SI Unit of kg/mol. ) Calculate the same quantities for H at 500 C. KE = 9640 J/mol = KE of O at 500 C RMS speed = 3RT = M Kg m /s 3 8.314 773 K mol K.0x10-3 Kg/mol = 9.64x10 6 m /s = 3.1x10 3 m/s 3) Note that RMS speed H = RMS speed O 3.1x10 3 m/s 1/x10-3 = 4.00 = 1/3x10-3 = 7.76x10 m/s 3 = 16 = 4 This is an example of Graham's Law of Diffusion ( or Effusion). Diffusion = motion of gas through space Effusion = escape of gas to a vacuum through a small opening Graham's law = relative rates of effusion ( or diffusion ) of two gases are inversely proportional to the square roots of their molar masses. rate 1 = speed rate speed 1 = 3RT/M 1 = M III. Real Gases. A. Compressibility of real gases. 3RT/M M 1 6
1. Compressibility factor, Z = PV nrt = PV RT. a. For an ideal gas, Z = 1.000 under all conditions. b. Can use experimental plots of Z vs. P at constant T to investigate the P-V-T-n relationships of real gases.. Plot of Z vs. P for some gases. H CO Z 1 ideal gas P, atm Only at very low pressures will Z = 1. At higher pressures, Z will deviate from ideal behavior. We must consider the interactions that lead to positive deviations (Z greater than 1) and negative deviations (Z less than 1) in terms of the Kinetic Molecular Theory. 3. Positive Deviations (Z > 1) a. At sufficiently high pressures all gases will show positive deviations. b. Due to the fact that real molecules have finite, nonzero volumes and there is a limit to how much a gas can be compressed. The total volume, V, can be thought of as being composed of a free volume, V f, which is the compressible volume and an excluded volume, V ex, which is incompressible. The free volume is the one that should be used in the gas equation. V f = V - V ex c. The excluded volume should be proportional to the number of moles of gas, n V ex n or V ex = bn where b is the proportionality constant. Therefore a better equation would be of the form 4. Negative deviations (Z<1). P(V - nb) = nrt. a. Is large for polar molecules and increases as the temperature decreases. 7
b. Due to attractive forces that exist between molecules. 1) These forces should tend to make the pressure of a real gas less than that of an ideal gas. ) Therefore, must add a correction to the pressure. Since the effect of these attractive forces will be greatest during a collision when the molecules are close together, the correction should be proportional to the collisions per second between gas molecules. It can be shown that this is proportional to the square of the concentration of gas molecules. Therefore, pressure correction (n/v) or = an /V. Where a is a constant that reflects the attractive forces that exist between gas molecules. 5. Van der Waals Equation. a. A better equation of state for a gas is (P + an )(V - nb) = nrt V b. This is called the van der Waals equation. The constants a and b will differ for each gas. They can be determined experimentally by careful P-V-n-T measurements. c. There are a large number of other equations of state for gases. Each one attempts to correct for the nonzero volumes of gas molecules (repulsion forces between gases) and forces of attraction. 6.There are a large number of equations relating P, V and T for real gases. They all attempt to correct for the nonzero volume of real gases and lone range attraction between gases. The nature of these forces will be discussed later. 8
Problems on Gases (See problems 1 18 in the Moles and Stoichiometry Unit) 1. Suppose a particular sample of a gas occupied a volume of 500 ml at 17 C and 800 torr. a) For this gas sample, calculate 1) the volume at 4 C and 700 torr. ) the pressure that would be exerted by the gas if the volume were 300 ml at a temperature of 0 C. 3) the temperature of the gas if its pressure were 900 torr at a volume of 800 ml. 4) the volume at STP. b) If the gas had a molar mass of 30 g, what would be its density, in g/liter at STP? What would be its density, in g/liter, at 17 C and 800 torr pressure?. Consider the following gases H He CO Ne CO C H 6 under similar conditions of temperature and pressure: a) Which gas would have the highest density? b) Which gas would have the lowest density? c) Which gas would have the slowest rate of diffusion? d) Which gas would have the highest average speed? e) Compare the average kinetic energies per mole of the different gases. 3. Hydrogen reacts with PbO to give Pb and H O. What volume of H, measured at 8 C and 1.05 atm pressure, would be required to produce 8.00 g of Pb by this reaction? 4. When.0 g of He (g) and an unknown amount of H (g) are placed in a 30.8 liter container at 7 C, the total pressure is 1. atm. a) Calculate the partial pressure of He. b) Calculate the number of grams of H present. 5. A 4.68 liter box contains 0.96 g of O and 0.63 g of N at 100 C. Calculate the total pressure. N and O are both gases under the above conditions. 6..00 g of a gas collected over water at 7 C and a total pressure of 747 torr occupies a volume of 500 ml. On analysis the substance is found to contain 7.7 % H and 9.3 % C by weight. What is the molecular formula of the substance? 9
7. The RMS speed of gaseous He is found to be 00 m/s. At what temperature must the He be? 8. Under equivalent conditions the rate or diffusion of a gaseous substance is 1.41 times as fast as that of O. The substance contains only the elements hydrogen and carbon. What is the molecular formula of the substance? 9. A cubical box 100 cm on an edge contains 0.30 g of the gas C H 6 at 17 C. a) How many molecules of gas are in the box? b) What is the pressure in the box? c) Calculate the RMS speed of the molecules in the box. 10. Write van der Waals equation for n moles of a real gas. For what does the van der Waals equation attempt to correct? 11. How does the ease of liquefaction of a real gas vary with the value of van der Waals constant"a"? 1. Consider a 3.5 mole sample of the gas CH 4 at 700 C. a. Calculate the average kinetic energy per mole of the gas. b. Calculate the RMS speed of the gas. c. At what temperature would the gas C H 6 have the same average kinetic energy per mole? d. At what temperature would the C H 6 have the same RMS speed as the CH 4? 13. A compound was analyzed and found to contain 9.3% C and 7.7% H by mass. In another experiment it was found that this gas had about the same rate of diffusion as did N under equivalent conditions. Calculate the molecular formula of the compound 10
Answers 1. a. 1) 44 ml, ) 1643 torr; 3) 447 C, 4) 359 ml b. 1.34 g/l; 0.96 g/l. a) CO b) H c) CO d) H e) all the same 3. 1.8 L 4. a. 0.40 atm b) pressure of H = 0.80 atm,.0 g of H 5. 61 torr 6. C 8 H 8 7. 6.4 K 8. CH 4 9. a) 6.0x10 1 b) moles = 1.0x10 - pressure = 50 torr c) 577 m/s 10. See Notes 11. increases as a increases 1. a) 1.1x10 4 m/s b) 1.3x10 3 m/s. c) 700 C d) 184 K 13. C H 11