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Topic 14: The Gaseous State of Matter LECTURE SLIDES States of Matter: Gas vs Liquid, Solid Pressure relationship to V, T, n Combined Gas Law Ideal Gas Law Gas state stoichiometry Kotz & Treichel, Chapter 12, Sections 12.1-4, 6

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The Gas State of Matter Chapter 12 Let us turn to the gas state, and learn how to measure and calculate amounts of matter when found in this high energy situation. Consider:

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Also, consider three lidded containers, holding respectively a solid, liquid and gas sample of material: Retains shape, volume Retains volume, takes shape of container Takes shape, volume of container

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Differences in MP’s and BP’s, and retention or loss of shape and volume, all are due to attractions between the molecules or ions or atoms which make up the sample. Let us look more closely at each state in these terms...

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Solids are made up of particles (molecules or ions or atoms) which are highly attracted to each other and packed as closely together as possible in some sort of “crystal lattice” arrangement which keeps them rigidly in place. Solid particles can vibrate in position but not flow past each other. They retain their shape and volume due to this internal attraction between particles. solid state volume doesn’t change with pressure due to close packing of particles, and volume can be calculated from mass and a reference density

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Liquid samples are made up of particles which are still strongly attracted to each other, but possess the energy to flow. They are still packed as close together as possible, but are not confined to a “crystal lattice”. Accordingly, liquids retain their volume when poured from one container to another, but assume the shape of the container. Like solids, liquids resist volume change under pressure due to “close packing”and volume can be calculated from mass and a reference density.

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Gases are made up of independent particles with the necessary energy to escape the attractions of neighboring particles. They move freely within the volume in which they are confined: they fill volume completely, assuming both the shape and the volume of the container. Pressure deeply affects gas volumes, as the particles are as far apart as the container allows: gases are readily compressed into smaller volumes. To correctly assess the amount of a gas sample, it is necessary to consider four factors: temperature, pressure, number of particles (“moles”), and volume.

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We will return to the nature of the attractions between particles which make some materials solids at room temperature, and others liquids or gases. We should have gathered from what we had just examined that high attraction leads to solids and low attraction to gases.... First however, we need to consider the inter- relationship of the four factors needed to define any sample in the gas state: P, pressure; T, temperature; V, volume; and n, number of moles.

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Solids: Definite shape, volume: know mass, calculate volume know volume, calculate mass Liquids: Definite volume, shape determined by container: know mass, calculate volume know volume, calculate mass D= mass/ volume Gases: Volume, “shape” determined by container To determine volume, mass, molar mass, P and T required.....

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GAS PRESSURE Gas samples are composed of independent particles moving in a straight line path until collision with the walls of the container (or another particle). Collision with the walls of the container results in a pressure, defined as a force per unit area. The pressure created depends on how frequently the collisions occur and how forceful they happen to be.

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Rapid random movement of particles results in frequent collisions with the walls of the container, creating pressure: Lidded container collisions

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Pressure Dependence on T, V, n Using the container as our model, let’s consider how P relates to the other variables, considering each as we hold the other two constant: We’ll hold number of moles (n) and T constant first, and decide how P changes with changing V.

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Pressure and Volume, constant n, T If the volume of the container is decreased, the particles hit the wall more frequently (they haven’t as far to travel!). The P goes up as the number of collisions increases. If the volume of the container is increased, collision will occur less frequently as the particles have further to travel and the pressure will go down.

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In a nutshell, n,T constant: as V increases, P decreases as V decreases, P increases In mathematical terms, P is “inversely proportional” to V at constant n, T. This relationship was studied first by Robert Boyle in the late 1600’s, and the law defining this relationship is termed “Boyle’s Law”. P 1 V C B P = V PV = C B Proportionality Constant

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C B is a “proportionality constant” obtained by graphing P vs. 1 / V. This constant, C B, is the slope of the straight line obtained. Boyle’s Law states that for a given sample of gas at a constant temperature, any product of the P times the V equal a constant: P 1 V 1 = C B = P 2 V 2 (n, T constant)

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Pressure and Temperature, constant n, V If the temperature of the sample is increased, the particles hit the wall harder and more frequently (they will be moving faster, with the added energy heat provides). The P goes up as the number and intensity of collisions increases. If the temperature of the sample is decreased, collision will occur less frequently and with less force as the less energetic particles move at a slower pace. The pressure will go down.

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So, in a nutshell, at constant V, n: as T increases, P increases as T decreases, P decreases P T P = C k T P 1 P 2 = C k = T 1 T 2 Pressure is directly proportional to T (in the absolute or Kelvin scale) as can be shown graphically. The straight line obtained plotting P vs. T yields the constant “C k ”.

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Summary, to date: Boyle’s Law: given sample of gas, constant n, T: P 1 V 1 = P 2 V 2 Charles’s Law: given sample of gas, constant n, V: P 1 P 2 = T 1 T 2

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Combined Gas Law We can combine the three factors which describe a “confined gas” (constant n, number of particles) as follows: P 1 / V, P T : P T P =C c T V V PV = C c AND P 1 V 1 = C c =P 2 V 2 T T 1 T 2

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If we hold the one factor constant for this “confined gas”, then : P 1 V 1 = P 2 V 2 T 1 T 2 V 1 = V 2 T 1 T 2 Constant T P 1 V 1 = P 2 V 2 Constant P Constant V P 1 = P 2 T 1 T 2

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Units Utilized in Gas Law Problems: PRESSURE*: generally done in reference to a column of mercury immersed in a dish of mercury open to atmospheric pressure. (CD ROM) Pressure as a FORCE PER UNIT AREA: English SI = 14.7 lbs/ft 2 (psi) = 101.3 kilopascal (k Pa) 760 mm Hg = 760 Torr = 1 atmosphere (atm) 1 Pa, pascal = 1 Newton / m 2 1 Newton = 1 kg. m / s 2 (SI unit of force)

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Volume*: Liters, L (most general) Temperature: Absolute or Kelvin Scale; K= o C +273.15 In all cases, K, Kelvin scale must be used! Where P, V = 0, T=0, absolute zero, - 273.15 o C * Any V or P unit OK for combined gas law, if used consistently Units Utilized in Gas Law Problems, continued:

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Since gas law problems always include lots of data, it is a standard practice to set up a table of all given data before proceeding: For the combined gas law, an appropriate table is below:

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A gas exerts a pressure of 735 mm Hg in volume of 2.50 L at a temperature of 73 o C. What pressure would it exert if the temperature were increased to 110 o C and the volume increased to 3.00 L?

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data formula solve

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A sample of CO 2 gas has a pressure of 56.5 mm Hg in a 125 mL flask. The gas is transferred to a new flask where it has a pressure of 62.3 mm Hg at the same temperature. What is the volume of the new flask?

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GROUP WORK 14.1 A sample of gas occupies 12.0 liters at 240. o C under a pressure of 80.0 kPa. At what temperature would the gas occupy 15.0 liters if the pressure were increased to 107 kPa?

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GW 14.1:

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VARIATION OF P with n at constant V, T If we keep the volume of our container constant and also the temperature, the introduction of more particles into the container will mean more collision with the wall of the container. Accordingly, we can say: P n P = C n n P 1 P 2 = C n = n 1 n 2

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Relating P to n, V, T P 1 / V, P T, P n : P n T P =R nT V V where R = gas constant used to interrelate the four factors. Rearranging: PV = nRT the “ideal gas equation” (No such thing as an “ideal gas” but if very low T’s and very high P’s are avoided, most gases can be described fairly accurately using the “ideal gas law”.)

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Evaluation of value of R If P, V, n, and T for a gas are experimentally determined, the value for this constant can readily be calculated, which is appropriate for use with any gas (if the conditions are not too extreme): PV / nT =R We can calculate R for one mole of any gas using what is described as “Standard Conditions”, a conventional P and T adopted as the norm for gases (“STP”): Standard Pressure = 1 atmosphere or 760 Torr (mm Hg) Standard Temperature = 0 o C = 273 K

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It was determined by Avogadro (1811) that equal volumes of any gas, at the same temperature and pressure, contain the same number of molecules. It has since been experimentally determined that at STP, 1.00 mole of any “ideal” gas occupies 22.414 L, called the standard molar volume. 1.00 mole gas, STP = 22.4 L

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Calculation of R, STP: Let us use experimentally determined molar volume of 1.00 mole of gas at STP to evaluate R. Note that our table of values changes form when using PV= nRT :

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Group Work 14.2: Use 1.00 mole gas at STP to Calculate value of R, changing pressure unit to atmospheres:

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Useful Gas Law Knowledge, to date: P 1 V 1 = P 2 V 2 T 1 T 2 V 1 = V 2 T 1 T 2 Constant T P 1 V 1 = P 2 V 2 Constant P Constant V P 1 = P 2 T 1 T 2 Combined Gas Law, constant n:

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PV = nRT the “ideal gas equation” 1.00 mole gas, STP = 22.4 L STP: Standard Pressure = 1 atmosphere or 760 Torr (mm Hg) Standard Temperature = 0 o C = 273 k R = 62.4 Torr L / K mol = 62.4 Torr L K -1 mol -1 R =.0821 atm L / K mol =.0821 atm L K -1 mol -1

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n = ? 25.0 g N 2 gas = ? moles 28.0 g N 2 = 1 mole N 2 25.0 g N 2 1 mol =.893 mol N 2 28.0 g 25.0 g =.893 mol 28.0 g mol -1 Alternate formula approach Calculation of Moles: Dimensional Analysis vs “Formula” Dimensional analysis

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Problems using the ideal gas equation: A hot air balloon holds 30.0 kg of helium. What is the volume of the balloon if the final pressure is 1.20 atm and the temperature 22 o C?

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Convert formula first!

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Group Work 14.3 The nitrogen gas in an air bag, with a volume of 65 L, exerts a pressure of 829 mm Hg at 25 o C. How many moles and how many g of N 2 are in the air bag? (Solve for moles using ideal gas equation, then solve for g) PV =nRT n=

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Another handy “formula” to utilize: The ideal gas law can be used to calculate molar mass if grams of gas, and P,V, T are known: 1. Calculate moles of gas, n = PV / RT 2. Use moles, n and mass, g to calculate M n, moles = mass, g molar mass, M = mass, g molar mass, M n, moles Widely used

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Gas Laws : Density, Molar Mass Chloroform is a common liquid which vaporizes readily. If the pressure of the vaporized liquid is 195mm Hg at 25 o C, and the density of the gas is 1.25 g/L, what is the molar mass of the chloroform? V= 1.00 L mass= 1.25 g P= 195 mm Hg T= 25 o C = 298 K R = 62.4 L mm Hg K -1 mol -1 M, molar mass = ? g/mol Procedure: 1) solve for n, # mol 2) M = #g / #mol density

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V= 1.00 L mass= 1.25 g P= 195 mm Hg T= 25 o C = 298 K R = 62.4 L mm Hg K -1 mol -1 M, molar mass = ? g/mol CHCl 3 = 12 + 1 + 3(35.4) = 119.2 g/mol

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Group Work 14.4 What is the molar mass of a gas which has a density of 1.83 g/L measured at 27.0 o C and 0.538 atm? R =.0821 L atm K -1 mol -1

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Gas laws, Stoichiometry Problem 49, text: A self contained breathing apparatus uses canisters containing potassium superoxide. The superoxide consumes the CO 2 exhaled by the person and replaces it with O 2. What mass of solid KO 2, in grams, is required to react with 8.90 L of CO 2 at 22.0 o C and 767 mm Hg?

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When gases are included in a balanced equation for a reaction, the number of moles or grams of the gas used can be computed by the Ideal gas equation, PV = nRT. Use Ideal gas equation in equation situation to compute moles of gas, then proceed normally to do problem. Word of Warning: never use grams of solid or liquid in gas law equation: it won’t give a correct answer...

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? grams V= 8.90 L T= 22.0 o C = 296.1 K P= 767 mm Hg R= 62.4 L mm Hg K -1 mol -1 n =? FIRST

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? grams 1 K =39.1 2 O= 32.0 71.1 g/mol n =.369 moles second

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Group Work 14.5 What volume of O 2, collected at 22.0 o C and 728 mm Hg would be produced by the decomposition of 8.15 g KClO 3, M = 122.5 g/mol? 2 KClO 3 (s) 2 KCl (s) + 3 O 2(g) 1. Calculate moles of O 2 from equation 2. Calculate V of O 2 from ideal gas law

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