1 3.2 – Modeling a gas
2 The mole and molar mass ▪ ▪ ▪The mole (symbol “mol”) is the amount of substance that contains the same number of particles (atoms/molecules) as kg of carbon-12. It is one of the seven SI base units ▪ The Avogadro constant NA = 6.021023 molecules 1 mole of a substance contains 6.02 x 1023 atoms or molecules. ▪ Molar Mass is the mass of one mole of a substance (kg/mol).
3 PROBLEM: Find the mass (in kg) of one mole of carbon.From the periodic table we see that it is just 1 mole C = grams = kg. PROBLEM: What is the gram atomic weight of oxygen? g = kg. PROBLEM: What is the molar mass of phosphorus in kilograms? From the periodic table molar mass of phosphorus = grams. = kg.
4 PROBLEM: Water is made up of 2 hydrogen atoms and 1 oxygen atom and has a molecular formula given by H2O. Find the gram atomic weight of water. (b) the mass in grams of 1 mole of water. (c) how many moles of hydrogen and oxygen there are in 1 mole of water. (a) 2( ) + 1( ) = g/mole. (b) Mass of 1 mole of H2O is g. (c) Each mole of H2O has 2H and 1O: there are 2 moles of H and 1 mole of O for each mole of water. PROBLEM: Suppose we have g of water in a DixieTM Cup? How many moles of water does this amount to? gram atomic weight of H2O is g/mole # of moles = g / g) = mol. Maintain your vigilance regarding significant figures!
5 PROBLEM: How many atoms of P are there in31.0 g of it? How many atoms of C are there in 12.0 g of it? NA = 6.021023 atoms of P in 31.0 g of it. NA = 6.021023 atoms of C in 12.0 g of it. PROBLEM: How many atoms of P are there in 145.8 g of it? # of atoms = (# of moles)(NA) = g / g)(6.021023 atoms / 1 mol) = 2.831024 atoms of P. PROBLEM: A sample of carbon has 1.281024 atoms as counted by Marvin the Paranoid Android. a) How many moles is this? b) What is its mass? a) # of moles = 1.281024 atoms/6.021023 atoms) = 2.13 mol. b) mass = (2.13 mol)( g / mol) = 25.5 g of C.
6 Example EXAMPLE: A sample of carbon has 1.281024 atoms as counted byMarvin the Paranoid Android. a) How many moles is this? b) What is its mass? a) (1.281024 atoms)(1 mol / 6.021023 atoms) = 2.13 mol. b) (2.13 mol)( g / mol) = 25.5 g of C. Example Molar mass of Oxygen gas is 32x10-3 kg mol-1 How many moles and how many molecules is 20g of Oxygen? 20 x 10-3 kg / 32 x10-3 kg mol-1 mol mol x 6.02 x 1023 molecules x 1023 molecules
7 Equation of state for an ideal gasAn ideal gas can be characterized by four state variables: pressure (P), volume (V), number of moles (n) and absolute temperature (T). ideal gas law: is the result of experimental observations about the behavior of gases. It describes how gases behave. PV = nRT R is the universal gas constant = 8.31 J mol-1 K-1 p is measured in Pascals or Nm-2. V is measured in m3. T is measured in K.
8 An ideal gas is kept in a container of fixed volume at a temperature of 300 C and a pressure of 6.0 atm. The gas is heated at constant volume to a temperature of 3300 C. new pressure temperature 3300C gas pressure 6.0 atm temperature 300C gas The new pressure of the gas is about A atm B. 3.0 atm C. 12 atm D. 66 atm. piVi = nRTi, pfVf = nRTf. T(K) = T(°C) + 273 ▪ Ti = = 303 K. pfVf piVi nRTf nRTi = ▪ Tf = = 603 K. pf = piTf / Ti ▪ Vi = Vf. pf = (6)(603) / 303 = 12
9 (i) n, the number of moles =?. GIVEN: p = 20106 Pa, V = 2.010-2 m3. From T(K) = T(°C) + 273 T(K) = = 290 K. pV = nRT ⇾ n = pV / (RT) n = (20106)(210-2) / [ (8.31)(290) ] n = 170 mol. (ii) N = n NA. 6.021023 atoms mol N = 1.01026 atoms. N = 170 mol
10 An Ideal Gas Real Gases Is a theoretical gas that obeys the gas lawsAnd thus fit the ideal gas equation exactly Real Gases Real gases conform to the gas laws under certain limited conditions But they condense to liquids and then solidify if the temperature is lowered Furthermore, there are relatively small forces of attraction between particles of a real gas This is not the case for an ideal gas
11 The Kinetic Theory of Gases"the theory of moving molecules"; Rudolf Clausius, 1857 The ideal gas equation is the result of experimental observations about the behavior of gases. It describes how gases behave. A gas expands when heated at constant pressure The pressure increases when a gas is compressed at constant temperature But, why do gases behave this way? What happens to gas particles when conditions such as pressure and temperature change? That can be explained with a simple theoretical model known as the kinetic molecular theory. The kinetic theory relates the macroscopic behaviour of an ideal gas to the microscopic behaviour of its molecules or atoms This theory is based on the following postulates, or assumptions.
12 The kinetic model of an ideal gasGases consist of tiny particles called atoms or molecules The total number of particles in a sample is very large The particles are in constant random motion The range of the intermolecular forces is small compared to the average separation The size of the particles is relatively small compared with the distance between them, so they are treated as points Collisions of a short duration occur between particles and the walls of the container Collisions are perfectly elastic No forces act between the particles except when they collide Between collisions the particles move in straight lines And obey Newton’s Laws of motion
13 Macroscopic BehaviourThe kinetic model of an ideal gas Macroscopic Behaviour The large number of particles ensures that the number of particles moving in all directions is constant at any time With these basic assumptions we can relate the pressure of a gas (macroscopic behaviour) to the behavior of the molecules themselves (microscopic behaviour).
14 Pressure The kinetic model of an ideal gas ℓ• Pressure is the result of collisions between molecules and the wall of the container • Focus on one molecule moving toward the wall and examine what happens when on molecule strikes this wall. ℓ Elastic collision – no loss of kinetic energy, so speed remains the same, only direction changes.
15 The kinetic model of an ideal gasIf you can imagine 3-D picture you can “see” that only the component of the molecule’s momentum perpendicular to the wall changes: from –mvx (moving in the negative x direction) to +mvx Thus the change in momentum for one collision is: ∆(mv) = mvx – (-mvx) = 2mvx • Change in momentum implies that there must be a force exerted by the wall on the particle. • That means that there is a force exerted on the wall by that molecule. ● After the bounce, the molecule travels to the other side of the container and back before bouncing off the same wall again. The time required for this round trip of length 2ℓ is △t = 2ℓ/vx
16 The kinetic model of an ideal gas▪ Newton’s second law: the average force exerted by the wall on the molecule is ▪ The average pressure exerted by this wall is the force divided by the area: In deriving this equation we considered a single molecule with a particular speed. Other molecules, of course will have different speeds. In addition, the speed of any given molecule changes with time as it collides with other molecules in the gas. What remains constant, however, is the overall distributions of speeds. ▪ © 2006 By Timothy K. Lund
17 The kinetic model of an ideal gas▪ To calculate the pressure due to all molecules in the box, we have to add the contributions of each (N all together). By definition: so → Since the velocities of the molecules in our gas are assumed to be random, there is no preference to one direction or the other (the particles are equally likely to be moving in any direction - all three directions are equivalent), so the average value of vx2 must be the same as that of vy2 or vz2; it follows that (v2)avg = 3(vx2)avg or (vx2)avg = ⅓(v2)avg Therefore the pressure on the wall is: Now, finally we have the pressure in a gas expressed in terms of molecular properties. This is a surprisingly simple result! The macroscopic pressure of a gas relates directly to the average kinetic energy per molecule.
18 The kinetic model of an ideal gas½ m(v2)avg = (KE)avg average kinetic energy of the molecules in the gas. So, pressure can be written as: So, using kinetic theory we’ve shown that the pressure of a gas is directly proportional to the number of molecules and inversely proportional to the volume, as well directly proportional to average kinetic energy of its molecules. We got key connection between microscopic behaviour and macroscopic observables. If we compare the ideal-gas equation of state: PV = nRT with the result from kinetic theory, we find 𝑃𝑉=𝑛𝑅𝑇= 2 3 𝑁 𝑚 𝑣 2 𝑎𝑣𝑔 The average translational kinetic energy of molecules in a gas is directly proportional to the absolute temperature. The higher the temperature, according to kinetic theory, the faster the molecules are moving on the average.
19 result from kinetic theory:𝑃𝑉=𝑛𝑅𝑇= 2 3 𝑛 𝑁 𝐴 𝑚 𝑣 2 𝑎𝑣𝑔 result from kinetic theory: kB = 1.3810-23 J K-1 We got key connection between microscopic behaviour and macroscopic observables. • The average translational kinetic energy of molecules in a gas is directly proportional to the absolute temperature. • The higher the temperature, according to kinetic theory, the faster the molecules are moving on the average. • At absolute zero they have zero kinetic energy. Can not go lower. • This relation is one of the triumphs of the kinetic energy theory. Reminder: Since ideal gases have no intermolecular forces, their internal energy is stored completely as kinetic energy.
20 Absolute Temperature Molecular Speed• The absolute temperature is a measure of the average kinetic energy of its molecules • If two different gases are at the same temperature, their molecules have the same average kinetic energy, but more massive molecules will have lower average speed. • If the temperature of a gas is doubled, the average kinetic energy of its molecules is doubled , but the speed would increase by 2 Molecular Speed • Although the molecules of gas have an average kinetic energy (and therefore an average speed) the individual molecules move at various speeds • Some are moving fast, others relatively slowly • At higher temperatures at greater fraction of the molecules are moving at higher speeds • For O2 molecules at 300 K, the most probable speed is 390 m/s. • When temperature increases to 1100 K the most probable speed increases to roughly 750 m/s. Other speed occur as well, from speeds near zero to those that are very large, but these have much lower probabilities.
21 PROBLEM: Temperature is a measure of the EK of the gas.Reducing the EK reduces the frequency of collisions. For perfectly elastic collisions (as in an ideal gas) contact time is zero regardless of EK.
22 Differences between real and ideal gasesFYI ▪ Real gases are often polyatomic (N2, O2, H2O, NH4, etc.) and thus not spherical. ▪ Ideal gases cannot be liquefied, but real gases have intermolecular forces and non-zero volume, so they can be liquefied.
23 PROBLEM: ▪ Under high pressure or low volume real gases’ intermolecular forces come into play. ▪ Under low pressure or large volume real gases’ obey the equation of state.
24 Kinetic Model of an Ideal GasApplication of the "Kinetic Molecular Theory" to the Gas Laws Macroscopic behavior of an ideal gas in terms of a molecular model.
25 Just as temperature was a measure of the random kinetic energy of molecules for solids and liquids, so it is for an ideal gas. Temperature is a measure of the average random kinetic energy of the molecules of an ideal gas. 𝐾𝐸 𝑎𝑣𝑔 = 3 2 𝑘𝑇= 3 2 𝑅 𝑁 𝐴 𝑇 • Increase in temperature is equivalent of an increase in average kinetic energy (greater average speed). This leads to more collisions and collisions with greater impulse. Thus resulting in higher pressure. low T medium T high T • Decrease in volume results in a smaller space for gas particles to move, and thus a greater frequency of collisions. This results in an increase in pressure. Also, depending on the speed at which the volume decreases, particles colliding with the moving container wall may bounce back at greater speeds. This would lead to an increase in average kinetic energy and thus an increase in temperature. An increase in volume would have an opposite effect.
26 Pressure Law (Gay-Lussac’s Law)Effect of a pressure increase at a constant volume Macroscopically: at constant volume the pressure of a gas is proportional to its temperature: PV = nRT → P = (const) T n is constant (a closed jar, or aerosol can, thrown into a fire will explode due to increase in gas pressure inside). Isochors: lines of constant volume Straight line projected back to 0K (-2730C) at zero pressure. It implies that if the gas could be cooled to -2730C it would have zero pressure, and at lower temperature a negative pressure, which makes no sense, of course. The concept of absolute zero temperature was first deduced from experiments with gases. Zero pressure would mean still world.
27 Pressure Law (Gay-Lussac’s Law)Effect of a pressure increase at a constant volume n is constant Microscopically: ∎ As T increases, KE of molecules increase ∎ That implies greater change in momentum when they hit the wall of the container ∎ Thus microscopic force from each molecule on the wall will be greater ∎ As the molecules are moving faster on average they will hit the wall more often ∎ The total force will increase, therefore the pressure will increase low T medium T high T low p medium p high p
28 The Charles’s law Effect of a volume increase at a constant pressureMacroscopically: at constant pressure, volume of a gas is proportional to its temperature: PV = nRT → V = (const) T n is constant Isobars: lines of constant pressure A football inflated inside and then taken outdoors on a winter day shrinks slightly. A slightly underinflated rubber life raft left in bright sunlight swells up The plunger on a turkey syringe thermometer pops out when the turkey is done V vs. T relationship is linear, with an intercept of absolute zero on the Kelvin scale. The slope depends on the pressure. The dashed lines on the graph are to represent the fact that you can not achieve absolute zero, because the volume would attain its lowest possible value – ZERO
29 Effect of a volume increase at a constant pressureV = (const) T Microscopically: ∎ An increase in temperature means an increase in the average kinetic energy of the gas molecules, thus an increase in speed ∎ There will be more collisions per unit time, furthermore, the momentum of each collision increases (molecules strike the wall harder) ∎ Therefore, there would be an increase in pressure ∎ If we allow the volume to change to maintain constant pressure, the volume will increase with increasing temperature small V medium V high p large V medium p low p
30 Boyle-Marriott’s Law Effect of a pressure decrease at a constant temperature Macroscopically: at constant temperature the pressure of a gas is inversely proportional to its volume: PV = nRT → P = (const)/V n is constant Each line on the graph is a line of constant temperature, and is called an isotherm (meaning literally –constant temperature). In a plot of P versus 1/V, we see that the isotherms are straight lines with constant slopes. Doubling the pressure on a gas halves its volume, as long as the temperature of the gas and the amount of gas aren't changed. The bubbles exhaled by a scuba diver grow as the approach the surface of the ocean. (The pressure exerted by the weight of the water increases with depth, so the volume of the bubbles increases as they rise.) Deep sea fish die when brought to the surface. (The pressure decreases as the fish are brought to the surface, so the volume of gases in their bodies increases, and pops bladders, cells, and membranes). Pushing in the plunger of a plugged-up syringe decreases the volume of air trapped under the plunger.
31 Boyle-Marriott’s Law Effect of a pressure decrease at a constant temperature Microscopically: ∎ Constant T means that the average KE of the gas molecules remains constant ∎ This means that the average speed of the molecules, v, remains unchanged ∎ If the average speed remains unchanged, but the volume increases, this means that there will be fewer collisions with the container walls over a given time ∎ Therefore, the pressure will decrease