2. 3.2 – Modeling a gas

3. mole The mole (symbol “mol”) is the amount of substance that contains the same number of particles (atoms/molecules) as 0.012 kg of carbon-12. 1 mole of a substance contains 6.02 x 10 23 atoms or molecules. It is one of the seven SI base units Molar Mass Molar Mass is the mass of one mole of a substance (kg/mol). N A = 6.02  10 23 molecules The Avogadro constant The mole and molar mass ▪ ▪ ▪

4. PROBLEM: Find the mass (in kg) of one mole of carbon. From the periodic table we see that it is just 1 mole C = 12.011 grams = 0.012011 kg. PROBLEM: What is the gram atomic weight of oxygen? 15.9994 g = 0.015994 kg. PROBLEM: What is the molar mass of phosphorus in kilograms? From the periodic table molar mass of phosphorus = 30.973762 grams. = 0.030973762 kg.

5. PROBLEM: Water is made up of 2 hydrogen atoms and 1 oxygen atom and has a molecular formula given by H 2 O. Find (a) 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(1.00794) + 1(15.9994) = 18.01528 g/mole. (b) Mass of 1 mole of H 2 O is 18.01528 g. (c) Each mole of H 2 O has 2H and 1O: there are 2 moles of H and 1 mole of O for each mole of water. PROBLEM: Suppose we have 12.25 g of water in a Dixie TM Cup? How many moles of water does this amount to? gram atomic weight of H 2 O is 18.01528 g/mole # of moles = 12.25 g / 18.01528 g) = 0.6800 mol.  Maintain your vigilance regarding significant figures!

6. PROBLEM: How many atoms of P are there in 31.0 g of it? How many atoms of C are there in 12.0 g of it? PROBLEM: How many atoms of P are there in 145.8 g of it? N A = 6.02  10 23 atoms of P in 31.0 g of it. N A = 6.02  10 23 atoms of C in 12.0 g of it. # of atoms = (# of moles)(N A ) = 145.8 g / 30.984 g)(6.02  10 23 atoms / 1 mol) = 2.83  10 24 atoms of P. PROBLEM: A sample of carbon has 1.28  10 24 atoms as counted by Marvin the Paranoid Android. a) How many moles is this? b) What is its mass? a) # of moles = 1.28  10 24 atoms/6.02  10 23 atoms) = 2.13 mol. b) mass = (2.13 mol)(12.011 g / mol) = 25.5 g of C.

7. EXAMPLE: A sample of carbon has 1.28  10 24 atoms as counted by Marvin the Paranoid Android. a) How many moles is this? b) What is its mass? a) (1.28  10 24 atoms)(1 mol / 6.02  10 23 atoms) = 2.13 mol. b) (2.13 mol)(12.011 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  0.625 mol  0.625 mol x 6.02 x 10 23 molecules  3.7625 x 10 23 molecules

8. Equation of state for an ideal gas An ideal gas can be characterized by four state variables: pressure (P), volume (V), number of moles (n) and absolute temperature (T). ideal gas law: experimental 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 m 3.  T is measured in K.

9. An ideal gas is kept in a container of fixed volume at a temperature of 30 0 C and a pressure of 6.0 atm. The gas is heated at constant volume to a temperature of 330 0 C. pressure 6.0 atm temperature 30 0 C gas new pressure temperature 330 0 C gas The new pressure of the gas is about A. 0.60 atmB. 3.0 atmC. 12 atmD. 66 atm. ▪ T i = 30 + 273 = 303 K. T(K) = T(°C) + 273 p i V i = nRT i, p f V f = nRT f. ▪ V i = V f. pfVfpiVipfVfpiVi = nRT f nRT i p f = p i T f / T i p f = (6)(603) / 303 = 12 ▪ T f = 330 + 273 = 603 K.

10.  GIVEN: p = 20  10 6 Pa, V = 2.0  10 -2 m 3.  From T(K) = T(°C) + 273 (i)  n, the number of moles =?. T(K) = 17 + 273 = 290 K.  pV = nRT ⇾ n = pV / (RT) n = (20  10 6 )(2  10 -2 ) / [ (8.31)(290) ] n = 170 mol. N = 170 mol  6.02  10 23 atoms mol N = 1.0  10 26 atoms. (ii)  N = n N A.

11. An Ideal Gas Is a theoretical gas that obeys the gas lawsIs a theoretical gas that obeys the gas laws And thus fit the ideal gas equation exactlyAnd thus fit the ideal gas equation exactly Real Gases Real gases conform to the gas laws under certain limited conditionsReal gases conform to the gas laws under certain limited conditions But they condense to liquids and then solidify if the temperature is loweredBut 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 gasFurthermore, there are relatively small forces of attraction between particles of a real gas This is not the case for an ideal gasThis is not the case for an ideal gas

12. 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.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 pressureA gas expands when heated at constant pressure The pressure increases when a gas is compressed at constant temperatureThe pressure increases when a gas is compressed at constant temperature But, why do gases behave this way?But, why do gases behave this way? What happens to gas particles when conditions such as pressure and temperature change?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.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 atomsThe 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.This theory is based on the following postulates, or assumptions.

13. Gases consist of tiny particles called atoms or moleculesGases consist of tiny particles called atoms or molecules The total number of particles in a sample is very largeThe total number of particles in a sample is very large The particles are in constant random motionThe particles are in constant random motion The range of the intermolecular forces is small compared to the average separationThe 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 pointsThe 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 containerCollisions of a short duration occur between particles and the walls of the container Collisions are perfectly elasticCollisions are perfectly elastic No forces act between the particles except when they collideNo forces act between the particles except when they collide Between collisions the particles move in straight linesBetween collisions the particles move in straight lines And obey Newton’s Laws of motionAnd obey Newton’s Laws of motion The kinetic model of an ideal gas

14. Gas consists of large numbers of tiny particles called atoms or Gas consists of large numbers of tiny particles called atoms or moleculesthat behave like hard, spherical objects in a state molecules that behave like hard, spherical objects in a state of constant, random motion. of constant, random motion. The size of the particles is relatively small compared with The size of the particles is relatively small compared with the distance between them, so they are treated as points the distance between them, so they are treated as points Collisions of a short duration occur between particles and Collisions of a short duration occur between particles and the walls of the container the walls of the container No Intermolecular forces act between the particles except when they collide, No Intermolecular forces act between the particles except when they collide, so between collisions the particles move in straight lines so between collisions the particles move in straight lines Collisions are perfectly elastic (none of the energy of a gas Collisions are perfectly elastic (none of the energy of a gas particle is lost in collisions) particle is lost in collisions) Energy can be transferred between molecules during collisions. Energy can be transferred between molecules during collisions. They all obey Newton’s Laws of motion They all obey Newton’s Laws of motion The kinetic model of an ideal gas

15. Macroscopic Behaviour  The large number of particles ensures that the number of particles moving in all directions is constant at any time 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 (macroscopic behaviour) to the behavior of the molecules themselves (microscopic behaviour). (microscopic behaviour). The kinetic model of an ideal gas

16. Pressure Pressure is the result of collisions between molecules Pressure is the result of collisions between molecules and the wall of the container and the wall of the container Focus on one molecule moving toward the wall and Focus on one molecule moving toward the wall and examine what happens when on molecule strikes this wall. examine what happens when on molecule strikes this wall. Elastic collision – no loss of kinetic energy, so speed remains the same, only direction changes. The kinetic model of an ideal gas ℓ

17. Change in momentum implies that there must be a Change in momentum implies that there must be a force exerted by the wall on the particle. force exerted by the wall on the particle. That means that there is a force exerted on the wall That means that there is a force exerted on the wall by that molecule. by that molecule. If you can imagine 3-D picture you can “see” that only the component of the molecule’s momentum perpendicular to the wall changes: from –mv x (moving in the negative x direction) to +mv x Thus the change in momentum for one collision is: ∆(mv) = mv x – (-mv x ) = 2mv x ∆(mv) = mv x – (-mv x ) = 2mv x The kinetic model of an ideal gas ● 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/v x

18. ▪ 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: The kinetic model of an ideal gas 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. ▪

19. 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 v x 2 must be the same as that of v y 2 or v z 2 ; it follows that (v 2 ) avg = 3(v x 2 ) avg or (v x 2 ) avg = ⅓(v 2 ) avg → Therefore the pressure on the wall is: pressure in a gas expressed in terms of molecular properties. 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.

20. The kinetic model of an ideal gas ½ m(v 2 ) avg = (KE) avg average kinetic energy of the molecules in the gas. So, pressure can be written as: We got key connection between microscopic behaviour and macroscopic observables. 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. PV = nRT If we compare the ideal-gas equation of state: PV = nRT with the result from kinetic theory, we find 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.

21. result from kinetic theory: The average translational kinetic energy of molecules in a gas The average translational kinetic energy of molecules in a gas is directly proportional to the absolute temperature. is directly proportional to the absolute temperature. The higher the temperature, according to kinetic theory, The higher the temperature, according to kinetic theory, the faster the molecules are moving on the average. the faster the molecules are moving on the average. At absolute zero they have zero kinetic energy. Can not go lower. At absolute zero they have zero kinetic energy. Can not go lower. This relation is one of the triumphs of the kinetic energy theory. This relation is one of the triumphs of the kinetic energy theory. We got key connection between microscopic behaviour and macroscopic observables. The absolute temperature is a measure of the The absolute temperature is a measure of the average kinetic energy of its molecules average kinetic energy of its molecules If two different gases are at the same temperature, their molecules If two different gases are at the same temperature, their molecules have the same average kinetic energy, but more massive molecules have the same average kinetic energy, but more massive molecules will have lower average speed. will have lower average speed. If the temperature of a gas is doubled, the average kinetic If the temperature of a gas is doubled, the average kinetic energy of its molecules is doubled energy of its molecules is doubled Absolute Temperature

22. The absolute temperature is a measure of the The absolute temperature is a measure of the average kinetic energy of its molecules average kinetic energy of its molecules If two different gases are at the same temperature, their molecules If two different gases are at the same temperature, their molecules have the same average kinetic energy, but more massive molecules have the same average kinetic energy, but more massive molecules will have lower average speed. will have lower average speed. If the temperature of a gas is doubled, the average kinetic If the temperature of a gas is doubled, the average kinetic energy of its molecules is doubled energy of its molecules is doubled Absolute Temperature Although the molecules of gas have an average kinetic energy (and therefore Although the molecules of gas have an average kinetic energy (and therefore an average speed) the individual molecules move at various speeds an average speed) the individual molecules move at various speeds Some are moving fast, others relatively slowly Some are moving fast, others relatively slowly At higher temperatures at greater fraction of the molecules At higher temperatures at greater fraction of the molecules are moving at higher speeds are moving at higher speeds For O 2 molecules at 300 K, the most probable speed is 390 m/s. For O 2 molecules at 300 K, the most probable speed is 390 m/s. When temperature increases to 1100 K the most probable speed increases 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 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. those that are very large, but these have much lower probabilities. Molecular Speed

23. Average kinetic/internal energy of an ideal gas  Since ideal gases have no intermolecular forces, their internal energy is stored completely as kinetic energy.  The individual molecules making up an ideal gas all travel at different speeds:  Without proof, the average kinetic energy E K of each ideal gas molecule has the following forms:  k B is called the Boltzmann constant. Topic 3: Thermal physics 3.2 – Modeling a gas E K = (3/2) k B T = 3RT / [ 2N A ] Average kinetic energy of ideal gas Where k B = 1.38  10 -23 J K -1.

25. The kinetic model of an ideal gas  An ideal gas is an imaginary gas that is used to model real gases, and has the following properties: Topic 3: Thermal physics 3.2 – Modeling a gas

26. The kinetic model of an ideal gas  Consider a syringe full of an ideal gas.  If we make the volume less we see that the temperature will increase.  Since the plunger exerts a force on the gas, and executes a displacement, it does work on the gas.  From the work-kinetic energy theorem we know that if we do work on the gas, its kinetic energy must increase.  Thus its speed will increase, which means its temperature increases.  On the other hand, if the process is reversed the gas will do the work, lose E K and cool. Topic 3: Thermal physics 3.2 – Modeling a gas

27. The kinetic model of an ideal gas  Temperature is a measure of the E K of the gas.  Reducing the E K reduces the frequency of collisions.  For perfectly elastic collisions (as in an ideal gas) contact time is zero regardless of E K. Topic 3: Thermal physics 3.2 – Modeling a gas

28. Differences between real and ideal gases  Recall the properties of an ideal gas:  The kinetic theory of gases is, of course, a model.  As such, it doesn’t apply perfectly to real gases. Topic 3: Thermal physics 3.2 – Modeling a gas

29. FYI  Real gases are often polyatomic (N 2, O 2, H 2 O, NH 4, 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. Differences between real and ideal gases  Here are the properties of a real gas. Topic 3: Thermal physics 3.2 – Modeling a gas

30. Differences between real and ideal gases  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. Topic 3: Thermal physics 3.2 – Modeling a gas

31. Constant pressure process – isobaric process  In an isobaric process, p does not change.  As an example of an isobaric experiment, suppose we take a beaker that is filled with an ideal gas, and stopper it with a gas-tight, frictionless cork and a weight, as shown.  The weight F causes a pressure in the gas having a value given by p = F / A, where A is the area of the cork in contact with the gas.  If we now heat up the gas it will expand against the cork, pushing it upward:  Since neither F nor A change, p remains constant. x ∆V∆V A Topic 3: Thermal physics 3.2 – Modeling a gas

32. FYI  If ∆V > 0 (gas expands) then W > 0.  If ∆V < 0 (gas contracts) then W < 0. Constant pressure process – work done by a gas  From the previous slide: p = F / A  F = pA.  From the picture note that  V = Ax.  Recall the work W done by the gas is just the force F it exerts on the weighted cork times the displacement x it moves the cork. Thus W = Fx = pAx = p  V. x ∆V∆V A F work done by expanding gas (constant p) W = p∆V Topic 3: Thermal physics 3.2 – Modeling a gas

33. FYI  The symbol  means “is proportional to.” EXAMPLE: Show that for an isolated ideal gas V  T during an isobaric process. SOLUTION: Use pV = nRT. Then V = ( nR / p )T.  Isolated means n is constant (no gas is added to or lost from the system).  Isobaric means p is constant.  Then n and P are constant (as is R). Thus V = ( nR / p )T = ( CONST )T V  T. ( isobaric process ) Constant pressure process – isobaric process Topic 3: Thermal physics 3.2 – Modeling a gas

34. Constant volume process – isovolumetric process  In an isovolumetric process, V does not change.  We have already seen an isovolumetric experiment when we studied the concept of absolute zero:  During an isovolumetric process the temperature and the pressure change.  Note that the volume was kept constant in this experiment. 0 10 20 30 p T (°C) -300 -200-100 0 100 200 300 Topic 3: Thermal physics 3.2 – Modeling a gas

35. FYI  Isovolumetric is sometimes called isochoric. EXAMPLE: Show that for an isolated ideal gas p  T during an isovolumetric process. SOLUTION:  Use pV = nRT. Then p = ( nR / V )T.  Isolated means n is constant (no gas is added to or lost from the system).  Isovolumetric means that V is constant.  Then n and V are constant (as is R). Thus p = ( nR / V )T = (CONST)T p  T. ( isovolumetric process ) Constant volume process – isovolumetric process Topic 3: Thermal physics 3.2 – Modeling a gas

36. EXAMPLE: A graduated syringe which is filled with air is placed in an ice bath and allowed to reach the temperature of the water. Demonstrate that p 1 V 1 = p 2 V 2. SOLUTION:  Record initial states after a wait: p 1 = 15, V 1 = 10, and T 1 = 0ºC.  Record final states after a wait: p 2 = 30, V 2 = 5, and T 2 = 0ºC. p 1 V 1 = 15(10) = 150. p 2 V 2 = 30(5) = 150.  Thus p 1 V 1 = p 2 V 2. Constant temperature process –isothermal process  In an isothermal process, T does not change. 0 10 20 30 Why do we wait before recording our values? Topic 3: Thermal physics 3.2 – Modeling a gas

37. PRACTICE: Show that for an isolated ideal gas p 1 V 1 = p 2 V 2 during an isothermal process. SOLUTION:  From pV = nRT we can write p 1 V 1 = nRT 1 p 2 V 2 = nRT 2.  Isolated means n is constant.  Isothermal means T is constant so T 1 = T 2 = T.  Obviously R is constant.  Thus p 1 V 1 = nRT = p 2 V 2. ( isothermal ) Constant temperature process –isothermal process Topic 3: Thermal physics 3.2 – Modeling a gas

38. FYI  We “built” the 3D sphere with layers of 2D circles.  We have transformed a 3D surface into a stack of 2D surfaces. Sketching and interpreting state change graphs  Perhaps you have enjoyed the pleasures of analytic geometry and the graphing of surfaces in 3D.  The three variables of a surface are x, y, and z, and we can describe any surface using the "state" variables x, y, and z:  The “equation of state” of a sphere is x 2 + y 2 + z 2 = r 2, where r is the radius of the sphere. y z Topic 3: Thermal physics 3.2 – Modeling a gas x

39. FYI  Each layer is an isotherm.  The 3D graph (above) can then be redrawn in its simpler 2D form (below) without loss of information. Sketching and interpreting state change graphs  The three state variables of a gas (if n is kept constant) are analogous.  We can plot the three variables p, V, and T on mutually perpendicular axes like this:  We have made layers in T. Thus each layer has a single temperature. V p T V p i s o t h e r m s T 1 T 2 T 3 T 4 Topic 3: Thermal physics 3.2 – Modeling a gas T 1 T 2 T 3 T 4

40. FYI  The purple line shows all three states changing. EXAMPLE: In the p-V graph shown, identify each process type as ISOBARIC, ISOTHERMAL, OR ISOVOLUMETRIC (isochoric). SOLUTION:  A  B is isothermal (constant T).  B  C is isobaric (constant p).  C  A is isochoric (constant V).  We will only have two states change at a time. Phew! Sketching and interpreting state change graphs  A thermodynamic process involves moving from one state to another state. This could involve changing any or even all of the state variables (p, V, or T). p i s o t h e r m i s o t h e r m A B C V Topic 3: Thermal physics 3.2 – Modeling a gas D

41. EXAMPLE: An internal combustion engine is an example of a heat engine that does useful work on the environment. A four-stroke engine is animated here.  This example illustrates how a gas can be made to do work and illustrates a thermodynamic cycle. http://www.animatedengines.com/otto.html http://chemcollective.org/activities/simulations/engine Sketching and interpreting state change graphs Topic 3: Thermal physics 3.2 – Modeling a gas

42. EXAMPLE: A fixed quantity of a gas undergoes a cycle by changing between the following three states: State A: (p = 2 Pa, V = 10 m 3 ) State B: (p = 8 Pa, V = 10 m 3 ) State C: (p = 8 Pa, V = 25 m 3 ) Each process is a straight line, and the cycle goes like this: A  B  C  A. Sketch the complete cycle on a p-V diagram. SOLUTION:  Scale your axes and plot your points… Sketching and interpreting state change graphs  A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. p V 2 8 1025 A B C Topic 3: Thermal physics 3.2 – Modeling a gas

43. EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example): (a) Find the work done during the process A  B. (b) Find the work done during the process B  C. SOLUTION: Use W = p  V. (a) From A to B:  V = 0. Thus the W = 0. (b) From B to C:  V = 25 – 10 = 15; p = 8. Thus W = p  V = 8(15) = 120 J. Sketching and interpreting state change graphs p V 2 8 1025 A B C Topic 3: Thermal physics 3.2 – Modeling a gas

44. EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example): (c) Find the work done during the process C  A. SOLUTION:  Observe that ∆V is negative when going from C (V = 25) to A (V = 10).  Observe that p is NOT constant so W  p∆V.  W = Area under the p-V diagram. = - [ (2)(15) + (1/2)(6)(15) ] = - 75 J. Sketching and interpreting state change graphs p V 2 8 1025 A B C Topic 3: Thermal physics 3.2 – Modeling a gas

45. FYI  Because W cycle is positive, work is done on the external environment during each cycle. EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example): (d) Find the work done during the cycle A  B  C  A. SOLUTION: (d) Just total up the work done in each process.  W A  B = 0 J.  W B  C = + 120 J.  W C  A = -75 J.  W cycle = 0 + 120 – 75 = + 45 J. Sketching and interpreting state change graphs p V 2 8 1025 A B C Topic 3: Thermal physics 3.2 – Modeling a gas

46. FYI  Reversing the cycle reverses the sign of the work. PRACTICE: Find the total work done if the previous cycle is reversed. SOLUTION:  We want the cycle A  C  B  A.  W A  C = Area = + [ (2)(15) + (1/2)(6)(15) ] = + 75 J.  W C  B = P∆V = 8(10–25) = - 120 J.  W B  A = 0 J (since ∆V = 0).  W cycle = 75 - 120 + 0 = - 45 J. Sketching and interpreting state change graphs p V 2 8 1025 A B C Topic 3: Thermal physics 3.2 – Modeling a gas

47.  Fixed mass and constant volume means n and V are constant. Thus  pV = nRT  p = (nR/V)T  p = (CONST)T. (LINEAR)  Since the t axis is in ºC, but T is in Kelvin, the horizontal intercept must be NEGATIVE… Sketching and interpreting state change graphs

48.  p = 0 at absolute zero.  From pV = nRT:  p = (1R / V )T: Topic 3: Thermal physics 3.2 – Modeling a gas Sketching and interpreting state change graphs

49. EXAMPLE: 2.50 moles of hydrogen gas is contained in a fixed volume of 1.25 m 3 at a temperature of 175  C. a) What is the average kinetic energy of each atom? b) What is the total internal energy of the gas? SOLUTION: T(K) = 175 + 273 = 448 K. a) E K = (3/2) k B T = (3/2)(1.38  10 -23 )(448) = 9.27  10 -21 J. b) From n = N / N A we get N = nN A. N = (2.50 mol)(6.02  10 23 atoms / mol) = 1.51  10 24 atm. E K = NE K = (1.51  10 24 )(9.27  10 -21 J) = 14000 J. Average kinetic/internal energy of an ideal gas Topic 3: Thermal physics 3.2 – Modeling a gas E K = (3/2) k B T = 3RT / [ 2N A ] Average kinetic energy of ideal gas Where k B = 1.38  10 -23 J K -1.

50. EXAMPLE: 2.50 moles of hydrogen gas is contained in a fixed volume of 1.25 m 3 at a temperature of 175  C. c) What is the pressure of the gas at this temperature? SOLUTION: T(K) = 175 + 273 = 448 K. c) Use pV = nRT: Then p = nRT / V = 2.50  8.31  448 / 1.25 = 7450 Pa. Average kinetic/internal energy of an ideal gas Topic 3: Thermal physics 3.2 – Modeling a gas E K = (3/2) k B T = 3RT / [ 2N A ] Average kinetic energy of ideal gas Where k B = 1.38  10 -23 J K -1.

51. Application of the "Kinetic Molecular Theory" to the Gas Laws Microscopic justification of the laws

52. 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 = NkT → P = (const) T example: a closed jar, or aerosol can, thrown into a fire will explode due to increase in gas pressure inside.

53. Microscopically: As T increases, KE of molecules increase As T increases, KE of molecules increase That implies greater change in momentum when they hit That implies greater change in momentum when they hit the wall of the container the wall of the container Thus microscopic force from each molecule on the wall Thus microscopic force from each molecule on the wall will be greater will be greater As the molecules are moving faster on average they As the molecules are moving faster on average they will hit the wall more often will hit the wall more often The total force will increase, therefore the pressure The total force will increase, therefore the pressure will increase will increase

54. The Charles’s law Effect of a volume increase at a constant pressure Macroscopically: at constant pressure, volume of a gas is proportional to its temperature: is proportional to its temperature: PV = NkT → V = (const) T PV = NkT → V = (const) T

55. An increase in temperature means an increase in the average kinetic energy of the gas molecules, thus an increase in speed An increase in temperature means an increase in the average kinetic energy of the gas molecules, thus an increase in speed Microscopically: There will be more collisions per unit time, furthermore, the momentum of each collision increases (molecules strike the wall harder) 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 Therefore, there would be an increase in pressure If we allow the volume to change to maintain If we allow the volume to change to maintain constant pressure, the volume will increase with increasing temperature

56. 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 = NkT → P = (const)/V

57. Microscopically: Constant T means that the average KE of the gas molecules remains constant 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 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 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 Therefore, the pressure will decrease