Kinetic Theory. Reading: Chapter 19. Ideal Gases. Ideal gas law:

Transcription

1 Reading: Chapter 19 Ideal Gases Ideal gas law: Kinetic Theory p nrt, where p pressure olume n number of moles of gas R 831 J mol -1 K -1 is the gas constant T absolute temperature All gases behae like ideal gases at low enough densities Work Done by an Ideal Gas at Constant Temperature Isothermal expansion: During the expansion, the temperature is kept constant Isothermal compression: During the compression, the temperature is kept constant Using the ideal gas law, an isotherm on a p- diagram is gien by p nrt 1

2 Work done by an ideal gas during an isothermal expansion from i to f : f W p d W i f i nrt d Since T is constant in an isothermal expansion, f d W nrt i nrt f [ ln ] i Since ln a ln b ln (a/b), W nrt ln f i For isothermal expansion, f > i, W is positie For isothermal compression, f < i, W is negatie Other thermodynamic processes: For constant-olume processes, W 0 For constant-pressure processes, W p( f i ) p

3 Examples 19-1 A cylinder contains 1 L of oxygen at 0 o C and 15 atm The temperature is raised to 35 o C, and the olume reduced to 85 L What is the final pressure of the gas in atmospheres? Assume that the gas is ideal Since p nrt, p T p T i i f f f i p p i i f f T T f i nr atm atm (ans) 3

4 19- One mole of oxygen (assume it to be an ideal gas) expands at a constant temperature T of 310 K from an initial olume i of 1 L to a final olume f of 19 L (a) How much work is done by the expanding gas? (b) How much work is done by the gas during an isothermal compression from i 19 L to f 1 L? f (a) W nrt ln i 19 (1)(831)(310) ln J (ans) i (b) W nrt ln f 1 (1)(831)(310) ln J (ans) Work is done on the gas to compress it 4

5 Pressure, Temperature and RMS Speed Consider n moles of an ideal gas in a cubical box of olume at temperature T The gas molecules are moing in the box When they collide with the walls, the collision is elastic and momentum is transferred to the walls Using Newton s second law, this change in momentum results in a force acting on the wall This force comtributes to the pressure of the gas In the x direction, the elocity of a molecule of mass m changes from x to x when it collides with the shaded wall Change in momentum p x m x The time between collisions t L/ x Hence the rate at which momentum is transferred to the shaded wall is: p t mx L / x x m L x Using Newton s law, this is the force exerted by the gas on the shaded wall Summing up the forces due to all gas molecules, 5

6 m1x + L+ mnx m F ( 1x + L+ Nx ) L L Pressure: F m p ( 1 ) 3 x + L + Nx L L To simplify this expression, let 1 x ( 1x + L + Nx ) N Also, L 3 Therefore, p mn x Since mn is the total mass of the gas, mn nm, where n number of moles of gas, M mass of 1 mole of gas (the molar mass of the gas) p nm x Since x + y + z, and x y z, we hae x 1 3 Hence 6

7 p nm 3 The root-mean-square speed rms for N molecules is defined by 1 rms ( 1 + L + N ) N This yields p nm 3 rms This tells us that the (macroscopic) pressure of the gas depends on the (microscopic) speed of the molecules Using the ideal gas law, p nrt, we hae rms 3RT M Example 19-3 Here are fie numbers: 5, 11, 3, 67, 89 (a) What is the aerage alue n of these numbers? (b) What is the rms alue n rms of these numbers? (a) n 40 8 (ans) (b) n rms 5 1 (ans) 5 7

8 Translational Kinetic Energy The translational kinetic energy of the gas: 1 K m m 1 rms 1 m Since rms 3RT / M, K 1 3RT m M Since M N A m, where N A is the Aogadro s number (number of molecules in 1 mole mol -1 ), This is generally written as K 3RT N A 3 K kt, where k is called the Boltzmann constant, gien by k N R A 31 Jmol JK 1 1 K mol Conclusion: At a gien temperature T, all ideal gas molecules no matter what their mass hae the same aerage translational kinetic energy, namely, kt When we measure the temperature of a gas, we are also measuring the aerage translational kinetic energy of its molecules 3 8

9 Mean Free Path Mean free path λ is the aerage distance traersed by a molecule between collisions A collision will take place if the centers of two molecules come within a distance d of each other Consider a single molecule traeling at constant speed and assume that all other molecules are at rest In time t, the center of the moing molecule sweeps out a cylinder of length t and radius d; any other molecule whose center lies in this cylinder will be collided olume of the cylinder (πd )( t) Number of molecules in the cylinder (πd )( t)(n/) Mean free path: length of path during t t 1 λ number of collisions in t πd tn / πd N / Howeer, since all molecules are moing, the expression should include an extra factor of 1 / That is, λ 1 πd N / To obtain the extra factor, let 1 and be elocity of two molecules Then the relatie elocity is estimated as 9

10 rel ( 1 ) ( 1 ) Aeraging both sides, and noting that 0, 1 rel It becomes reasonable to estimate that rel Maxwell Distribution of Molecular Speeds Maxwell found that when the probability of finding a molecule at an energy state E is E P( E) exp kbt z In the state space, the probability x of finding a molecule in a cube of sides d x, d y, d z at elocity ( x, y, z ) is y P (,, ) d x y z x d y d m( x + y + z ) Aexp dxd kbt m Aexp dxdydz kbt z In spherical coordinates, the olume with elocity between and + d is 4π d Hence the probability of finding a molecule with elocity between and + d is y d z 10

11 11 4 exp ) ( d T k m A d P B π A is a constant satisfying 1 ) ( 0 d P The mathematical result is 3/ T k m A B π Final result: exp 4 ) ( 3/ T k m T k m P B B π π Fraction of molecules with speed between 1 and 1 ) ( frac d P Aerage speed 8 8 ) ( 0 M RT m T k d P B π π

12 Mean square speed 0 P( ) d 3kBT m rms speed 3kB T rms m 3RT M Most probable speed kbt m P RT M 1

13 Internal Energy Consider a monatomic ideal gas (eg helium, neon, argon) Internal energy: translational kinetic energy of the atoms no rotational kinetic energy (because monatomic) no potential energy (because no intermolecular force) For n moles of the gas, number of molecules nn A aerage kinetic energy of a molecule total internal energy: 3 kt E int 3 ( nn A ) kt Since N A k R, 3 Eint nrt (monatomic gas) The internal energy E int of an ideal gas is a function of the gas temperature only; it does not depend on its pressure or density 13

14 Molar Specific Heat at Constant olume Using the first law of thermodynamics, E int Q W Since the olume is fixed, W 0 Furthermore, E int 3 nr T Therefore, 3 nr T Q Molar specific heat at constant olume: C Q n T, 3 C R (monatomic gas) Its alue is C 15 Jmol 1 K 1 14

15 General Kinds of Gases Polyatomic molecules possess both translational and rotational kinetic energy Hence their internal energy and specific heat are greater than those of monatomic gases In general, if the molar specific heat at constant olume is C, then and E E int nct, int nc T A change in the internal energy E int of an ideal gas at constant olume depends on the change in the gas temperature only; it does not depend on what type of process produces the change in temperature path 1: constant olume path : constant pressure path 3: adiabatic compression 15

16 Molar Specific Heat at Constant Pressure Using the first law of thermodynamics, E int Q W For ideal gases, p nrt Thus for work done at constant pressure, Furthermore, Therefore, W p nr T E int nc T nc T Q nr T, Q nc T + nr T Molar specific heat at constant pressure: C P Q, CP C + R n T 16

17 Equipartition of Energy (Maxwell) If a kind of molecule has f independent ways to store energy, then it has f degrees of freedom Each degree of freedom has an aerage energy of 1 kt per molecule (or 1 RT per mole) associated with it eg helium: 3 translational + 0 rotational E nr T, and C int 3 3 eg oxygen: 3 translational + rotational E nr T, and C int 5 5 R R eg methane: 3 translational + 3 rotational E 3 nr T, and C 3 R int In general, f E nr T, and C int f R 17

18 Example 19-7 A bubble of 5 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase T of 0 o C at constant pressure As a result, the bubble expands (a) How much energy is added to the helium as heat during the increase and expansion? (b) What is the change E int in the internal energy of the helium during the temperature increase? (c) How much work W is done by the helium as it expands against the pressure of the surrounding water during the temperature increase? (a) C P C + R 3 R + R 5 Q ncp T n R C T (5)(5)(831)(0) 080 J (ans) 3 (b) Eint nc T n R C T (5)(15)(831)(0) 150 J (ans) (c) Using the first law of thermodynamics, W Q E int J (ans) 5 R 18

19 19-8 We transfer 1000 J to a diatomic gas, allowing it to expand with the pressure held constant The gas molecules rotate but do not oscillate How much of the 1000 J goes into the increase of the gas s internal energy? Of that amount, how much goes into K tran (the kinetic energy of the translational motion of the molecules) and K rot (the kinetic energy of their rotational motion)? Q 1000 J A diatomic gas has 5 degrees of freedom Therefore, C 5R/ and C P 7R/ At constant pressure, Q T 7nR / 5 Q 5 E nc T n R Q int 7nR / J (ans) 7 There are 3 translational degrees of freedom and rotational degrees of freedom Thus, 3 3 Q 3 K tran n R T n R Q 7nR / J (ans) 7 Q K n R T nr Q rot 7nR / J (ans) 7 19

20 A Hint of Quantum Theory Below about 80 K, C /R 15 3 translational degrees of freedom As the temperature increases, C /R increases to 5, two additional degrees of freedom These are the rotational degrees of freedom, and this motion requires a certain minimum amount of energy In quantum theory, an energy quantum is required to excite the rotation This energy is not aailable at low temperatures Aboe 1000 K, C /R increases to 35, because the oscillatory degrees of freedom are excited The energy quantum required to excite oscillations is not aailable at lower temperatures 0

21 The Adiabatic Expansion of an Ideal Gas For adiabatic processes, Q 0 Using the first law of thermodynamics, adiabatic compression causes heating, adiabatic expansion causes cooling We can proe: p γ constant, where γ C P /C, the ratio of the molar specific heats Or, p 1 γ 1 p γ Using the ideal gas equation, p nrt, nrt T γ γ constant, 1 constant or T 1 γ 1 1 T γ 1 Example: Fog formation when a cold can of soft drink is opened 1

22 Proof: First law of thermodynamics for a small expansion: deint dq dw, where de int nc dt, dq 0 for adiabatic processes, dw pd From the ideal gas equation, p nrt, so that dw nrt d Substituting, nc dt nrt 0 d, we arrie at the differential equation d dt C R T Solution (erify by substitution): C / R constant T R / C, T constant Since R/C (C P C )/C γ 1, we get the adiabatic laws

23 Free Expansion No work is done: W 0 No heat is transferred: Q 0 Hence no change in internal energy: E int Q W 0 For ideal gases, this implies no change in temperature: T i T f or p i i p f f 3

24 Example mol of oxygen expands adiabatically from an initial olume of 1 L at 310 K to a final olume of 19 L (a) What would be the final temperature of the gas? Oxygen is diatomic and here has rotation but not oscillation (b) What is the initial and final pressure of the gas? (c) If, instead, the gas had expanded freely from the initial olume to the final olume, what would be the final temperature and pressure? (a) Oxygen has 5 degrees of freedom, therefore 5 7 CP 7 C R, CP R, γ 14 C 5 γ Since T 1 constant, γ i 1 T f Ti K γ 1 (ans) f 19 (b) Using ideal gas law, initial pressure: nrti ( 1)(831)(310) p i Nm 1 atm 3 i 1 10 Final pressure: nrtf p f Nm 111atm 3 f (c) For free expansion, T i T f and p i i p f f final temperature 310 K final pressure i 1 p Nm f pi 134 atm f 19 (ans) 4

25 Phase Transitions A phase transition is a discontinuous change in the properties of a substance when its enironment changes infinitesimally For example water can change from solid to liquid at 0 or from liquid to gas at 100 But other factors, such as pressure, can also affect a phase transition besides temperature The phase diagram show the phases (solid, liquid, gas) as a function of T and P liquid/gas coexistence (eaporation) solid/liquid/gas coexistence Solid/gas coexistence(sublimation) The lines in the phase diagram indicate the conditions that two phases can coexist in equilibrium (See dry ice bomb in YouTube) In the regions separated by the phase lines, respectiely, ice, water and steam are the most stable phases Howeer, metastable phases can still exist Example: supercooled water can exist at the left of the solid/liquid coexistence line Example: Superheated water can exist at the right of the liquid/gas coexistence line See superheating and supercooling in YouTube 5

26 Compare the phase diagram for H O aboe to CO below A ery important difference is that the solid-liquid phase boundary for CO has a positie slope (melting temperature increases when pressure increase), which is the usual behaior For H O, the solid-liquid boundary, actually has a negatie slope (melting temperature decreases with increasing pressure), which is unusual That is why we can skate on ice but not on any other solid surface! At the phase boundary between liquid and gas, the liquid coexists at some temperature with gas at an appropriate pressure called the apour pressure the apor pressure The liquid-gas phase boundary has positie slope 1 At higher temperature, a greater pressure must be applied to keep the liquid and gas in coexistence At higher pressure, the gas becomes more dense 3 At the critical point, the liquid and dense gas becomes indistinguishable - then we call the system a fluid (See critical point of benzene in YouTube) 6

27 apor Pressure The apor pressure of a apor is its pressure when it is in thermodynamic equilibrium with its condensed phase in a closed system All liquids hae a tendency to eaporate, and some solids can sublimate into a gaseous form ice ersa, all gases hae a tendency to condense back to their liquid form, or deposit back to solid form The equilibrium apor pressure is an indication of a liquid's eaporation rate It relates to the tendency of particles to escape from the liquid (or a solid) A substance with a high apor pressure at normal temperatures is often referred to as olatile Obserations 1 The atmospheric pressure boiling point of a liquid (also known as the normal boiling point) is the temperature at which the apor pressure equals the ambient atmospheric pressure With any incremental increase in that temperature, the apor pressure becomes sufficient to oercome atmospheric pressure and lift the liquid to form apor bubbles inside the bulk of the substance Bubble formation deeper in the liquid requires a higher pressure, and therefore higher temperature, because the fluid pressure increases aboe the atmospheric pressure as the depth increases 3 The boiling point of water at high altitude is lower than 100 o C due to the lower atmospheric pressure 7

28 4 When the temperature is lowered, the apor pressure of water apor decreases Condensation of water takes place at the dew point 5 The relatie humidity is the density of water apor diided by saturated apor density It is also equal to the apor pressure of water apor diided by the saturated apor pressure When the relatie humidity reacahes 100%, condensation takes place 6 The apor pressure that a single component in a mixture contributes to the total pressure in the system is called partial apor pressure For example, air at sea leel, saturated with water apor at 0 C has a partial pressures of 3 mbar of water, and about 780 mbar of nitrogen, 10 mbar of oxygen and 9 mbar of argon Surface Tension (Reference: Wikipedia) Surface tension is a property of the surface of a liquid that allows it to resist an external force It is reealed, for example, in floating of some objects on the surface of water, een though they are denser than water, and in the ability of some insects (eg water striders) and een reptiles (basilisk) to run on the water surface This property is caused by cohesion of like molecules, and is responsible for many of the behaiors of liquids Surface tension has the dimension of force per unit length, or of energy per unit area The two are equialent but when referring to energy per unit of area, people use the term surface energy which is a more general term in the sense that it applies also to solids and not just liquids See Wikipedia for the examples and pictures 8

29 Two Definitions dx L F Let γ be the surface tension of the liquid interface Suppose the liquid interface is stretched by dx Applied force: F γl Work done: dw Fdx γldx Increase in surface energy: de dw γldx de Surface energy per unit area: γ Ldx This shows that the surface tension can be interpreted as the surface force per unit length, as well as the surface energy per unit area Remark: If the interface in the aboe experiment is a film, eg a soap film, then there is an interface aboe the film and interface below the film The force, work done, and surface energy has to be multiplied by Surface Curature and Pressure F dθ R F L Let p be the excess pressure inside the cylindrically cured interface 9

30 Force on the cured segment due to the excess pressure p( LRdθ ) This is balanced by the ertical components of the surface tension dθ γlsin γldθ Equating the two forces, p ( LRdθ ) γldθ γ p R If the surface is cured in both the x and y directions, 1 1 p γ + R x R y Remark: This result is applicable to interfaces of a bulk droplet, eg a water drop For interfaces of a film droplet, eg a soap bubble, the result should be multiplied by Capillary Action Let r be the radius of the capillary tube Upward force due to surface tension ( γ cosθ ) πr Weight of the liquid column ( ρπr h) g Equating the forces, ( γ cosθ )πr ( ρπr h) g γ cosθ h ρgr θ h γ 30