Chapter 1 - The Properties of Gases. 2. Knowledge of these defines the state of any pure gas.

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1 Chapter 1 - The Properties of Gases I. The perfect gas. A. The states of gases. (definition) 1. The state variables: volume=v amount of substance, moles = n pressure = p temperature = T. Knowledge of these defines the state of any pure gas. 3. These variables do not vary independently, but are connected by an equation of state. In general, pressure p = f(t,v,n) e.g., simplest: pv = nrt (perfect or ideal gas) 4. Therefore, knowledge of any three plus the equation of state defines the state of the gas. 5. V and n are extensive properties, proportional to size of system. T and p are intensive properties, independent of size of system. 6. Pressure p = force per unit area that a system exerts on its surroundings. Two major sources of pressure: i) downward gravitational force, and ii) impacts of molecules on surroundings. Pressure exerted by a gas is a manifestation of incessant impact of particles on their surroundings. Above system is in mechanical equilibrium with its surroundings when the force per unit area of system particles pushing the piston out equals force per unit area pushing in. p system = p surroundings Piston stops moving when this is achieved. 1

2 7. Units of pressure: SI unit of force = N (Newton) = kg m/s area = m (meters ) so SI unit of pressure = N/m 1 Pascal (Pa) = 1 N/m 8. Other more familiar and useful units of pressure: 1 atmosphere (atm) = the pressure required to support a column of Hg 760 mm high at 0 C. 1 atm = 760. mm Hg = 760. Torr 1 atm = kpa (kilopascals) 1 bar = 100 kpa so 1 atm ~ 1 bar 9. Definition of standard pressure for reporting thermodynamic data: = 1 bar = Torr 10. Measurement of pressure - The Torricelli barometer: P grav = gdh g = acceleration of gravity = 9.81 m/s d = density of Hg = 13.6 g/cm 3 or 13.6 x 10 3 kg/m 3 h = 760 mm = m 3 Hg rises until the forces balance and atmospheric p = p grav This height h is 760. mm at sea level on an average day.

3 11. Temperature T: A measure of thermal energy per unit sample of matter, and a manifestation of molecular motion, difficult to define in words. It is the property that tells which direction heat will flow. 1. Zeroth Law of thermo = if A is in thermal equilibrium with B, and B with C, then C is also in thermal equilibrium with A. 13. Measurement of temperature Thermometer. Simplest are based on thermal expansion of a liquid in a capillary (Hg or alcohol). Celsius scale: 0 mark in system where water and ice are in equilibrium 100 mark in system where liquid water is boiling at sea level (in equilb with steam) Above is the typical Hg or alcohol thermometer. Ambiguity: Hg and alcohol don't expand perfectly linearly with T. Better thermometer uses expansion of an ideal gas (The ideal gas thermometer): since p T at fixed V 14. Temperature scales (K = Kelvin) T/K = θ/ C (θ is symbol used for temperature in Celsius) 3

4 B. The gas laws. 1. The ideal gas law (or perfect gas law): PV = nrt R = universal gas constant = J K -1 mol -1 = L atm K -1 mol -1 = cal K -1 mol -1 Note: 1 J = 1 Pa-m 3 Works well for most gases at room T and 1 atm. Works well for all gases at lower pressures or higher temperatures.. The response to pressure - Boyle's Law (166): p 1/V (at constant n, T) or p 1 V 1 = p V (at constant n, T) or pv = constant (at constant n, T) p-v isotherms: lines of constant T 4

5 3. The response to temperature: V T (at constant n, p) - Charles' Law V 1 /T 1 = V /T V-T isobars (lines of constant Pressure) Q: How would this plot look if one plotted vs Celsius temperature? p T (at constant n, V) - Gay-Lussac's Law p 1 /T 1 = p /T p-t isochores (lines of constant volume) 5

6 4. Avogadro's principle: V n (at constant p, T) molar volume = V m = volume occupied per mole of molecules = V n. is independent of type of molecule. P V m = RT V m = RT/P Problem: Calculate V m in liters (L) for an ideal gas at 0 C and 1.00 atm pressure. V m = R(0 C )/1.00atm [must use Kelvin!] What R to use? L atm K -1 mol -1 V m = * 73.15K/1.00 atm V m =.415 L 5. Standard conditions for reporting data. STP - old definition, standard temp. and pressure: 0 C, 1 atm SATP - standard ambient temp. & pressure: 98.15K (5 C), 1 bar 6. Mass density of ideal gases. Let ρ = mass density = m/v = mass/volume moles n = m/m = mass/molar_mass pv = nrt pv = (m/m) RT m/v = Mp/RT ρ= Mp/RT Therefore ρ M 6

7 7. Combined plot of equation of state at fixed n: Combined gas laws for fixed number of moles n p! V! T! = p!v! T! Combined gas laws for variable number of moles p! V! n! T! = p!v! n! T! 8. Variation of atmospheric pressure with altitude h given by the barometric formula: h /H p = p o e where Here : H = RT /Mg M = average molar mass of air g= acceleration of gravity T is varying with altitude 7

8 9. Dalton's law - mixtures of gases. The pressure exerted by a mixture of perfect gases is the sum of partial pressures exerted by the individual gases. total pressure p = nrt/v = (n A + n B +...)RT/V p = n A RT/V + n B RT/V +... p = p A + p B +... p A and p B... are partial pressures, n = n A + n B Mole fractions and partial pressures. mole fraction of component J defined as: x J = n J /n so x A + x B +... = 1; therefore: proof: p J = x J p p J = n J RT/V = (n J /n) (n RT/V) = (n J /n) P so p J = x J p 8

9 II. Kinetic Theory of Gases. Molecular (microscopic & classical) theory of gases which predicts: a) equilibrium thermodynamic properties - Pressure b) dynamical properties - rates of processes: effusion through openings, viscosity of gases, collision frequencies (used in collision theory of chemical reactions) A. Model of Perfect Gas. 1. Elementary kinetic theory assumptions. a. Gas sample is mostly empty space (molecules themselves occupy small volume). b. No attractions or repulsions between molecules in free flight (i.e. travel in straight lines) c. Perfectly elastic collisions between molecules (no energy transferred into internal degrees of freedom: rot, vib, elec) d. Pressure exerted by gas = time-averaged force per unit area exerted by many individual impacts of molecules on a surface. e. Classical variables describe gas. - position x of molecule - velocity v of molecule - v x, v y, v z are Cartesian components of velocity. Molecular Speed. molecular speed v = v = scalar quantity, not vector v = magnitude of the velocity vector of an individual molecule v = v x + v y + v z v = v x + v y + v z 9

10 3. Root-mean-square (rms) speed of sample = c = v 1/... denotes average (c whole sample; v single molecule) v = N i=1 v i N But v i = v xi + y yi + v zi v = = i v + v + v ( xi ) yi zi v xi N N + v y N + v z = v x mean-square speed in x-direction + v y (v xi veloc in x-dir for molecule i) i + v z But since motion is random: N i v x = v y = v z So: v = 3 v x root-mean-square speed: v or v has units speed or velocity 10

11 3. Pressure exerted by N molecules. Container volume V = a x b x c follow 1 typical molecule and its motion in x-direction time interval between collisions with face A = Δt Δt = time required to travel distance a velocity x time = distance v x Δt = a Δt = a v x To calculate pressure P, need force per unit area 1 st, find time-averaged force F exerted by 1 molecule avg force in time Δt (during which we have 1 collision) F = force of 1 collision F = momentum change in 1 collision Δt F = mv x Δt = mv x a P 1 = pressure due to 1 molecule = Force Area ( P 1 = mv / a x ) bc P 1 = mv ( x) abc = mv x V where area of face A = bc where V is container volume Now the pressure due to N molecules in volume V: P = Nm v x V 11

12 But we already showed that v = 3 v x So: P = Nm v 3V key result Recognize N V = N = number density (not mass density) P = N m v 3 Could also write: PV= 1 3 Nmc Now, avg translational (kinetic) energy E t of a single molecule of mass m is: E t = 1 mv = 1 m v So: PV = 3 NE t But we also know that: or PV= nrt PV= Nk B T (where n = moles and N = molecules) So: NE = Nk T 3 t B E t = 3 k B T Equipartition theorem result for translational energy So avg k.e. of a molecule depends only on T, not on mass of particle 1

13 Problem: Calculate v and v 1/ of He atoms and Cl molecules at 300 K. PV = Nm v 3 = Nk B T So: m v 3 = k B T v = 3k B T m c = v! = 3k T $ B # " m & %! or = 3RT $ # & " M % v He = JK 1 300K 4.00 g mol 10 3 kg = m 1 mol g s c He = v = 1,367 m He s v = v 4.00 Cl He = m s c Cl = v Cl s = 35 m 13

14 B. Distribution of Molecular Velocities. Although every identical gas molecule in a sample may have same average speeds and average Cartesian velocity components (averaged over period of time), at any instant of t there is a wide range of velocities possible. Let f(v x )dv x = fraction of molecules having x-component of velocity in range v x to v x +dv x. Also = prob that a single given molecule has x-comp of velocity in rangev x to v x +dv x. f(v x ) = probability density that a single given molecule has x-comp of velocity in range v x to v x +dv x. 1. f(v x ) dv x v x This distribution is the Maxwell-Boltzmann distribution, which can be derived from the Boltzmann equation ( ) = $ πk T B f v x " $ # m % ' & e mv /k x B T Have similar probability densities in y and z directions So: F(v x, v y, v z ) = prob density that a given molecule has v x, v y, v z components 14

15 Max-Boltz in 3D ( ) = $ πk T B F v x,v y,v z " $ # m where v x + v ( y + v z ) = v % ' & 3/ distribution of molecular velocities e m v +v ( x y +v z ) /k B T F(v x, v y, v z )dv x,dv y,dv z = a probability or fraction, is normalized Meaning that dv x dv y dv z F ( v x,v y,v z ) = 1 dv x e mv /k x B T % πkt ( ' * & m ) dv y e mv /k y B T % πkt ( ' * & m ) dv z e mv /k z B T % πkt ( ' * & m ) =1 =1 =1 Calculation of averages using M-B distribution v x = dv x dv y dv z v x F ( v x,v y,v z ) dv x v x e mv /k x B T % πkt ( ' * & m ) e mv /k y B T dv y % πkt ( ' * & m ) e mv /k z B T dv z % πkt ( ' * & m ) 1 =1 =1 integral over v x is integral of type x e αx dx = where α = m k B T π α 3/ 15

16 So: v x π = m 3/ " % " πkb T % 1 $ # k B T ' $ & # m ' & ( ) ( 1 ) v x = k B T m v = 3 v x = 3k B T m = 3RT M c = v! c = 3k T $ B # " m & =! 3RT $ # & % " M % C. Maxwell Distribution of Speeds. Let F(v)dv = fraction of molecules having speed in range v to v+dv or = prob that a given molecule has speed in range v to v+dv (v= length of the velocity vector) F(v)dv = F(v x,v y,v z ) 4πv dv Maxwell Boltzmann distribution of speeds F(v)dv = " πk B T % $ # m ' & 3/ e mv /k B T 4πv dv v e αv RMS speed v which = c v = dv F(v)v v 0! = 3RT $ # & " M % 16

17 v! = 3RT $ # & " M % = c v v c *! 3RT $ # & " M % " 8RT % $ ' # πm &! RT $ # & " M % Speed of sound ~ average speed in gas D. Types of Average Speeds. 1. Most probable speed c* maximum of F(v) occurs at df(v)/dv = 0 Solve for v! c* = RT $ # & " M %. Mean speed v or = c v = 0 dv F(v)v " v = πkt % $ ' # m & 3/ is integral of type dx x 3 0 dv e mv /kt 4πv v e αx " v = 8RT % $ ' # πm & ".546 RT % = $ ' # M & 3. Mean relative speed = c rel The mean speed at which a molecule approaches another molecule. Collision rates between molecules depends on mean relative speed c rel = c 17

18 and for the case of two dissimilar molecules: " c rel = 8RT % $ ' # πµ & where µ is reduced mass Proof:! $ mean relative speed = c rel = # v 1 + v & " % Now since v i " = 8k T % B $ # πm ' i & mean speed mean speed of 1 of We have: " c rel = 8k T B $ # πm 1 + 8k B T πm % ' & " = 8k T % B $ # π ' & " % $ # m 1 m ' & " c rel = 8k T % B $ ' # & πµ 1 =1/µ 1 reduced mass µ 1 = m 1 m m 1 + m " Note: if m 1 = m c rel = 8k T % B $ # πm ' 1 & E. Kinetic Energy Distribution. = c F E (E)dE = fraction of molecules with K.E. value in interval E to E+dE Given E = 1 mv = 1 m v + v ( x y + v z ) F E (E)dE = e E/k B T π E de ( πk B T) 3/ 18

19 Problem: Gas phase reaction has activation energy = E a. What fraction f of gas sample has kinetic energy greater than some required activation energy E a for chemical reaction? fraction f = E a de π E e E/k B T ( πk B T) 3/ F. Collision rates with a wall or opening in a wall. Use: calculation of pressure, escape rate (effusion) calculate z w = # of collisions unit Area x unit time = "collision frequency" consider small time interval = dt patch of area = A # of collisions due to molecules with z-veloc will be number of molecules within distance = v z dt of patch (only those will strike) That will = number in cylinder of volume Av z dt Which will = NAv z dt where N is number density of molecules (# per unit vol) # of collisions due to molecules with v z velocity = N Av z dt f v z Total # = N Adt v z 0 ( ) dv dv z x f v z f v x ( ) f ( v x) f ( v y) dv dv dv x y z ( ) dv y =1 =1 ( ) f v y z w = z w = N total # coll Adt = N dv z z w = N ( ) dv z v z f v z RT πm # πk B T & % $ m ( ' v z e mv /kt z 19

20 " or using c = v = 8RT % $ ' # πm & z w = N c 4 G. Collisions between molecules. Need to consider this in order to determine: -mean free path -diffusion coefficient -heat conductivity -viscosity 1. Collision cross section σ and effective target σ = area the molecule sweeps out as it travels, σ = πd, where d = hard sphere diameter, no long-range attractive forces. Collision frequency z of one molecule z = σc rel N = σc rel p k B T Collision cross-sections 0

21 3. Mean free path λ = avg distance traveled by a molecule between collisions λ = c rel z = k B T σp note λ 1 P Problem A requirement for effusive flow through an opening is that radius of opening must be small compared to λ. O has effective spherical diameter d =3.61Å. At 5 C at what pressure will the mean free path be equal to a hole of radius = 1 mm? require λ = 1 mm = 10-3 m λ = k B T σp P = k B T σλ σ = πd P = J K 98K ( ) 10 3 m π m = 10.0 pascals x 1 atm 10135Pa = atm 4. Graham s Law of Effusion already have z w = collisions per unit time per unit area with wall or opening A = area of opening " Effusion rate R = z w A = A k T % B $ # πm' & N # density 1

22 Compare gases R 1 R = m m 1 = M M 1 molec weights R 1 M Requirement for effusion: λ > diameter of opening Then. molecules pass independently through the opening rather than creating a hydrodynamic flow. 5. Collision frequency between molecules. Let z 1() = # of collisions per unit time of one molecule of type 1 with all molecules of type # of collisions in = vol of this cylinder N time dt number density of type molecules = c rel dtπ ( r 1 + r ) N let d 1 = r 1 + r = collision diameter for molec 1 & types z 1() = c rel πd 1 N " z 1() = 8k T % B $ ' # & πµ 1 πd 1 N Now, total rate of collisions per unit volume between molecules of type 1 and those of type is: Z 1 = z 1() N 1 = 8k T 1 " % B $ ' πd 1 N 1 N # & πµ 1

23 Collision theory of gas phase reaction rates uses this collision frequency. Rate = Z 1 x (Probability of successful rxn upon collision) e E a /k B T f E a = activation energy f = geometric factor Rate = " 8k B T % $ # πµ ' & π d 1 fe E a /k B T N 1 N ~rate constant k gives concentrations = k c 1 c ( nd order rate law) If type 1 and type molecules are identical: Z 1 Z 11 = v πd N 1 where d is diam of 1 " v is 8k T % B $ # πm ' 1 & 3

24 III. Real gases. A. Molecular Interactions. 1. In general - gases deviate from ideal behavior due to interactions between molecules. Repulsive forces - exist over very short range, a molecular diameter. Called excluded volume forces. Causes deviation from ideality only when the density is so high that the physical volume of the molecules begins to become comparable to V. Attractive forces - exist over several molecular diameters.. The compression factor Z = a dimensionless quantity expressing deviation from ideality. Z pv nrt = 1 for gas obeying ideal gas law. Since V n = V m Z = pv m RT Also, Z = V m V m o where V m o is ideal gas molar volume = RT/p 4

25 5

26 3. Virial expansions: Since ideality pertains at low p, this suggests that we mathematically expand Z in the pressure: Z = 1 + B'p + C'p +... B', C' are called the nd and 3rd virial coefficients. Virial coefficients are not just curve-fitting parameters, but can be calculated from molecular properties by statistical mechanics. Alternately, the virial expansion can be expressed in powers of 1/V m 4. Condensation: Z = (1+ B V m + C V m +...) Decreasing the volume of a confined gas (increasing its p and density) at constant T below a certain T called the critical temperature Tc causes condensation. p-v isotherms of CO

27 5. Critical constants: Every gas has a critical point with a critical pressure P c critical volume V c critical temperature T c T c = temperature above which a gas can not be made to undergo a condensation phase transition no matter how much it is compressed. Always one phase. Synoptic Table 1C. Critical constants of gases B. The van der Waals equation - an intuitively appealing approximate equation of state for a real gas. 1. Constructing the equation. a) First account for repulsive forces. Replace V in pv=nrt by the effective available volume the molecules have to move in: Modified equation becomes r P(V-nb) = nrt Excl volume Where b = excluded volume vdw parameter Excluded volume effect is to increase the P at fixed n, T. 7

28 Derive relation between b and the physical volume of a particle for the hard sphere gas case: Answer: b = 4V particle *N avo For example: if molecule is 1.0 Å in radius b = 4* (4/3)π (1 Å) 3 * 6.0 x 10 3 =1.009 x 10-5 m 3 /mol ~ 1.0 x 10 - L/mol b) Second account for attractive forces. Should be proportional to the frequency of collisions in the sample: collision frequency (n/v) = (molar conc) (P + a n /V )(V-nb) = nrt Where a = vdw attractive forces parameter. Attractive term effect is to decrease the observed pressure p, everything else remaining fixed. Every gas is characterized by parameters a and b. Synoptic Table 1C.3 van der Waals coefficients So what is the effective hard sphere radius of an Argon atom? b = 4V Argon * N avo L = 4 $ 4 ' mol & 3 πr3 ) N avo % ( Solve for R. (Top Hat question) 8

29 . van der Waals isotherms: 3. Relation of a and b to critical constants: Since vdw equation predicts the critical point, derive relationship between critical point and parameters a and b. At critical point: dp dv = 0 d p d V = 0 Use n=1 and differentiate vdw equation: V c = 3b a P c = 7b T c = 8a 7bR 9

30 Table 1C.4 Selected equations of state 4. Relation of a and b to virial coefficients: By expanding p in a power series, we identify B = b - a/rt (second virial coefficient) C. The principle of corresponding states. Define reduced variables (dimensionless variable): Pr = P/P c Vr = V m /V c Tr = T/T c Observation that real gases behave virtually identically if expressed in terms of reduced (dimensionless) variables. I.e., at same reduced V & T they exert approximately the same reduced pressure. Works very well for spherical molecules. Why does this work so well? To see this rewrite vdw equation in terms of reduced variables. Start with (P + an /V )(V-nb) = nrt 30

31 set n = 1 (P + a/v m )( V m -b) = RT Make substitutions: (P r P c + a/v m )(V r V c - b) = RT r T c Now sub in V c = 3b, P c = a/(7b ), T c = 8a/(7Rb) Now reorganize, noting that a's + b's disappear, leaving: p r = 8T r 3V r 1 3 V r Since no a or b remain in the equation, the equation of state in reduced variables is identical for every type of gas (as long as vdw equation is valid). Example: H at 1 atm and 98 K behaves identically to NH 3 at 8.7 atm and 3638 K. How did we determine this? Look up the critical constants of the gases in question and use the principle of corresponding states. 31

32 Notes: 3

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