Lecture 5: Kinetic Theory. Where does pressure come from? Mean free path Molar heat capacity Equipartition Theorem

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2 Lecture 5: Kinetic Theory Where does pressure come from? Mean free path Molar heat capacity Equipartition Theorem

3 Kinetic Theory of Gases. The Model. Identical molecules in random motion Molecules obey Newton's laws of motion Molecules make ELASTIC collisions with each other and with walls of container No force on molecules except during collision Number of molecules N is large but NV m << V, where V m = volume occupied by each molecule and V = volume of container new quantity: ν=n/v = number density or concentration What are the principal results of this model?

4 Pressure. Micro-derivation. z Single collision y x Force on wall from one particle Number of particles hitting area A (through time t) L=v x t A #=νv/2=νal/2

5 So, where does the ideal gas law come from? p = 1 3 ρ v2 < v 2 >= 3kT m p = 1 3 ρ v2 = 1 Nm 3 V = NkT V 3kT m

6 Maxwell Speed Distribution. Given an ideal gas with molecules of mass m at temperature T, what is the probability P(v) of finding a molecule with a speed in the range:? [ v, v + dv] m P(v) = 4π 2πkT 3 / 2 v 2 e mv 2 2 kt T = 80K T = 300K P(v)dv =? 0 Average of some quantity

7 Maxwell Speed Distribution. Average speeds. Average speed of a molecule: v = v vp(v)dv = 8kT πm 0 RMS speed of a molecule: 2 = 0 v 2 P( v) dv Most probable speed: = 3kT m v MP < v AVG < v RMS dp dv = 0 v MP = 2kT m

8 Molecular speeds. Sample problem. By how much would the speed increase if I were to increase room temperature by 20 C? v v T + 20 = = = 3.3% Troom increase

9 Internal Energy, Kinetic Energy and Temperature Average KE of a molecule: KE = 1 2 m v2 = 1 2 m 3kT m Total KE of gas = Internal energy U = 3 2 NkT = 3 2 kt Note: so far, we're only considering TRANSLATIONAL kinetic energy. But the statement above is more general

10 Mean Free Path. Definition. Ideal gas of N molecules, each of diameter d,, in a container of volume V. How far does a molecule in an ideal gas move (on average) before it collides? Call this distance the MEAN FREE PATH λ = 1 2πd 2 N V

11 Mean Free Path. Sample problem. Estimate the mean free path of O 2 molecules in this lecture room. Assume molecular diameter ~ 0.30 nm. λ = 1 2πd 2 N V = 1 2πd 2 p kt Answer: 0.1mm or 380 molecular diameters

12 Molar heat capacity The heat required to raise the temperature of 1 mole of gas by 1 K = "Molar heat capacity C process := Q process /( T T n) n Two kinds of molar heat capacity: C V = molar heat capacity during an isochoric process (constant volume) C p = molar heat capacity during an isobaric process (constant pressure) Determine a general expression for C V for an ideal gas Q = U + nc v T Q = U = 3 2 W nr T C v = 3 2 R

13 C p for an Ideal Gas Given n moles of an ideal gas at some initial temperature T 1 Increase the temperature of this gas to T 2 via two different paths: Isochore (vertical line) Isobar (horizontal line) Change in internal energy is the same in both cases (why?) p T 1 A T 2 C B V U AC = nc v T U AB = nc p T p V nc v T = nc p T p V nc v T = nc p T nr T C v = C p R C p = C v + R = 5 2 R

14 2 Predictions from Kinetic Theory C V = 3 2 R He GAS C P J/mol.K 20.8 C V J/mol.K 12.5 C p -C v J/mol.K 8.33 γ 1.67 Ar C P = C V + R = 5 2 R H O C P H 2 O = γ = 5 C 3 CH V 4 Note: (3/2)R = 12.5 J/mol.K (5/2)R = 20.8 J/mol.K γ = 1.67 Works well for monatomic gases Deviations for more complex gases

15 The Equipartition Theorem Identify "degrees of freedom = number of independent ways for a system to store energy A point mass (or very small sphere) has 3 degrees of freedom: motion along x, y and z axes Equipartition theorem: each degree of freedom contributes (1/2)kT( per molecule to the internal energy U = (3/2) NkT So far: we've only looked at TRANSLATIONAL kinetic energy -- i.e. molecules were small spheres with negligible moment of inertia What if the molecule is more complicated? Are there additional degrees of freedom?

16 Diatomic Gases 3 translation degrees of freedom 2 ROTATIONAL degrees of freedom z (Why only 2 rotational degrees of freedom?) 5 degrees of freedom total U = NkT = 5 2 NkT OR U = 5 2 nrt So: C V =? C P =? γ =? x y

17 Diatomic gases: 5 degrees of freedom GAS C p J/mol.K C V J/mol.K C p -C v J/mol.K g H O C V = 5 2 R = 20.8J /mol.k C P = C V + R = 7 2 R = 29.1J /mol.k C P C V = γ = 7 5 = 1.40

18 Polyatomic Gases 3 translation degrees of freedom z 3 ROTATIONAL degrees of freedom 6 degrees of freedom total U = NkT = [ 3NkT ] OR U = 3nRT Any other possible degrees of freedom?? So: C V =? C P =? γ =? x y

19 Polyatomic gases: 6 degrees of freedom GAS C p J/mol.K C V J/mol.K C p -C V J/mol.K g H 2 O CH C V = 3R = 25J / mol. K C P = C V + R = 4R = 33.3J / mol. K C C P V = γ = 4 3 = 1.33

20 Limitations of Kinetic Theory The kinetic theory explains a lot of experimental data but there are still significant discrepancies For instance, the molar heat capacity of a gas can vary with temperature: different degrees of freedom are "turned on" with increasing temperature Limitations due to "quantum mechanics" -- effects not included in classical mechanics

21 Adiabatic Process. Equation. An isotherm on a p-v diagram is described by p ~ 1/V OR pv = constant Is there a generic mathematical description for an adiabatic process? pv γ = Recall: constant Pressure γ = C p C v Volume

22 Adiabatic process. Sample problem. 2 moles of an ideal diatomic gas expand adiabatically from 100 K to 300 K. Calculate the work done by the gas. Q = U + W W = U = 0 5 W = ncv T = nr T 2 W = - (2.5)(2 mol)(8.314 J/mol.K)( )K = J

23 Adiabatic process. Sample problem. An ideal gas initially at pressure p 0 undergoes a free expansion until its volume is tripled. It is then slowly and adiabatically compressed back to its original volume. The final pressure is 3 1/3 p 0. Is the gas monatomic, diatomic or polyatomic? "Free expansion": gas expands RAPIDLY into a vacuum Does temperature change in a free expansion?

24 Adiabatic process. Sample problem. A-->B: free expansion B-->C: adiabatic compression Free expansion: p A V A = p B V B p 0 V 0 = p B (3V 0 ) p B = p 0 /3 (continued) p p f p 0 T 1 A C B T 2 Adiabatic compression: p B V Bγ = p C V C γ V 0 3V 0 V p γ 1 = 3 3 [ ] γ = 3 3p [ V ] γ V 0 γ = 4 3 Gas is Monatomic!

25 Adiabatic expansion. Fire Extinguishers As the extinguisher expels (adiabatically): Volume increases Pressure decreases Temperature decreases pv γ or TV γ 1 =const

26 NEXT LECTURE: Ch. 20 ENTROPY! 2nd law of Thermodynamics Efficient Engines