Lecture18 - Kine.c theory of gases Gas in box When...


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Unformatted text preview: Kine.c theory of gases Gas in box When gas molecules move around in a box, they occasionally collide with the walls of the box. What happens when they collide with the walls ? They transfer momentum to the wall ! How much ? x ­component: Δpx = p fx − pix = ( − m0 vx ) − ( m0 vx ) = 2 m0 vx The molecules will hit the same wall of the box every t=2L/vx. Therefore the rate of momentum transfer is: 2 Δpx 2 m0 vx m0 vx = = Δt 2 L / vx L Chapter 21 From Newton’s second law, what is the rate of momentum transfer equal to ? dp = ΔI mpulse = Force of collision = m0 a dt Gas in a box (contd…) From Newton’s second law, rate of momentum transfer equal to force: 2 Δpx 2 m0 vx m0 vx F1 = = = Δt 2 L / vx L Now, pressure P is force divided by unit area (‘per unit area’): Adding considera.ons from all molecules in the gas: P= ⎞ F 1⎛ Δpx ⎞ m0 ⎛ 2 = 2⎜ ∑ = 3 ⎜ ∑ vxj ⎟ ⎠ A L ⎝ molecules Δt ⎟ L ⎝ molecules ( j ) ⎠ We can use the average velocity of the molecules: (add All up, divide by N) 2 2 vx ≡ vx , avg = 1⎛ 1⎛ 2⎞ 2⎞ vx , j ⎟ = vx , j ⎟ ∑ ∑ ⎝ ⎝ N ⎜ molecules ⎠ nN A ⎜ molecules ⎠ Now for all 3 ­D molecules v2 = vx2+vy2+vz2 = 3 vx2 and molar mass M = m0NA: nm0 N A 2 nM 2 nM 2 vx = vx = v V V 3V nM 2 P= v 3V P= RMS speed The square root of the <v2> is called the root ­mean ­square (rms) speed. It is usually denoted by vrms. In terms of rms speed, the pressure is given by: P= nM 2 vrms ...........(1) 3V This equa.on relates a microscopic quan.ty, vrms, to macroscopic quan..es p & V. This is the essence of kine7c theory and sta7s7cal mechanics. How fast are the molecules moving at some temperature T ? Use ideal gas equa.on pV = nRT, in (1) we get vrms = 3RT M Average kine.c energy What is the average kine.c energy of gas molecules at temperature T ? 1 1 K = m0 v 2 = m0 vr2ms 2 2 We know: Using M = NA m0 we ﬁnd: Recall kB = R/NA: K= 1 3RT 3 RT m0 = 2 M 2 NA 3 K = k BT 2 This is the average transla7onal kine.c energy of a molecule in the gas at temperature T. This shows that the temperature is a measure for the average kine2c energy (since the two are propor.onal) The total transla.onal kine.c energy of all the N molecules is then: K trans = ⎛1 ⎞ 3 3 Nk BT = nRT = N ⎜ m0 v 2 ⎟ 2 2 ⎝2 ⎠ Where n = N/NA is the number of moles. Since for ideal gas par.cles’ interac.on is negligible, this is the total internal energy of the gas. Thus for an ideal gas internal energy is a func.on of temperature only. 3 3 Eint = 2 Nk BT = 2 nRT Mean free path Although molecules move at speeds of up to 1 mile/second at room temperature, it takes a while before a fragrance will reach our nose. This is because the molecules will collide with each other frequently and change direc.on. On average, molecules will only move a small distance before colliding. This distance is called mean free path. 1 λ= 2π d 2 N / V cross ­sec.onal area of molecules number density of molecules Example: What is the mean free path for oxygen molecules at T = 300 K and pressure p = 1 atm ? (diameter 0.29 nm) Ideal gas law: The mean free path is about 1/1000 the width of a human hair. How o`en do molecules collide ? The RMS speed of oxygen molecules at 300 K is 480 m/s. So they collide every Distribu.on of molecular speeds We know that the rms speed of molecules is given by But some molecules will be slower and some will be faster. The distribu.on of speeds in a gas at temperature T was ﬁgured out by James Clark Maxwell in 1852: where P(v) dv is the probability that molecules have a speed v and v+dv. P(v) is called a probability distribu.on. This distribu.on is called Maxwell ­Boltzmann distribu.on. Calcula.ng average values For a probability distribu.on like P(v) we have: “the probability that a molecule has any velocity (between 0 or inﬁnity) is equal to 1.” Also, average of a quan.ty A (e.g. A may be energy E) that depends on v is given by: A = ∫ A(v)P(v) dv ∞ RMS velocity: Average velocity: v2 3RT = ∫ v 2 P(v ) dv = ⇒ vrms = M 0 ∞ 0 v2 = 3RT M Most probable velocity: Molar speciﬁc heat of a gas When you heat up n moles of a sample, the amount of heat needed to raise its temperature by ΔT is given by: C is the ‘molar speciﬁc heat’ of the sample. C is measured in J/(mol K) There are two molar speciﬁc heat: molar speciﬁc heat at constant volume (CV) and molar speciﬁc heat at constant pressure (CP) CV for a monatomic gas Internal energy = Kine.c energy (since there are no signiﬁcant contribu.ons of poten.al energy in individual molecules) Now heat the gas, but keep its volume constant (W = 0): 3 3 Eint = N K = (nN A )( kT ) = nRT 2 2 1st law of thermodynamics: ΔEint = Q − W = Q ⇒ ΔEint = nCV ΔT From above: Compare & ﬁnd: 3 R 2 CV = 12.5 J/(molK) CV = Speciﬁc heat at constant pressure We just calculated the molar speciﬁc heat for a monatomic gas heated at constant volume, CV = 3/2 R. What if we let the gas expand while hea.ng it at constant pressure ? Would you expect that it takes more or less heat to raise the temperature in this case compared to the ﬁxed volume case ? Heat required to raise temperature at constant pressure: Internal energy change from before: Work done by gas, using ideal gas law: 1st law of thermodynamics: Solve for Cp: For monatomic gas: C p = CV + R Cp = 5 R 2 More complicated gases Most gases are not monatomic. For example hydrogen (H2), oxygen (O2), nitrogen (N2) are diatomic, and CO2 , NH3 (ammonia) or CH4 (methane) are polyatomic. These more complicated molecules can also have rota.onal & vibra.onal contribu.ons to the kine.c energy. Single atoms do not have rota.onal kine.c energy. Each independent type of mo.on is called a degree of freedom (dof) of the molecule (Examples: 3 transla.ons: x, y, z, any number of rota.ons, vibra.ons etc.) . Equipar..on Theorem Every kind of molecule has a certain number f of degrees of freedom. Each degree of freedom contributes on average ½ k T of energy per molecule to the total internal energy (or ½ R T per mole). Example: O2 molecules: Because of symmetry, O2 has two rota.on ­axes (about the other axis, the molecule is symmetric & thus wouldn’t change when rotated). Thus the number of degrees of freedom is: 3 (transla.ons ­x,y,z) + 2 (rota.ons) = 5. Thus, per mole, we have an internal energy of: And, CV for diatomic molecule is: ‘ Turning on’ degrees of freedom In reality, degrees of freedom such as rota.on and vibra.on do not contribute to the speciﬁc heat un.l some minimum temperature is reached. Example: for H2 we have the following diagram. The reason is, that in order to ini.ate rota.on or vibra.on of a molecule, some minimum energy is required, called an energy ‘quantum’. This is explained by quantum mechanics. Adiaba.c Process If a system is well insulated or a process happens so rapidly that no heat is exchanged with the environment (e.g., piston moves very fast). No heat added. First law gives: Only work changes the internal energy. Temperature will change since Eint changes. Therefore p, V, and T change. How do they change? γ We can ﬁnd (see textbook, 21.3) that: pV = a constant (Compare this with isothermal case pV = constant.) The constant γ = 5/3 for a monatomic ideal gas. TV γ −1 = constant Note that in pV diagram, an adiaba2c is steeper than an isotherm. Adiaba.c Process (contd…) Compare adiaba.c process to isothermal process: Adiaba.c Isothermal Example: Air at 20 oC in the cylinder of a diesel engine is compressed from an ini7al pressure of 1 atm, and ini7al volume = 800 cm3 to a ﬁnal volume = 60 cm3. Assume air has γ = 1.40, and compression is adiaba7c. Find the Pf and Tf of the air. γ ⎛V ⎞ pV γ = const ⇒ p f V γ = p f V γ ⇒ p f = pi ⎜ i ⎟ i i ⎝ Vf ⎠ ⎛ 800 ⎞ p f = (1atm) ⎜ ⎝ 60 ⎟ ⎠ 1.4 = 37.6 atm ...
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This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.

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