PHY 171. Lecture 14. (February 16, 2012)

Transcription

1 PHY 171 Lecture 14 (February 16, 212) In the last lecture, we looked at a quantitative connection between acroscopic and icroscopic quantities by deriving an expression for pressure based on the assuptions of the kinetic theory of gases. We found that PV = 1 3 N v 2 In this for, the acroscopic quantities, pressure P and volue V are on the left, whereas the icroscopic quantities nuber of olecules N, ass of a olecule, and average speed (squared) of all olecules v 2, are on the right. By coparing this equation to the Ideal Gas Law, we were able to show that the average kinetic energy K is equal to K = 3 2 k BT where k B is the Boltzann constant, and T is the absolute teperature (in Kelvins). We interpreted this equation to ean that the higher the teperature, the faster the olecules are oving on the average. We will now calculate the root ean square (rs) speed, which gives us an idea of how fast the gas olecules are oving on average. Root ean square speed Let us put back K = 1 2 v2 into the equation for K above, which then becoes Rearranging ters, we get 1 2 v 2 = 3 2 k BT v 2 = 3k BT Look carefully at the left hand side. We have the square of the velocity, and then we have written its ean (average) value. The square root of this quantity is called the root ean square (rs) speed, v rs. That is, v rs = v 2 It gives one way to characterize the typical speeds of gas olecules; we will look at two other ways below. Therefore, we find that the root ean square velocity v rs is given by v rs = 3kB T (L14.1) In eq. (L14.1) above, T is the absolute teperature (in K), is the ass of a single olecule, and k B is the Boltzann constant.

2 Exaple Calculation: Let us calculate the rs speed for oxygen olecules at roo teperature, usually taken to be 27 C, so that we get a rounded nuber for the Kelvin teperature of 3 K. Of course, 27 C (or 81 F) is a little higher than you d want your roo to be, but the rounded nuber of 3 K is too good to sacrifice for soe iaginary discofort! It is a case of straightforward substitution in equation (L14.1), of course, but be careful about writing the ass in the denoinator. The easiest way is to ultiply the atoic ass of oxygen by 2, and then by the ass of a proton to get in kg. So ( = 16 2 )( kg ) This will give us: v rs = 3 ( J/K ) 3K 32 ( kg ) = 482/s This value is quite large, about 11 iles per hr. One ay well ask if the olecules are oving so fast, why does it take so long to sell a perfue opened across the roo (i.e., why doesn t the scent travel instantaneously across the roo)? The answer is that this is due to collisions. Collisions between perfue olecules and air olecules do not allow olecules to ove straight through the roo. The perfue olecules suffer any collisions before reaching the person s nose at the other end of the roo. In fact, you read in the war-up assignent how one can define a ean free path λ as the average distance between such collisions. If a olecule undergoes N coll collisions as it travels distance L, then the average distance between collisions is given by the ean free path: λ = L N coll (L14.2) Page 543 of your text shows how N coll can be deterined fro a siple odel of two colliding spheres. Note that the rs speed only tells us how fast the gas olecules are oving on average. Soe olecules will be oving faster, others will be slower. In fact, even if you start out with all olecules having the sae speed v, collisions will quickly change this state of affairs. We will deonstrate this below. In reality, the speeds of olecules are distributed according to well known laws. Their distribution was worked out by Maxwell and Boltzann. Page 2 of 7

3 Distribution of Speeds: Maxwell-Boltzann Distribution Think of the Maxwell-Boltzann distribution as a probability distribution of the speeds of olecules in a gas. That is, if we pick a velocity v, the Maxwell-Boltzann distribution tells us what fraction of olecules has velocity between v and (v +dv). The distribution function looks coplicated, but we will write it down because it reveals one very iportant feature. The Maxwell-Boltzann distribution of the velocities of gas olecules looks like: ( ) 3/2 [ ] v f(v) = 4π v 2 2 exp (L14.3) 2πk B T 2k B T It is clear fro eq. (L14.3) above that, for a given gas (i.e., once we have fixed the ass ), the Maxwell-Boltzann distribution depends only on the absolute teperature of the gas. A plot of the distribution in eq. (L14.3) is shown below. The distribution function f(v) is plotted along the vertical axis, whereas the speed v is plotted along the horizontal axis. What does the distribution function tell us? If we look at the blue rectangle of width dv, then its area f(v)dv gives the fraction of olecules whose speeds lie in the interval v to (v+dv). Another way to think of f(v) is to say that it is equal to N(v)/N, where N is the total nuber of olecules, and N(v) is the nuber of olecules with speed v. Page 3 of 7

4 Based on our discussion above, we find that the fraction of olecules with speeds between v 1 and v 2 is v2 v 1 f(v)dv (L14.4) Since all olecules ust have speeds between and infinity, suing up the fraction f(v) over this entire range of possible velocities ust be equal to 1. That is f(v)dv = 1 (L14.5) If we write f(v) = N(v) as stated above, then another way of thinking about eq. (L14.5) is to say N that where N is the total nuber of olecules. N(v)dv = N Looking again at the plot of f(v) vs. v, note how the distribution has very low tails on both sides. This akes sense: the distribution of velocities is established by collisions, and we do not expect too any olecules to have speeds uch greater or uch less than v rs, because that would require an unlikely series of preferential collisions. As stated above, if all the olecules of the gas had the sae speed v, this situation would not last very long due to collisions. For a deonstration, look at the applet at cop.uark.edu/~jgeabana/ol_dyn enter 3 in the top square and click Set, then click Run in the lower left corner notice how the olecules quickly settle into the Maxwell-Boltzann distribution, even when you start with an ordered distribution of particles. Average speed, Most Probable speed, and RMS speed In addition to the rs speed, two other speeds are used to characterize the velocity distribution of olecules in a gas. They are the average speed, and the ost probable speed. The ost probable speed v p is the speed at which f(v) is a axiu. In other words, at any given teperature the nuber of olecules in a given speed interval increases with increasing speed up to a axiu given by v p, then decreases asyptotically toward zero. Recall fro calculus that such a axiu value is found by setting df/dv =, and then solving for v. If we do this, we will get 2kB T v p = (L14.6) The ost probable speed v p is arked on the plot of the Maxwell-Boltzann distribution shown above. Note that the distribution is not syetrical about v p. Why? It is because the lowest speed ust be zero, but there is no classical liit to the upper speed a olecule can attain. Page 4 of 7

5 The average speed v is found by ultiply v by the distribution f(v), and integrating over all velocities; that is v = vf(v)dv If we substitute eq. (L14.3) and integrate, we will find for the average speed that (L14.7) v = 8kB T π (L14.8) Clearly, the average speed (L14.8) is greater than the ost probable speed (L14.6); see the plot of the Maxwell-Boltzann distribution above where both velocities are arked. This is because the distribution is not syetrical about the ost probable speed, as we noted above. There are ore olecules with speeds greater than the ost probable speed, so the average shifts to a higher value than v p. We have already defined the rs speed v rs above. Another way to think of it is to define it via the integral over v 2, naely v rs = v 2 = v 2 f(v)dv (L14.9) As we have seen above, this gives us the expression v rs = 3kB T (L14.1) Since it is based on the average of the squares of the speeds, v rs will be greater than v. Putting it all together v p < v < v rs This is shown on the plot of the Maxwell-Boltzann distribution above. Maxwell-Boltzann Distribution: Effect of Teperature Increase We will now exaine the effect of increase in teperature on the Maxwell-Boltzann distribution. As the teperature increases, the rs speed increases (as do the ean speed, and the ost probable speed), as indicated by its forula above. Fro the figure above, it is clear that the nuber of olecules that have speeds greater than soe given speed increases as the teperature increases. This explains any phenoena, such as the rates of cheical reactions with rising teperature. Page 5 of 7

6 The range of typical speeds is now greater, so the distribution broadens. Since the area under the distribution curve gives the total nuber of olecules in the saple, it ust reain the sae. Therefore, the distribution ust also flatten as the teperature rises. All of this is shown in the figure on the right. The brown curve is for T = 8 K, whereas the green curve is for a higher T = 3 K. The distribution of speeds of olecules in liquids also resebles the M-B distribution. This explains evaporation, in which soe olecules in a liquid can escape through the surface even when the teperature is very uch below the boiling point. The olecules that are escaping are the fast ones at the high-velocity tail of the distribution. Since the olecules with high kinetic energy have escaped, the average kinetic energy of the reaining olecules drops, eaning that there is a drop in the teperature of the liquid. This explains the cooling effect of evaporation. (Note that if the surface area exposed is sall, this cooling effect is often offset by the liquid absorbing heat energy fro its surroundings to war back up again). The authors present a discussion of this in explaining rain. They also present a discussion of the proton-proton chain in the Sun that causes nuclear reactions. Usually protons would repel each other, since they have the sae (positive) charge; but high velocity protons in the tail are able to coe together and cause the reaction. Notice also in the M-B speed distribution forula that there is a ass dependence. The saller the ass, the larger the proportion of high-speed olecules at a given teperature. You can also realize this by calculating rs speed it is uch greater for hydrogen than for oxygen. This explains why the Earth s atosphere has little hydrogen in it, although the priordial Earth ust have had quite a bit of it. Internal Energy Recall that we learned that teperature (in K) is a easure of the average kinetic energy of individual olecules. Below, we will see that the total energy of all the olecules in the object is called the internal energy. Internal Energy of an Ideal Gas Consider anidealonatoicgas. TheinternalenergyE int ofthegasisthesuofthetranslational kinetic energy of all the atos. But this is just equal to the average kinetic energy per olecule ties the total nuber of olecules N. [ 1 E int = N 2 v 2 ] Page 6 of 7

7 Now, recall that we have obtained previously: 1 2 v 2 = K = 3 2 k BT Therefore E int = 3 2 Nk BT or E int = 3 2 nrt where, as you should know fro earlier discussions, N is the nuber of olecules k B is the Boltzann constant n is the nuber of oles R is the Universal Gas Constant T is the absolute teperature in kelvin (K). Therefore Internal energy of an ideal gas depends only on teperature and the nuber of oles of gas. If gas olecules contain ore than one ato, we need to take into account rotational and vibrational energies also then E int will be greater at a given T than for a onatoic gas, but it will still be a function of teperature for an ideal gas. Likewise for real gases, but when they deviate fro ideal gas behavior, E int also depends soewhat on P and V. Thingsgetorecoplicatedforsolidsandliquids, becayse E int ust includeelectrical potential energy associated with the forces (or cheical bonds) between atos and olecules. Soe parts of this lecture were delivered on Wednesday (Feb 15), but are written in these notes for continuity. The lecture today was shorter because a significant aount of class tie was spent on the worksheet with concept questions. Page 7 of 7