MP203 Statistical and Thermal Physics. Jon-Ivar Skullerud and James Smith

Transcription

1 MP203 Statistical and Thermal Physics Jon-Ivar Skullerud and James Smith October 27, 2017

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3 Contents 1 Introduction Temperature and thermal equilibrium The zeroth law of thermodynamics The ideal gas law The ideal gas law and absolute temperature What is an ideal gas? Microscopic model of an ideal gas Equipartition of energy Mean Free Path The First Law of Thermodynamics Heat and Work Compression Work Isothermal and Adiabatic Compression

4 Chapter 3 The First Law of Thermodynamics 3.1 Heat and Work Until the mid-19th century, heat was seen as a substance (caloric fluid that was transferred in the process of heating or cooling objects. It was measured in calories. 1cal = the amount of heat required to increase temperature of 1g of water from 145 C to 155 C at 1atm pressure. Note: 1 calorie in food is actually 1kcal = 1000cal. James Joule showed that you could achieve the same temperature increase by doing mechanical work, without any heat. This established that heat is not a substance, but a form of energy, or rather a form of energy transfer (like work. In thermodynamics, we define heat and work as Heat: Spontaneous flow of energy from one abject to another, caused by a temperature difference between them. Work: All other types of energy transferred to or from a system (mechanical, electrical, chemical, etc There are three types of heat: 1. Conduction - Heat transfer between nearby molecules 2. Convection - Bulk motion of a fluid 3. Radiation - Emission of electromagnetic waves The first law of thermodynamics states that the total energy is conserved in any process and can be represented mathematically as The First Law of Thermodynamics U = Q + W (3.1 14

5 Figure 3.1: Work done by a piston compressing a gas. In this equation, U = change in internal energy of a system Q = heat transferred to the system W = work done on the system Comments (a W is often taken to be the amount of work done by the system. In that case, the first 1st law reads U = Q W. (b According to the definitions above, it makes no sense to talk about the amount of heat in a system. We can talk about the amount of energy in the system, but it is never stored at heat or work. This is clearly not the same as the everyday meaning of heat, just as work also has a different meaning in physics. 3.2 Compression Work The most important type of work done on a system(typically a gas is compressing it, e.g. pushing a piston. Unless you slam the piston really fast (faster than the speed of sound, the force will equal the pressure force of the gas. F = pa (3.2 Where A equals the area of the piston. The work done pushing the piston a small distance x is W = F X = pa x (3.3 But A x is just the reduction in the volume of the gas, = A x = W = p (3.4 15

6 In general, the pressure (and hence the force can change as volume is reduced, and we have W = x2 x2 F dx = padx = x 1 x 1 = W = pd (3.5 p( d (3.6 If we know the pressure as a function of volume, we can use this to calculate the work. Example 3.1 1mol of ideal gas is compressed from an initial volume of 20.0L at atmospheric pressure, while keeping the temperature constant, to a final volume of 15.0L. Find the amount of work done. Answer:We know that Hence, p = nrt = p( = nrt nrt 2 W = d = nrt = nrt [ln( 2 ln( ] = nrt ln for constant T. (3.7 ( 1 = W = p 1 ln 2 d ( 1 2 (3.8 (3.9 (3.10 ( 20.0m = ( N/m 2 ( m 3 3 ln 10.0m 3 (3.11 = 2020J 0.288= 581J (3.12 This is an example of isothermal compression. For an ideal gas, we also know that the total energy is given by the equipartition theorem, U = f 2 Nk bt where f is the number of degrees of freedom. So if N and T do not change, the gas loses 581J of heat to the environment. 16

7 Example 3.2 An ideal gas is made to do undergo the cyclic process shown on the right. For each step A, B, C, determine whether the following are positive, negative or zero: (i the work done on the gas, (ii the change in the energy content of the gas, (iii the heat added to the gas. Finally, determine the sign of each of these for the whole cycle. Answer: A. (i Here we have expansion at constant pressure. The work done is then since 2 > W = pd = p( 2 < 0 (3.13 (ii From the Ideal Gas Law, p = Nk b T, if increases while p, N are constant, the temperature T must increase. Since the energy U Nk b T for an ideal gas, this implies U > 0. (iii The gas loses energy through work, but gains energy overall, so heat must be added i.e Q > 0. B. (i We are now at constant volume ( 3 = 2 so no work is done, i.e. W = 0. (ii For the same reason as in step 1., T must increase and therefore U > 0. (iii U = Q + W < 0 but W > 0, so Q > 0. C. (i Now we have compression along a line where p, say p = b. Then 1 1 [ 1 ] W = pd = b d = b 2 1 = b( > 0. (3.14 (ii In this case, p and both decrease, so p = Nk b T decreases, and U > 0. (iii U = Q + W < 0 but W > 0, so Q < 0. 17

8 Total: (i We have W tot = W A + W C = p 1 ( b( (3.15 since 3 = 2. But we also have p 1 = b, so W = 1 2 b( b ( 2 (3.16 = 1 2 b 2 2 b b 2 1 (3.17 = 1 2 ( 2 2 > 0 (3.18 (ii Since we get back to where we started, U is the same, i.e. U = 0 (iii U = Q + W = 0 = Q = W, i.e. heat escapes from the gas. The net result result of this process is that work is done on the gas and is converted into heat. 3.3 Isothermal and Adiabatic Compression We will consider two important special cases of compression work: Isothermal: The temperature does not change (slow Adiabatic: No heat flow in or out (faster We already looked at isothermal compression in Example 3.1. For an ideal gas, p = Nk b T, so if T (and N is constant, we have p( = Nk bt. W = = Nk b T p( d = Nk b T [ ] 2 ln = Nk b T ln ( 1 2 d (3.19 (3.20 Also, for an ideal gas, U Nk b T = constant U = 0. Therefore ( 1 Q = U W = W = Nk b T ln 2 (3.21 If > 2 (compression, W > 0 (work done on the gas and Q < 0 (heat leaves the gas Now consider adiabatic compression, i.e. Q = 0. In this case, U = Q + W = W. According to the equipartition theorem, U = f 2 Nk bt (f is the number of degrees of freedom. 18

9 Clearly, the temperature will have to change, as will the pressure. Consider an infinitesimal compression d, with temperature change dt. The change in internal energy is thus du = f 2 Nk bdt = W = pd (3.22 f 2 Nk bdt = Nk bt d f dt 2 T = d (3.23 (3.24 using the fact that p = Nk bt. Integrating on both sides, we get f T2 dt 2 T 1 T = f ( 2 ln T2 d = T 1 = ln ( 2 (3.25 Now, exponentiate both sides, using e ln(x = x, and a ln(x = ln(x a ( T1 T 2 f/2 = 2 T f/2 1 = 2 T f/2 2 or T f/2 = constant (3.26 For any given adiabatic compression, from to 2, this gives us the change in temperature and therefore the energy, U = f 2 Nk b T = W. We may also find an expression for the pressure p γ = constant, γ = f + 2 f = adiabatic exponent (3.27 Deriving this is left as an exercise. Example mol of Oxygen gas (O 2 has temperature 310K and volume 12l. (a What is the final temperature if the gas expands adiabatically? (b How much work is done by the gas in the process? Oxygen gas at this temperature has rotations, but no oscillation/vibrations. Answer: (a The number of degrees of freedom is f = 3(translation + 2(rotation = 5. For adiabatic expansion, we have ( T f/2 1 = 2 T f/2 1 2/f 2 T 2 = T1 = 2 ( 12 2/5 310K= 258K (

10 (b Here we can use that W = U for an adiabatic process, and U = f 2 Nk bt = f 2 n mrt (3.29 U 1 = 5 2 n mrt 1 = 5 1mol 8.31J/(mol K 310K = 6440J 2 (3.30 U 2 = 5 2 n mrt 1 = 5 1mol 8.31J/(mol K 258K = 5360J 2 (3.31 W = U 2 U 1 = 1080J (3.32 This is the work done on the gas. The work done by the gas is 1080J. We can also consider work done at constant pressure. In this case the expression for the work is quite simple: W = pd = p d = p( 2 = p (3.33 Summary First law of thermodynamics: the change in internal energy = heat transferred to the system + work done on the system. U = Q + W Compression work: W = pd Isothermal compression/expansion: (T =constant p(v = Nk bt Constant volume: ( 1 W = Nk b T ln, U = 0, Q = W 2 W = 0 Q = U = f 2 Nk b T Constant pressure: W = pd Adiabatic expansion/compression: Q = 0 T f/2 = constant, p γ = constant, γ = f + 2 f W = U = f 2 Nk b(t 2 T 1 Underlined expressions hold for ideal gases. but not in general. 20