Final Review Solutions

Transcription

1 Final Review Solutions Jared Pagett November 30, 206 Gassed. Rapid Fire. We assume several things when maing the ideal gas approximation. With inetic molecular theory, we model gas molecules as point particles bouncing around in a container. These point particles have the following properties. a) Gas molecules are noninteracting that is, there are no intermolecular forces. It follows that we assume that phase changes which are due to such forces) do not occur. Additionally, it follows that gas molecules move randomly in their container. b) Collisions between gas molecules and the container walls are elastic. c) Gas molecules are point particles thus, they occupy negligible volume compared to their container. In reality, of course, no gas is ideal by these measures. Many gases, though, such as N 2, O 2, H 2, and the noble gases, behave at reasonable pressure and temperature lie an ideal gas. It is only when the separations between gas molecules become small in the ballpar of less than 0 times the size of the molecules though it varies from gas to gas) that intermolecular forces become important, and the gas ceases to behave ideally. 2. The equation of state for an ideal gas is the ideal gas law: pv = nrt. 3. As said above, the ideal gas law ceases to apply when intermolecular forces become important. This occurs for very high pressures when the separations between molecules become small and for very low temperatures when the inetic energy of the gas molecules becomes comparable to the interaction energies of intermolecular forces. 4. An ideal gas is described by the ideal gas law: pv = nrt. If the number of molecules that is, the number of moles n) remains constant, then we may rearrange the ideal gas law so that only constant terms appear on one side. That is, pv/t = nr = constant. Thus we have that for a quasi-static process involving an ideal gas, pv/t remains constant. 5. Density of any substance is given by ρ = m/v. Rewriting mass as that of all the gas molecules, m = nm, we have ρ = nm/v. Considering the ideal gas law, we rearrange pv = nrt into n/v = p/rt. Thus the density of an ideal gas is given by ρ = pm/rt 6. From the derivation of pressure using inetic molecular theory, we obtain that p = 2 N V 2 m v2 x = 2 N V K x where K x is the average inetic energy due to motion in the x direction. Because there is no force that would mae the inetic energy greater in any one direction than any other, we conclude that the total

2 energy of the gas is divided equally among all of its f available degrees of freedom. Thus, we have that f K x = K avg,tot. We may then replace K x in the equation for pressure above with /f)k avg,tot. Doing so, we have p = 2 N V f K avg,tot But from the ideal gas law, p = nrt/v = NT/V. Equating these two expressions for pressure, NT = 2 N V V f K avg,tot Canceling common factors and solving for K avg,tot, we obtain K avg,tot = f 2 T That is, each degree of freedom contains /2)T of inetic energy. For a monatomic gas, there are three degrees of freedom three translational, one in each of the directions the gas molecule can move so that f monatomic = 3. Therefore, for a monatomic gas, K avg,tot,monatomic = 3 2 T For a diatomic gas, there are too the three translational degrees of freedom. In addition, the diatomic gas molecule s extent allows it to rotate. We neglect both the degrees of freedom of the molecule s rotation about the axis connecting the molecules as the energy of this rotation is negligible) and the vibrational degrees of freedom in the bond between the molecules as these are only excited at very high temperatures), and so are left with only two rotational degrees of freedom. There are thus five degrees of freedom for a diatomic molecule that is, f diatomic = 5, and so K avg,tot,diatomic = 5 2 T 7. This question was actually supposed to read of a monatomic gas molecule, but I suppose this maes for a more interesting question. The tricy bit here is that not all of the inetic energy of a diatomic molecule is translational. Only the translational inetic energy should be taen into account when computing the rms translational speed. Since each of the three translational degrees of freedom contain /2)T of inetic energy, we have that K avg,translational = 2 m v2 = 3 2 T 3T v 2 = v rms = m.2 Some Quic Examples. No. We derived earlier that the density of an ideal gas is ρ = pm/rt thus, if temperature is constant, the density of the gas still depends on pressure. 2. As discussed above, P V/T is a constant. That is, P i V i T i = P f V f T f P iv i = P f 3V i ) T i 2 T P f = i) P i 6 3. As derived before, v rms T. Thus, if we wish to increase v rms by a factor of 3, we must increase T by a factor of Since we are dealing with an ideal gas, we may apply pv = nrt. If there is one mole of gas, we have n =. Thus, V molar = RT p At STP T = 0 K and P = atm), we obtain the familar Standar Molar Volume, V molar,st P = 22.4 L. As long as the gas can be treated as approximately ideal, this quantity is independent of the gas s composition. 2

3 .3 True or False? Helium and Argon. False. There is an equal mass of each gas. Since helium is less massive than argon, there must be more helium atoms than argon atoms. With more helium atoms in the container, at the same volume and temperature as the argon atoms, the helium atoms will exert more pressure on the walls of the container than will the argon atoms. 2. False. The rms speed of a gas molecule is inversely proportional to the mass of the molecule, and so the helium atoms with a relatively lower mass) will have a greater rms and thus, greater average) speed. 3. False. See point. 4. True. The average inetic energy of a gas molecule is only dependent on the gas s temperature. 5. False. Though the average inetic energy of each gas molecule is equal, there are more helium atoms than argon atoms, and so the total inetic energy of helium atoms is greater than that of the argon atoms. Three Vessels This Time. True. Since the gases occupy the same volume and are at the same temperature and pressure, the ideal gas law tells us that there are the same number of molecules in each container. 2. False. Again, the rms speed of a gas molecule is inversely proportional to its mass. Thus, the more massive gases will have lower rms and thus, lower average) speeds. 3. True. The average inetic energy of a gas molecule is only dependent on the gas s temperature. 4. False. There are the same number of molecules in each container, occupying the same volume, but the masses of each molecule differ from gas to gas. Thus, there is a different mass of gas in each container, and so the densities of each gas will differ. As the heaviest gas, the container filled with O 2 will have the greatest density. 2 Everything Thermodynamics 2. Some Brief Conceptual Questions. I will address each equation from left to right. a) The first equation, the first law of thermodynamics, is a statement of energy. A gas can change internal energy by either absorbing heat Q) or by performing wor W ). The former will increase the internal energy hence the +Q) and the latter will decrease the internal energy as it has been used to perform wor, hence the W ). Combining both of these methods by which the gas may change in internal energy, we obtain U = Q W. More generally, for small changes dq and dw, the differential form of the first law is du = dq + dw. b) The second expression can be in somewhat of a jan fashion) obtained from the fact that the total inetic energy of an ideal gas is given by the sum of energies of each individual molecule of average inetic energy K tot,avg. That is, U = NK tot,avg = N [3/2)T ] = n3/2)rt = nc v T. This gives U = nc v T or, for infinitesimal changes, that du = nc v dt ). c) By considering a cylinder of gas at gauge pressure p, and allowing the gas to displace the cylinder head of area A a small distance dx, we see that the wor dw done by the gas by the force F = pa is dw = F dx = padx = pdv. For large displacements, the pressure will generally change, and so W = pdv. 3

4 2. This expression follows from the first law. Only when there is no wor performed, so that W = 0, that U = nc v T = Q. 3. Since the temperature of an ideal gas only depends on P and V, and U depends only on the temperature, U is uniquely determined by the endpoints of a process and so is path independent. W, however, is given by the area under the P V curve, and so is clearly dependent on the path taen through P V space. 4. The following terms imply the following about a process: a) Isothermal: Performed at constant temperature. implies T = 0, U = 0. b) Isobaric: Performed at constant pressure. Since U T, and constant temperature c) Adiabatic: Performed without heat exchange that is, Q = 0, which implies by the first law U = W. d) Isochoric: Performed at constant volume. Since volume is constant, dv = 0, and so W = 0, implying U = Q. 5. Process ) is isochoric and process 2) is isobaric. As for processes 3) and 4), we must discuss isothermal versus adiabatic processes. An adiabatic compression, with W > 0 U < 0, results in a decrease in temperature. Thus, an adiabatic process will transition between points on different isotherms, and so will be a steeper curve than would be an isothermal process which is constrained to one isotherm. Therefore process 3) is the adiabatic process, and process 4) is the isothermal process. 6. See the explanations above. Isothermal processes eep T constant, isochoric processes eep V constant, and isobaric processes p constant. As for adiabatic processes, P V γ, T V γ, and P γ T γ are all constant see section 2.3 below for explanation. 2.2 PV Boot Camp For a diatomic gas, we plug in that C v = 5 2R. I have neglected to so.. mostly because I forgot to and don t want to update everything.. The wor done is simply the area under the curve, which in the case of the isobaric process can be computed geometrically: W = P 0 V 0 α ) The change in internal energy can be found by computing the temperature at each point using the ideal gas law, then applying U = nc v T. Then U = nc v T B T A ) = C v R P 0V 0 α ) From W and U, we may obtain Q from the first law. Q = U + W = + C V R )P 0V 0 α ) Note that we have W > 0, Q > 0, and U > 0. The gas is thus simultaneously doing wor and absorbing heat. It absorbs more heat than it does wor, so that the internal energy and thus, the temperature) of the gas increases. Rewriting Q in terms of the temperature at points A and B, we see that it taes the form Q = nr + C v ) T nc p T This defines the heat capacity at constant pressure, C p = C v + R. 4

5 2. There is no change in volume, and thus W 2 = 0. We then have, by the first law, that U 2 = Q 2. We compute U 2 by using the ideal gas law to obtain the temperature at each point. Thus Q 2 = U 2 = nc v T B T A ) = C V R P 0V 0 β) Note that we have W = 0, Q < 0, and U < 0. The gas is thus releasing heat to the environment while doing no wor, and so is decreasing in internal energy. 3. For the adiabatic process, there is no heat exchange, and so Q 3 = 0. The first law then reads that U 3 = W 3. In this case, the temperatures are far easier to compute than the wor integral, and so we first compute U 3. We have nowledge of the pressures at each point and the temperature at point A, and desire nowledge about the temperature at point B. We thus mae use of the relation P γ T γ = constant. P γ A T γ A = P γ B T γ B T PA B = T A P B ) γ γ = T A β γ γ = βp 0V 0 nr β γ γ = P 0V 0 nr β γ We then have that the change in internal energy and negative the wor) are equal to W 4 = U 4 = nc v T B T A ) = C v R P 0V 0 β γ 4. In the isothermic process, temperature is constant, and so U 4 T 3 = 0. The first law then reads that Q 4 = W 4. Note that if T is constant, we have by the ideal gas law that P and V are inversely proportional. This means that when the pressure changes by a factor /β, the volume changes by a factor β. Thus V B = βv 0. We compute the wor and because of their equality, the heat) from the following integral: Q 4 = W 4 = B A p dv = βv 0 V 0 nrt = nrt ln β) = βp 0 V 0 lnβ) V where in the last step we used the ideal gas law to replace T in terms of the temperature and pressure. Note that W 4 > 0, Q 4 > 0, and U 4 = 0, meaning that the gas does wor and absorbs heat simultaneously in such a balanced way to eep the temperature and thus, the internal energy) fixed. 2.3 Worth Mention: Adiabatic Processes. Adiabatic implies that no heat is transferred during the process that is, that Q = Adiabatic processes usually occur in situations where the process occurs very quicly, in which case heat transfer does not have time to tae place while the process is happening. 3. P V γ, T V γ, and P γ T γ are all constant. This quality arises because dq = 0, so that the first law reads du = dw nc v dt = pdv Using the ideal gas law to replace p = nrt/v, the above becomes nc v dt = nrt V dv dt T = R dv C v V Since we define γ = C p /C v, and it can be obtained from the study of isobaric processes that C p = C v + R, we identify that R/C v = γ. Maing this substitution, and performing the indefinite integral, ln T = γ ) ln V + constant Rearranging, we find that T V γ = constant 4. Starting with T V γ = constant, we may use the ideal gas law T = P V/nR to rewrite T V γ = P V/nR)V γ P V γ = constant. We can play a similar game to obtain the third relation. 5

6 2.4 Feel the Heat... Engines Fridenginatorators A heat engine taes in heat to allow the system to perform wor, and a refrigerator uses wor done on the system to extract heat. The characterization of whether a cyclic process is a heat engine or a refrigerator is thus only up to whether wor is done by the system on on the system, respectively. Processes ) and 3), with W net > 0, are thus heat engines, and process 2), with W n et < 0, is a refrigerator. What s Your Sign A cyclic process, because it returns to the same to the same temperature, has U = 0. Thus whatever sign W possesses, Q also possesses. Processes ) and 3) have W > 0 and thus Q > 0. Process 2) has W < 0 and thus Q < 0. For the upper half of the process in 2), note that the temperature at the left point is lower than the temperature at the right point, and so U T < 0. Additionally, because we are integrating from a larger volume to a smaller volume, W < 0. We thus conclude that Q = U + W < 0. The PV Game for an Engine The wor done by the system in a complete cycle is merely the area under the curve that is, the area of the triangle. The base of the triangle is V 0 β ) in width, and the height is P 0 α ), and so the area the net wor done in the cycle is given by W = 2 P 0V 0 α )β ). Efficient AF For a cyclic process, U = 0 as described above). This implies that Q net = Q in + Q out = W net, which allows us to rewrite the numerator of e as shown. Don t Be A Square One would find by analyzing each portion of the P V diagram of ) that the only heat taen in that is, where Q > 0) is absorbed during the top side and the left side of the square process the isobaric expansion at pressure βp 0 from volume V 0 to volume αv 0. The wor done in this process is given by W top = P 0 V 0 βα ), and the change in internal energy is given by, using the ideal gas law to obtain the temperature at each point, U top = nc v T f T i ) = nc v V 0 P 0 nr β α ) = C v R V 0P 0 β α ) The heat absorbed during this process is then Q in, = U top + W top = C v R V 0P 0 β α ) + P 0 V 0 βα ) = R + V 0 P 0 βα ) From our wor on the isobaric process in PV Boot Camp, we now that at constant pressure, the heat absorbed is given by Q in,2 = nc p T f T i ) = nc v + R) P 0V 0 β ) = nr R + V 0 P 0 β ) Now all we need is the net wor, given by the area enclosed by the whole process. We obtain this simply by computing the area of the square, so that W net is given by So the net heat in is given by Q in, + Q out,2 = Thus the efficiency is as follows: W net = V 0 P 0 α )β ) R + V 0 P 0 βα ) + R + V 0 P 0 β ) Q in = R + V 0 P 0 [βα ] e = W net R α )β ) = Q in C v + R βα 6

7 The Carnot Cycle: The Most Efficient Heat Engine Note that the heat exchange occurs on the isothermal processes, AB and CD. Since U T for these processes is equal to zero, we have for these processes W = Q. Since it is on the path AB that W > 0, the heat is absorbed by the system during AB. Thus, Q in = Q AB = W AB. Additionally, for the entire cycle A A, we have that U net = 0, and so W net = Q net = Q AB + Q CD since the adiabatic processes BC and DA have Q BC = Q DA = 0). From our analysis of isothermal processes, we now that the wor done is given by VB Q AB = W AB = nrt H ln Our efficiency is thus e carnot = W net Q in Q CD = W CD = nrt C ln VD V C = Q AB + Q T C ln VD CD V C = + Q AB T H ln VB We now need only relate V B and, and V D and V C. We may do so by analyzing the adiabatic processes BC and DA. Recall that T V γ = constant. Thus T H V γ B = T C V γ C V C V B = T C V γ D = T H V γ A V D = Equating the left hand sides of these two equalities, we have that V C = V D = V D VA ln = ln V B V B V C The above gives that V B ln VD V C ln VB ) = TH T C TH T C VA V B ) ) γ ) γ ) = ln VB Plugging in this expression to our efficiency in place of the natural logs, we obtain 3 Harmonic Oscillators 3. Rapid Fire SHO e carnot = T C T H = T H T C T H. At points i) and v), the force on the bloc F = x) is largest in magnitude, and so the acceleration is greatest at these points. 2. At point iii), where all of the system s energy is in the form of inetic energy, and so the speed of the bloc is greatest. 3. The total energy of the system at any displacement x is given by 2 mv2 + 2 x2 Since there are no dissipative forces acting on the system, we may equate the energy of the system at its maximum displacement when v = 0 and x = D) to the energy of the system at point iii) where x = 0). Doing so and solving for v, we obtain 2 mv2 = 2 D2 v = D m ) 7

8 3.2 Meatier SHO Questions. The period of the bloc s oscillation depends only on the mass of the system, given by T = 2π/ω = 2π m/ that is, T m. Thus an increase in mass will result in an increase in period, and so in both cases the period of oscillation increases. a) If the clay is added at point iii), imposing momentum conservation yields that the speed of the system must decrease, and so the total energy of the system which is at iii) entirely inetic) will decrease. This means that there is less energy available to go into stretching the spring, and so the amplitude of oscillation will decrease. b) If the clay is added at point v), we note that all of the energy of the system is stored in spring potential energy, which is independent of mass. Thus the total energy of the system remains unchanged, and so the amplitude of oscillation remains the same. 2. Adding mass to the bloc at point iii) actually, at any point but i) and v) but we ll just use iii) since it s the easiest to deal with) will decrease the amplitude of oscillation, and adding mass to the bloc at any point will increase the period and thus decrease the frequency. Thus, both cases are possible. a) For the case where we wish to halve the amplitude of oscillation, we will need to reduce the speed of the bloc at iii) from it s value due to an initial displacement D, given in 3. part 3, to the value which will give a inetic energy necessary to stretch the spring to a displacement D/2. We may solve for this speed v by equating the inetic energy of the combined clay-bloc system of mass m + M, where M is the mass of the clay) with the spring potential energy for a maximum displacement D/2. Doing so, 2 m + M)v 2 = 2 D 2 ) 2 v = D 2 M + m We now enforce momentum conservation in the horizontal direction for the clay-bloc collision, which should reduce the speed of the bloc from the initial value v = D /m to the new value v. mv = m + M)v md m = m + M)D 2 M + m m = M + m 2 Solving for the mass of the bloc, we obtain that M = 3m. b) To halve the frequency, we must double the amplitude of oscillation. Since T m, to do so requires that we quadruple the mass, so that m = m + M = 4m. This gives that M = 3m. 3. The general solution for a harmonic oscillator is xt) = A cos ωt + φ), where ω /m. 4. We have two undetermined constants, A and φ, present in our equation of motion. To obtain these constants, we need two initial conditions the familiar two unnowns need two equations. We have two such initial conditions, given in the problem setup: at t = 0, the box is displaced to xt = 0) = D, and released from rest that is, vt = 0) = dx/dt t=0 = 0). Note that our general equation of motion gives the velocity the time derivative of xt). xt) = A cosωt + φ) and vt) = dx dt = Aω sin ωt + φ Applying our initial conditions, we obtain the following system of equations. D = xt = 0) = A cos φ and 0 = vt = 0) = Aω sin φ Taing the second equation, we see that 0 = sin φ. We are free to choose the smallest valid value of φ which satisfies this expression: φ = 0. With this, we have the first equation as D = A cos 0 D = A. With both constants determined, we thus have the specific equation of motion xt) of the box as xt) = D cosωt) 8

9 5. To obtain the time at which the system has equal inetic and spring potential energy, we simply equate the two energies at the position xt = t 0 ) and velocity vt = t 0 ). We thus obtain: 2 mvt 0) 2 = 2 xt 0) 2 Using the equations of motion found for xt) and vt) in the previous part, we have for the above equation md 2 ω 2 sin 2 ωt 0 ) = D 2 cos 2 ωt 0 ) Note that since ω = /m, mω 2 =. Canceling the common factors of and D 2, moving the sine squared to the left side, we have 0 = cos 2 ωt 0 sin 2 ωt 0 = cos 2ωt 0 where in the last equality we used the double angle identity for cosine. Since the first value of x for which cosx = 0 is x = π/2, we have 2ωt 0 = π 2 t 0 = π 4 t 0 = π 4 m ω 9