Thermodynamics Molecular Model of a Gas Molar Heat Capacities

Transcription

1 Thermodynamics Molecular Model of a Gas Molar Heat Capacities Lana Sheridan De Anza College May 3, 2018

2 Last time modeling an ideal gas at the microscopic level rms speed of molecules equipartition of energy heat capacities for monatomic ideal gases

3 Overview constant-pressure heat capacity for monatomic gases heat capacities for diatomic ideal gases adiabatic processes

4 Quick Recap For all ideal gases: K tot,trans = N K trans = 3 2 Nk BT = 3 2 nrt and E int = nc V T For monatomic gases: E int = K tot,trans = 3 2 nrt and so, C V = 3 2 R

5 (19-46) Heat Capacity for Constant Pressure Processes nstant pressure. This e C V, because energy the gas Inbut a constant also for pressure process, the work done on the gas is: Q T 19-11a. W = P V Thermal reservoir the first law of ther- (a) (19-47) substitute from Eq. e first note that since p V.Then we note (19-48) ugh by n T, we find (b) Pressure From the ideal gas equation: (19-49) iment, not only for their density So, is low for a monatomic gas p V i p V The temperature increase is done without changing the pressure. Volume Fig (a) The temperature of an ideal gas is raised from T to T Tin a constant-pressure P V process. = nr Heat Tis added and work is done in lifting the loaded piston. (b) The process on a p-v diagram.the work p V is given by the shaded area. f T + T V + V W = nr T T

6 Heat Capacity for Constant Pressure Processes First law: E int = Q + W rearranging: E int Q = W nc V T nc P T = nr T

7 Heat Capacity for Constant Pressure Processes First law: E int = Q + W rearranging: E int Q = W nc V T nc P T = nr T because Q = nc P T. Dividing by n T : C P C V = R For a monatomic gas: C P = 5 2 R

8 Kinetic Energy and Internal Energy The total kinetic energy of an ideal monatomic gas of N particles is the total translational K.E. K tot,trans = 3 2 Nk BT = 3 2 nrt In a monatomic gas these are the three translational motions are the only degrees of freedom. We can choose ( ) 3 E int = K tot,trans = N 2 k BT = 3 2 nrt (This is the thermal energy, so the bond energy is zero if we liquify the gas the bond energy becomes negative.)

9 Equipartition Consequences in Diatomic Gases Reminder: Equipartition of energy theorem Each degree of freedom for each molecule contributes an and additional 1 2 k BT of energy to the system. A monatomic gas has 3 degrees of freedom: it can have translational KE due to motion in 3 independent directions. A diatomic gas has more ways to move and store energy. It can: translate rotate vibrate

10 21.3 The Equipartition Consequences of Energy in Diatomic Gases Predictions based our model for molar specific heat agree quite well with the behavior Contribution of monatomic gases, but tonot internal with the behavior energy: of complex gases (see Table 21.2). The value predicted by the model for the quantity C P 2 C V 5 R, however, is the same for all gases. This similarity is not surprising because this difference is the result of the work done on the gas, which is independent of its molecular structure. To clarify the variations in C V and C P in gases more complex than monatomic 3 2 k BT (3 directions of motion) gases, let s explore further the origin of molar specific heat. So far, we have assumed the sole contribution to the internal energy of a gas is the translational kinetic energy of the molecules. The internal energy of a gas, however, includes contributions from the translational, vibrational, and rotational motion of the molecules. The rotational and vibrational motions of molecules can be activated by collisions and therefore are coupled to the translational motion of the molecules. The branch of physics known as statistical mechanics has shown that, for a large number of particles obeying the laws of Newtonian mechanics, the available energy is, on average, shared equally by each independent degree of freedom. Recall from 2 2 k BT (rotations about x and z axes) Section 21.1 that the equipartition theorem states that, at equilibrium, each degree of freedom contributes 1 2 k B T of energy per molecule. Let s consider a diatomic gas whose molecules have the shape of a dumbbell (Fig. 21.5). In this model, the center of mass of the molecule can translate in the x, y, and z directions (Fig. 21.5a). In addition, the molecule can rotate about three mutually perpendicular axes (Fig. 21.5b). The rotation about the y axis can be neglected because the molecule s moment of inertia I y and its rotational energy 1 2 I y v 2 about this axis are negligible compared with those associated with the x and z axes. (If the two atoms are modeled as particles, then I y is identically zero.) Therefore, there are five degrees of freedom for translation and rotation: three associated with the 2 2 k BT (KE and E int 5 3N PE1 1 2k B of T2 1harmonic 2N 1 1 2k B T Nk oscillator) B T 5 5 2nRT translational motion and two associated with the rotational motion. Because each degree of freedom contributes, on average, 1 2k B T of energy per molecule, the internal energy for a system of N molecules, ignoring vibration for now, is We can use this result and Equation to find the molar specific heat at constant volume: 1 de 1 d 21.3 The Equipartition of Energy 635 x x Translational motion of the center of mass a Rotational motion about the various axes b z Vibrational motion along the molecular axis c z y y

11 (Classical) Equipartition Prediction for Diatomic Gases In total: E int = N We can also write the internal energy: ( ) 7 2 k BT E int = 7 2 nrt And so, C V = 7 2 R This is what we would expect for a diatomic gas based on the equipartition theorem. It is not quite what is observed, however.

12 What Actually Happens in Diatomic Gases Prediction: C V = 7 2 R For most diatomic gases, such as H 2 and N 2, C V = 5 2 R at moderate temperatures (around room temperature). And at low temperatures for these gases C V = 3 2 R.

13 What Actually Happens in Diatomic Gases Prediction: C V = 7 2 R For most diatomic gases, such as H 2 and N 2, C V = 5 2 R at moderate temperatures (around room temperature). And at low temperatures for these gases C V = 3 2 R. It is almost as if degrees of freedom become activated once there is enough energy...

14 olecules are oscillating and C V /R 3.5. (In Fig , the plotted curve at What 3200 K because Actually there the Happens atoms of a hydrogen in Diatomic molecule oscillate Gases so that they overwhelm their bond, and the molecule then dissociates into two ate atoms.) Hydrogen gas: 21.4 Adiabatic Processes for an Ideal Gas 637 rational 4quantum states of a diatomic molecule. The lowest allowed state d the ground state. The black lines show the energies allowed for the molotice that allowed vibrational states are separated by larger energy gaps 7/2 e rotational 3 states. Oscillation w temperatures, the energy a molecule gains in collisions with its neighbors ally not large enough to raise it to the first excited state of either rotation 5/2 or n. Therefore, 2 even though rotation and vibration Rotationare allowed according to l physics, they do not occur in reality at low temperatures. All molecules are ground state for rotation and vibration. The only contribution to 3/2 the molaverage energy 1 is from translation, and the specific heat is that predicted by n Translation Vibrational e temperature is raised, the average energy of the molecules increases. In states llisions, a 0 molecule may have enough energy transferred to it from another le to excite 20the first 50 rotational state. 500 As the 1000temperature is 10,000 raised further, olecules can be excited to this Temperature state. The result (K) is that rotation begins to ute to the internal energy, and the molar specific heat rises. At about room ature Fig. in Figure , Cthe V /Rsecond versus plateau temperature has been for (diatomic) reached and hydrogen rotation confully gas. At to the Because 3200K molar rotational specific the molecules heat. and The oscillatory molar begin specific motions toheat begin is now at certain equal to the redicted energies, by Equation only translation is possible at very low temperatures.as e is no dissociate. the contribution temperature at increases, room temperature rotational from motion vibration can begin.at because still the molee still in the ground vibrational state. The temperature must be raised even higher temperatures, oscillatory motion can begin. to excite the first vibrational state, which happens in Figure 21.6 between and K. Left At 10 diagram, 000 K on Halliday, the right side Resnick, of the Walker; figure, vibration right diagram is con- Serway & Jewett C V /R ENERGY The rotational states lie closer together in energy than do the vibrational states. Rotational states Rotational states Figure 21.7 An energy-level diagram for vibrational and rotational states of a diatomic molecule.

15 Question Quick Quiz The molar specific heat of a gas is measured at constant volume and found to be 11R/2. Is the gas most likely to be (A) monatomic, (B) diatomic, or (C) polyatomic? 1 Serway & Jewett, page 637.

16 Question Quick Quiz The molar specific heat of a gas is measured at constant volume and found to be 11R/2. Is the gas most likely to be (A) monatomic, (B) diatomic, or (C) polyatomic? 1 Serway & Jewett, page 637.

17 A Useful Ratio The quantity γ is defined as: γ = C P C V For a monatomic gas: γ = 5 3 What is γ for a diatomic gas near room temperature?

18 A Useful Ratio The quantity γ is defined as: γ = C P C V For a monatomic gas: γ = 5 3 What is γ for a diatomic gas near room temperature? γ = 7 5

19 Summary of Results For ideal gases at moderate temperatures (around room temperature): monatomic gas diatomic gas C V 3 2 R 5 2 R C P 5 2 R 7 2 R γ = C P C V

20 Adiabatic Process in Ideal Gases For an adiabatic process (Q = 0): PV γ = const. and: TV γ 1 = const.

21 Adiabatic Process in Ideal Gases For an adiabatic process (Q = 0): PV γ = const. and: TV γ 1 = const. (Given the first one is true, the second follows immediately from the ideal gas equation, P = nrt V.)

22 Adiabatic Process in Ideal Gases Where this relation comes from: E int = W Considering a small volume change in time that produces a small change in temperature: nc V dt de int = P dv = dw (1)

23 Adiabatic Process in Ideal Gases Where this relation comes from: E int = W Considering a small volume change in time that produces a small change in temperature: nc V dt The ideal gas law derivative: de int = P dv = dw (1) P dv dp +V n dt = nr dt = 1 ( P dv ) dp +V R Substitute n dt into our energy equation (1).

24 Adiabatic Process in Ideal Gases C V R ( C V n dt ) ( P dv ) dp +V V dp = P dv = P dv ) = (1 + RCV P dv

25 Adiabatic Process in Ideal Gases C V R ( C V n dt ) ( P dv ) dp +V V dp = P dv = P dv ) = (1 + RCV P dv Notice: γ = 1 + R C V then, dividing by PV : 1 P dp = γ V dv

26 Adiabatic Process in Ideal Gases C V R ( C V n dt ) ( P dv ) dp +V V dp = P dv = P dv ) = (1 + RCV P dv Notice: γ = 1 + R C V then, dividing by PV : Integrating both sides: 1 P dp = γ V dv ln P = γ ln V + c

27 Adiabatic Process in Ideal Gases ln P = γ ln V + c Implies: PV γ = const. This equation characterizes an adiabatic process in an ideal gas, along with this one, which follows from PV = nrt : TV γ 1 = const.

28 Summary equipartition of energy molar heat capacities adiabatic processes Quiz in class, Friday. Collected Homework due Monday, May 7. Test on Thermodynamics, Monday, May 14. Homework Serway & Jewett: set yesterday: Ch 21, CQs: 7; Probs: 15, 17, 21, 23, 25, 29, 31, 47, 59, 65