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Announcements 1. Please bring your laptops to lab this week. 2. Physics seminar this week – Thursday, Nov. 20 at 4 PM – Professor Brian Matthews will…
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Announcements 1. Please bring your laptops to lab this week. 2. Physics seminar this week – Thursday, Nov. 20 at 4 PM – Professor Brian Matthews will discuss relationships between protein structure and function 3. Schedule – Today, Nov. 18th: Examine Eint especially for ideal gases Discuss ideal gas law Continue discussion of first law of thermodynamics Thursday, Nov. 20th: Review Chapters 15-20 Tuesday, Nov. 25th: Third exam 6/23/2016 PHY 113 -- Lecture 20 1 From The New Yorker Magazine, November 2003 6/23/2016 PHY 113 -- Lecture 20 2 First law of thermodynamics DEint = Q - W Vf W   PdV Vi For an “ideal gas” we can write an explicit relation for Eint. What we will show: n N ( ideal gas ) Eint  RT  k BT γ-1 γ-1 g is a parameter which depends on the type of gas (monoatomic, diatomic, etc.) which can be measured as the ratio of two heat capacities: g=CP/CV. 6/23/2016 PHY 113 -- Lecture 20 3 Thermodynamic statement of conservation of energy – First Law of Thermodynamics DEint = Q - W Work done by system Wif depends on path Heat added to system Qif depends on path “Internal” energy of system DEint = Eint(f)-Eint(i)  independent on path 6/23/2016 PHY 113 -- Lecture 20 4 How is temperature related to Eint? Consider an ideal gas  Analytic expressions for physical variables  Approximates several real situations Ideal Gas Law: PV = n R T temperature (K) volume (m3) gas constant (8.31 J/(moleK)) pressure (Pa) number of moles 6/23/2016 PHY 113 -- Lecture 20 5 Ideal gas – P-V diagram at constant T PV = n R T Pi Pf Pi Vi = Pf Vf Vi Vf 6/23/2016 PHY 113 -- Lecture 20 6 6/23/2016 PHY 113 -- Lecture 20 7 Microscopic model of ideal gas: Each atom is represented as a tiny hard sphere of mass m with velocity v. Collisions and forces between atoms are neglected. Collisions with the walls of the container are assumed to be elastic. 6/23/2016 PHY 113 -- Lecture 20 8 What we can show is the pressure exerted by the atoms by their collisions with the walls of the container is given by: 2N 1 2N P 2 m v avg  2 K avg 3V 3V Proof: Force exerted on wall perpendicular to x-axis by an atom which collides with it: Dp 2m v Fix  - ix  i ix Dt  2d / vix Dt Dt -vix 2mi vix mi vix2  Fix   vix 2d / vix d x 2 number of atoms Fix mi vix N P   mi vx2 d i A i dA V volume average over atoms 6/23/2016 PHY 113 -- Lecture 20 9 Ideal gas law continued: Recall that N 1N 2 N  1R 2  P  m v 2  Macroscopi c relation x m v 2  : PV  nRT  N m Tv  Nk BT V 3V 3 V  2N A  2 1  Microscopi c model : PV  N  m v 2   Nk BT 3 2  Therefore:  1 2  3  m v   k BT 2  2 3k BT  vrms  v 2  m Also: 1 m v 2   3 k T   B  2  2  1 2  3  Eint  N  m v   N k BT 2  2 6/23/2016 PHY 113 -- Lecture 20 10 Big leap! Internal energy of an ideal gas: Eint  N  m v 2   N k BT  1 3 N n k BT  RT 2  2 γ-1 γ-1 derived for monoatomic ideal gas more general relation for polyatomic ideal gas Gas g (theory) g(exp) He 5/3 1.67 N2 7/5 1.41 H2O 4/3 1.30 6/23/2016 PHY 113 -- Lecture 20 11 Determination of Q for various processes in an ideal gas: n Eint  RT γ-1 n DEint  RDT  Q - W γ-1 Example: Isovolumetric process – (V=constant  W=0) n DEint i  f  RDTi f  Qi f γ-1 n In terms of “heat capacity”: Qi  f  RDTi  f  nCV DTi  f γ-1 R CV  γ-1 6/23/2016 PHY 113 -- Lecture 20 12 Example: Isobaric process (P=constant): n DEint i f  RDTi  f  Qi f - Wi f γ-1 In terms of “heat capacity”: RDTi  f  Pi (V f - Vi   n n Qi  f  RDTi  f  nRDTi  f  nC P DTi  f γ-1 γ-1 R γR  CP  R γ-1 γ-1 Note: g = CP/CV 6/23/2016 PHY 113 -- Lecture 20 13 More examples: Isothermal process (T=0) n Eint  RT γ-1 n DEint  RDT  Q - W γ-1 DT=0  DEint = 0  Q=W Vf dV Vf V f  W   PdV  nRT  nRT ln  Vi Vi V  Vi  6/23/2016 PHY 113 -- Lecture 20 14 Even more examples: Adiabatic process (Q=0) DEint  -W n RDT  - PDV γ-1 PV  nRT DPV  PDV  nRDT nRDT  -(γ-1PDV  DPV  PDV DV DP -γ  V P  V fγ   Pf     - ln γ  ln   PiVi γ  Pf V fγ V   Pi   i  6/23/2016 PHY 113 -- Lecture 20 15 PiVi g Adiabat : P  g V PiVi Isotherm : P  V 6/23/2016 PHY 113 -- Lecture 20 16 Peer instruction question Suppose that an ideal gas expands adiabatically. Does the temperature (A) Increase (B) Decrease (C) Remain the same PiVi γ  Pf V fγ Ti PiVi  nRTi  Pi  nR Vi TiVi γ-1  T f V fγ-1 γ -1  Vi  T f  Ti    V f  6/23/2016 PHY 113 -- Lecture 20 17 Review of results from ideal gas analysis in terms of the specific heat ratio g  CP/CV: n R DEint  RDT  nCV DT ; CV  γ-1 γ-1 γR CP  γ-1 For an isothermal process, DEint = 0  Q=W Vf V f  V f  W   PdV nRT ln   PiVi ln  Vi  Vi   Vi  For an adiabatic process, Q = 0 PiVi γ  Pf V fγ TiVi γ-1  T f V fγ-1 6/23/2016 PHY 113 -- Lecture 20 18 Extra credit: Show that the work done by an ideal gas which has an initial pressure Pi and initial volume Vi when it expands adiabatically to a volume Vf is given by: γ -1 Vf PiVi   Vi    W   PdV  1-   γ - 1  V f   Vi   6/23/2016 PHY 113 -- Lecture 20 19 Peer instruction questions Match the following types of processes of an ideal gas with their corresponding P-V relationships, assuming the initial pressures and volumes are Pi and Vi, respectively. 1. Isothermal 2. Isovolumetric 3. Isobaric 4. Adiabatic (A) P=Pi (B) V=Vi (C) PV=PiVi (D)PVg=PiVig 6/23/2016 PHY 113 -- Lecture 20 20 Examples process by an ideal gas: Pf B C AB BC CD DA P (1.013 x 105) Pa Q Vi ( Pf - Pi ) γPf (V f - Vi ) - V f ( Pf - Pi ) -γPi (V f - Vi ) γ -1 γ -1 γ -1 γ -1 W 0 Pf(Vf-Vi) 0 -Pi(Vf-Vi) A D Pi DEint Vi ( Pf - Pi ) Pf (V f - Vi ) - V f ( Pf - Pi ) -Pi (V f - Vi ) γ -1 γ -1 γ -1 γ -1 Efficiency as an engine: Vi Vf e = Wnet/ Qinput 6/23/2016 PHY 113 -- Lecture 20 21 6/23/2016 PHY 113 -- Lecture 20 22
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