Physics 1501: Lecture 36 Today ’ s Agenda

1. Physics 1501: Lecture 36Today’s Agenda • Announcements • Homework #12 (Dec. 9): 2 lowest dropped • Midterm 2 … in class Wednesday • Honors’ students: Wednesdat at 3:30 in my office. • Today’s topics • Chap. 17: ideal gas • Kinetic theory • Ideal gas law • Diffusion • Chap.18: Heat and Work • Zeroth Law of thermodynamics • First Law of thermodynamics and applications • Work and heat engines

2. Kinetic Theory of an Ideal Gas • Pressure is

3. Concept of temperature • Does a Single Particle Have a Temperature? • Each particle in a gas has kinetic energy. On the previous page, we have established the relationship between the average kinetic energy per particle and the temperature of an ideal gas. • Is it valid, then, to conclude that a single particle has a temperature?

4. Example: Speed of Molecules in Air • Air is primarily a mixture of nitrogen N2 molecules (molecular mass 28.0u) and oxygen O2 molecules (molecular mass 32.0u). • Assume that each behaves as an ideal gas and determine the rms speeds of the nitrogen and oxygen molecules when the temperature of the air is 293K. • For nitrogen

5. Internal energy of a monoatomic ideal gas • The kinetic energy per atom is • Total internal energy of the gas with N atoms

6. Kinetic Theory of an Ideal Gas: summary • Microscopic model for a gas • Goal: relate T and P to motion of the molecules • Assumptions for ideal gas: • Number of molecules N is large • They obey Newton’s laws (but move randomly as a whole) • Short-range interactions during elastic collisions • Elastic collisions with walls • Pure substance: identical molecules • Temperature is a direct measure of average kinetic energy of a molecule

7.  Kinetic Theory of an Ideal Gas: summary • Theorem of equipartition of energy • Each degree of freedom contributes kBT/2 to the energy of a system (e.g., translation, rotation, or vibration) • Total translational kinetic energy of a system of N molecules • Internal energy of monoatomic gas: U = Kideal = Ktot trans • Root-mean-square speed:

8. a) x1.4 b) x2 c) x4 a) x1 b) x1.4 c) x2 Lecture 36: ACT 1 • Consider a fixed volume of ideal gas. When N or T is doubled the pressure increases by a factor of 2. 1) If T is doubled, what happens to the rate at which a single molecule in the gas has a wall bounce? 2) If N is doubled, what happens to the rate at which a single molecule in the gas has a wall bounce?

9. Diffusion • The process in which molecules move from a region of higher concentrationto one of lower concentration is called diffusion. • Ink droplet in water

10. Why is diffusion a slow process ? • A gas molecule has a translational rms speed of hundreds of meters per second at room temperature. At such speed, a molecule could travel across an ordinary room in just a fraction of a second. Yet, it often takes several seconds, and sometimes minutes, for the fragrance of a perfume to reach the other side of the room. Why does it take so long? • Many collisions !

11. Comparing heat and molecule diffusion • Both ends are maintained at constant concentration/temperature

12. A Th Tc Energy flow L Fick’s law of diffusion • For heat conduction between two side at constant T conductivity temperature gradient between ends • The mass m of solute that diffuses in a time t through a solvent contained in a channel of length L and cross sectional area A is diffusion constant concentration gradient between ends SI Units for the Diffusion Constant: m2/s

13. Chap. 18: Work & 1st Law The Laws of Thermodynamics 0)If two objects are in thermal equilibrium with a third, they are in equilibrium with each other. 1) There is a quantity known as internal energy that in an isolated system always remains the same. 2) There is a quantity known as entropy that in a closed system always remains the same (reversible) or increases (irreversible).

14. T2 T1 = If objects A and B are separately in thermal equilibrium with a third object C, then objects A and B are in thermal equilibrium with each other. U2 U1 Zeroth Law of Thermodynamics • Thermal equilibrium: when objects in thermal contact cease heat transfer • same temperature B C A

15. 1st Law: Work & Heat • Two types of variables • State variables: describe the system (e.g. T, P, V, U). • Transfer variables: describe the process (e.g. Q, W). • =0 unless a process occurs •  change in state variables. • Work done on gas • W = F d cos = -F y = - PA y = - P V • valid only for isobaric processes (P constant) • If not, use average force or calculus: W = area under PV curve PV diagram

16. 1st Law: Work & Heat • Depends on the path taken in the PV-diagram • Same for Q (heat) • Work:

17. a)> |W1| b)= |W1| c)< |W1| Lecture 36: Act 2Work • Consider the two paths, 1 and 2, connecting points i and f on the pV diagram. • The magnitude of the work, |W2|, done by the system in going from i to f along path 2is f 2 i 1 p V

18. work done “on” the system heat flow “in” (+) or “out” (-) variation of internal energy First Law of Thermodynamics • First Law of Thermodynamics U = Q + W • Independent of path in PV-diagram • Depends only on state of the system (P,V,T, …) • Energy conservation statement  only U changes • Isolated system • No interaction with surroundings • Q = W = 0  U = 0. • Uf = Ui : internal energy remains constant.

19. Other Applications • Cyclic process: • Process that starts and ends at the state • Must have U = 0  Q = -W . • Adiabatic process: • No energy transferred through heat  Q = 0. • So, U = W . • Important for • expansion of gas in combustion engines • Liquifaction of gases in cooling systems, etc. • Isobaric process: (P is constant) • Work is simply

20. Other Applications (continued) • Isovolumetric process: • Constant volume  W =0. • So U = Q  all heat is transferred into internal energy • e.g. heating a “can” (no work done). • Isothermal process: • T is constant • Using PV=nRT, we find P= nRT/V. • Work becomes : • PV is constant. • PV-diagram: isotherm.

21. p V Lecture 36: Act 3Processes • Identify the nature of paths A, B, C, and D • Isobaric, isothermal, isovolumetric, and adiabatic D A T1 C T2 B T3 T4

22. Heat Engines • We now try to do more than just raise the temperature of an object by adding heat. We want to add heat to get some work done! • Heat engines: • Purpose: Convert heat into work using a cyclic process • Example: Cycle a piston of gas between hot and coldreservoirs*(Stirling cycle) 1)hold volume fixed, raise temperature by adding heat 2)hold temperature fixed, do work by expansion 3)hold volume fixed, lower temperature by draining heat 4)hold temperature fixed, compress back to original V

23. 1 1 2 Gas Gas Gas Gas P T=TC T=TH T=TH T=TC 1 2 TH 3 4 TC V Va Vb 3 4 Heat Engines We can represent this cycle on a P-V diagram: • Example: the Stirling cycle *reservoir: large body whose temperature does not change when it absorbs or gives up heat

24. P 1 2 TH 3 4 TC V Va Vb Heat Engines • Identify whether • Heat is ADDED or REMOVED from the gas • Work is done BY or ON the gas for each step of the Stirling cycle: 2 1 step 3 4 ADDED ADDED ADDED ADDED HEAT REMOVED REMOVED REMOVED REMOVED BY BY BY BY WORK ON ON ON ON