1. Paperwork Mastering Physics Course # DRKIDD880131 Assignment Due tonight New Assignment up soon… Lab Reports due next Tuesday 5pm

2. Schedule Short Term Today – derive pressure/height, calculate Friday – Begin Chapter 19 Monday – Chapter 19 / Solving Tuesday – Lab #2 –Quiz#2 [Chapter 18, Labs] Thursday HMWK due 11pm

3. Pressure vs. Height Example 18.4 Thin object, mass m Force = pA Force = pA + (dp)A For an object in a fluid Pressure on sides of object is the same, so cancels (Book on desk is stationary) Assume pressure felt by top is slightly different than bottom (p+dp) dy

4. Pressure vs. Height Example 18.4 Thin object, mass m Force = pA Force = pA + (dp)A For an object in a fluid Pressure on sides of object is the same, so cancels (Book on desk is stationary) Assume pressure felt by top is slightly different than bottom (p+dp) dp can be +, - or even zero. Just much smaller than p for thin object Let’s say this object is stationary – floating in the fluid. What is sum of all forces on object? What are all forces on object? What if “Object” was just a portion of the fluid itself? dy

5. Pressure vs. Height Example 18.4 mass =  V =  A(dy) Force = pA Force = pA + (dp)A  F = 0 = pA - [pA + (dp)A] – mg 0 = pA – pA – (dp)A –  Vg (dp)A = -  Vg (dp)A = -  (Ady)g (dp/dy) = -  g Implications? dy

6. Pressure vs. Height Example 18.4 mass =  V =  A(dy) Force = pA Force = pA + (dp)A  F = 0 = pA - [pA + (dp)A] – mg 0 = pA – pA – (dp)A –  Vg (dp)A = -  Vg (dp)A = -  (Ady)g (dp/dy) = -  g dy For Ideal Gas  = m/V = pM/(RT) (dp/dy) = -  g

7. Pressure vs. Height Example 18.4 mass =  V =  A(dy) Force = pA Force = pA + (dp)A dy For Fluid that is an Ideal Gas  = m/V = pM/(RT) Pressure vs. Height Any Fluid (dp/dy) = -  g (dp/dy) = - pgM/(RT)

8. Pressure vs. Height (dp/dy) = - pgM/(RT) Now need to set up equation to solve (dp/p) = -(gM/RT)(dy) –Assume a constant temperature (?)

9. Pressure vs. Height (dp/dy) = - pgM/(RT) Now need to set up equation to solve (dp/p) = -(gM/RT)(dy) –Assume a constant temperature (?)

10. Pressure vs. Height

11. Let’s say integration was from sea level (p0=p 0, y0 = 0) To a point pF = p, yF = y Need to have known endpoints Then can derive equation for air pressure as a function of height above sea level Happy Equation: Should Check Accuracy Implications? Check at sea level.

12. Molecular Motion From Derivation in Section 18.3 –KE AVG =(3/2)Nk B T (N is # molecules) –pV = (2/3) KE AVG –pV = Nk B T Where does 3 come from? Other implications? Why do people care about root mean square? [v rms ] Hint: zero?

13. Mean Free Path Lifetime (Mean Free Time) Very Important Concept –Vacuum conditions –Behavior of electrons in solids –Any interacting (or not) particles Mean distance traveled before collision MFP = v t MFP Lifetime (mean free time) is time between collisions Why important for electrical conduction?

14. Lots of Averages… Gas at temperature 500K Average KE of One Mole KE AVG = (3/2)nRT = (3/2)RT –8.31 J / (mole K) KE AVG = 6232 J Say gas is N 2 (MM = 0.028kg / mole) KE AVG = (1/2)m(v 2 ) AVG What is max speed?

15. Phase Diagram Describes P&T conditions for materials Interesting Points, What is this at STP?

16. Phase Diagram Supercritical Fluid? Neat – can dissolve things like a liquid & Diffuse through solids like a gas…

17. Gas # molecules = n 0 Temperature = T 0 pressure = p 0 Volume = V 0 Discussion Q18.10 Start

18. Gas Initial State # molecules = n 0 Temperature = T 0 pressure = p 0 Volume = V 0 Discussion Q18.10 “Sudden” Hole in wall Gas Final State # molecules = ? Temperature = ? pressure = ? Volume = ? What Happens here?

19. Schedule Short Term Today – derive pressure/height, calculate Friday – Begin Chapter 19 Monday – Chapter 19 / Solving Tuesday – Lab #2 –Quiz#2 [Chapter 18, Labs] Thursday HMWK due 11pm