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Paperwork. Mastering Physics Course # DRKIDD880131 Assignment Due tonight New Assignment up soon… Lab Reports due next Tuesday 5pm. Schedule Short Term. Today – derive pressure/height, calculate Friday – Begin Chapter 19 Monday – Chapter 19 / Solving Tuesday – Lab #2
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Paperwork • Mastering Physics • Course # DRKIDD880131 • Assignment Due tonight • New Assignment up soon… • Lab Reports due next Tuesday 5pm
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
Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy Thin object, mass m Force = pA 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)
Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy Thin object, mass m Force = pA 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?
Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy mass = rV = rA(dy) Force = pA SF = 0 = pA - [pA + (dp)A] – mg 0 = pA – pA – (dp)A – rVg (dp)A = -rVg (dp)A = -r(Ady)g (dp/dy) = - rg Implications?
Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy mass = rV = rA(dy) Force = pA SF = 0 = pA - [pA + (dp)A] – mg 0 = pA – pA – (dp)A – rVg (dp)A = -rVg (dp)A = -r(Ady)g (dp/dy) = - rg For Ideal Gas r = m/V = pM/(RT) (dp/dy) = - rg
Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy mass = rV = rA(dy) Force = pA Pressure vs. Height Any Fluid (dp/dy) = - rg For Fluid that is an Ideal Gas r = m/V = pM/(RT) (dp/dy) = - pgM/(RT)
Pressure vs. Height • (dp/dy) = - pgM/(RT) • Now need to set up equation to solve • (dp/p) = -(gM/RT)(dy) • Assume a constant temperature (?)
Pressure vs. Height • (dp/dy) = - pgM/(RT) • Now need to set up equation to solve • (dp/p) = -(gM/RT)(dy) • Assume a constant temperature (?)
Pressure vs. Height Let’s say integration was from sea level (p0=p0, 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.
Molecular Motion • From Derivation in Section 18.3 • KEAVG=(3/2)NkBT (N is # molecules) • pV = (2/3) KEAVG • pV = NkBT • Where does 3 come from? • Other implications? • Why do people care about root mean square? [vrms] • Hint: zero?
Mean Free PathLifetime (Mean Free Time) • Very Important Concept • Vacuum conditions • Behavior of electrons in solids • Any interacting (or not) particles • Mean distance traveled before collision • lMFP = v tMFP • Lifetime (mean free time) is time between collisions • Why important for electrical conduction?
Lots of Averages… • Gas at temperature 500K • Average KE of One Mole • KEAVG = (3/2)nRT = (3/2)RT • 8.31 J / (mole K) • KEAVG = 6232 J • Say gas is N2 (MM = 0.028kg / mole) • KEAVG = (1/2)m(v2)AVG What is max speed?
Phase Diagram Describes P&T conditions for materials Interesting Points, What is this at STP?
Phase Diagram Supercritical Fluid? Neat – can dissolve things like a liquid & Diffuse through solids like a gas…
Discussion Q18.10Start Gas # molecules = n0 Temperature = T0 pressure = p0 Volume = V0
Discussion Q18.10“Sudden” Hole in wall Gas Initial State # molecules = n0 Temperature = T0 pressure = p0 Volume = V0 Gas Final State # molecules = ? Temperature = ? pressure = ? Volume = ? What Happens here?
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