Paperwork Mastering Physics Course # DRKIDD Assignments should be up Need to be de-enrolled from Physics I

Schedule Short Term Today – Equations for #2? Monday – Off Tuesday – Lab #1 –Copyworks –Quiz#1 [Chapter 17] Wed HMWK due 11pm –Finish Chapter 18

Equation of State Relationship between –p, pressure –V, volume –T, temperature –m or n (mass or # moles) Related by Molar Mass (MM)

Equation of State Relationship between –p, pressure –V, volume –T, temperature –m or n (mass or # moles) Related by Molar Mass (MM)

Equation of State for Solid Volume –Related to mass & density – V = m/  For a given volume V 0 : –Relate to changes in temperature & pressure V = V 0 [ 1+b(T-T 0 ) – k(p-p 0 ) ] –Examine this equation for a solid If T = T 0 & p = p0? What happens if T not T 0, p not p 0 ?

Equation of State for Gas pV=nRT Identify Equation components Units of pV? “Better” Version –pV = Nk B T k B T = Thermal Energy, more “Physicsy” Notice

Gas Density at given Parameters pV=nRT  = m/V m = nM (M is Molar Mass) Algebra to isolate m/V n = m/M pV = (m/M)RT pV/(RT) = m/M pM/(RT) = m/V =  Gas density equation. Examine

Gas Density at given Parameters pV=nRT  = m/V m = nM (M is Molar Mass) Algebra to isolate m/V n = m/M pV = (m/M)RT pV/(RT) = m/M pM/(RT) = m/V =  Gas density equation. Examine density is amount of mass per unit volume (dm/dV)

Isolated System pV=nRT Examine a closed system Mass cannot enter or escape –Balloon? Gas Tank? Examine at different parameters –p,V,T can change. R & n constant p 1 V 1 /T 1 = nR : case 1 p 2 V 2 /T 2 = nR : case 2 Example, what happens to a balloon that gets hot?

Isolated System pV=nRT p 1 V 1 /T 1 = nR : case 1 p 2 V 2 /T 2 = nR : case 2 Example, what happens to a balloon that gets hot? What is pressure felt by balloon? Warm balloon by some method. Does pressure change? What happens to balloon? –Approximation for weak rubber casing.

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

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

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

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

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)

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=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.

Schedule Short Term Today – Equations for #2? Monday – Off Tuesday – Lab #1 –Copyworks –Quiz#1 [Chapter 17] Wed HMWK due 11pm –Finish Chapter 18