Paperwork
• Mastering Physics
• Course # DRKIDD880131
• 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 V0:– Relate to changes in temperature & pressure
• V = V0 [ 1+b(T-T0) – k(p-p0) ]– Examine this equation for a solid
• If T = T0 & p = p0?
• What happens if T not T0, p not p0?
Equation of State for Gas
• pV=nRT• Identify Equation components
• Units of pV?
• “Better” Version– pV = NkBT
• kBT = 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 SystempV=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
• p1V1/T1 = nR : case 1
• p2V2/T2 = nR : case 2
• Example, what happens to a balloon that gets hot?
Isolated SystempV=nRT
• p1V1/T1 = nR : case 1
• p2V2/T2 = 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 fluidPressure 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 fluidPressure 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 objectLet’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] – mg0 = 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] – mg0 = 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. HeightAny 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 (?)
0 0
1( )
pF yF
p y
Mgdp dy
p 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 (?)
0 0
1( )
pF yF
p y
Mgdp dy
p RT
0 0
1( )
pF yF
p y
Mgdp dy
p RT
Pressure vs. Height
0 0
1( )
pF yF
p y
Mgdp dy
p RT
ln ln( 0) ( ) 0
ln ( ) 00
MgpF p yF y
RT
pF MgyF y
p RT
0 0
1( )
pF yF
p y
Mgdp dy
p RT
Pressure vs. Height
Let’s say integration was from sea level (p0=p0, y0 = 0)To a point pF = p, yF = yNeed to have known endpointsThen can derive equation for air pressure as a function of height above sea level
0
( )
0
( )
0
ln ln( 0) ( ) 0
ln ( ) 00
ln ( )
Mgy
RT
Mgy
RT
MgpF p yF y
RT
pF MgyF y
p RT
p Mgy
p RT
pe
p
p p e
Happy Equation: Should Check AccuracyImplications? 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