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