Paperwork
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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/r
• 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
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pV=nRT
r = 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 = r
Gas density equation. Examine
Gas Density at given Parameters
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pV=nRT
r = 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 = r
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
• p1V1/T1 = nR : case 1
• p2V2/T2 = nR : case 2
• Example, what happens to a balloon that gets
hot?
Isolated System
pV=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
Force = pA + (dp)A
Thin object, mass m
dy
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
Thin object, mass m
dy
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
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?
dy
Pressure vs. Height
• Example 18.4
Force = pA + (dp)A
mass = rV = rA(dy)
dy
Force = pA
SF = 0 = pA - [pA + (dp)A] – mg
0 = pA – pA – (dp)A – rVg
(dp)A = -rVg
(dp)A = -r(Ady)g
For Ideal Gas
(dp/dy) = - rg
r = m/V = pM/(RT)
(dp/dy) = - rg
Pressure vs. Height
• Example 18.4
Force = pA + (dp)A
mass = rV = rA(dy)
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 (?)
1
p 0  p  dp 
pF
yF
Mg
y 0 ( RT )dy
Pressure vs. Height
• (dp/dy) = - pgM/(RT)
• Now need to set up equation to solve
• (dp/p) = -(gM/RT)(dy)
– Assume a constant temperature (?)
1
p 0  p  dp 
pF
yF
Mg
y 0 ( RT )dy
1
Mg
p 0  p  dp  ( RT ) y0 dy
pF
yF
Pressure vs. Height
1
p 0  p  dp 
pF
yF

y0
(
Mg
)dy
RT
1
Mg
p 0  p  dp  ( RT ) y0 dy
pF
yF
Mg
ln  pF   ln( p0)  (
)  yF  y 0 
RT
 pF 
Mg
ln 
)  yF  y 0 
  (
RT
 p0 
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
ln  pF   ln( p0)  (

ln 


ln 

Mg
)  yF  y 0 
RT
pF 
Mg


(
)  yF  y 0 

p0 
RT
p
Mg
) y
  (
p0 
RT
Mg
(
) y 
p
RT
e
p0
p  p0 e
(
Mg
) y 
RT
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