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Paperwork Mastering Physics Course # DRKIDD880131 Assignments should be up Need to be de-enrolled from Physics I
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Schedule Short Term Today – Equations for #2? Monday – Off Tuesday – Lab #1 –Copyworks –Quiz#1 [Chapter 17] Wed HMWK due 11pm –Finish Chapter 18
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Equation of State Relationship between –p, pressure –V, volume –T, temperature –m or n (mass or # moles) Related by Molar Mass (MM)
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Equation of State Relationship between –p, pressure –V, volume –T, temperature –m or n (mass or # moles) Related by Molar Mass (MM)
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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 ?
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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
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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
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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)
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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?
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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.
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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
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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
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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
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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
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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)
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Pressure vs. Height (dp/dy) = - pgM/(RT) Now need to set up equation to solve (dp/p) = -(gM/RT)(dy) –Assume a constant temperature (?)
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Pressure vs. Height (dp/dy) = - pgM/(RT) Now need to set up equation to solve (dp/p) = -(gM/RT)(dy) –Assume a constant temperature (?)
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Pressure vs. Height
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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.
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Schedule Short Term Today – Equations for #2? Monday – Off Tuesday – Lab #1 –Copyworks –Quiz#1 [Chapter 17] Wed HMWK due 11pm –Finish Chapter 18
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