adiabatic processes. pressure/temp and vol/temp adiabatic compression if i compress air at...
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Adiabatic Processes
constantPV
V
P
c
c
5
7air
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Pressure/Temp and Vol/Temp
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Adiabatic Compression
If I compress air at atmospheric pressure and room temperature by a factor of 10 the temperature will go up by1. Less than 10 degrees C2. Between 10 and 50 degrees C3. Between 50 and 100 degrees C4. More than 100 degrees C
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Blow on your hand
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Malachi 2:10
Have we not all one father? hath not one God created us? why do we deal treacherously every man against his brother, by profaning the covenant of our fathers?
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The first law of thermodynamics
∆Eint = Q + W
The internal energy of an ideal gas depends only on the temperature of the gas.
Change of internal energy = heat put into system + work done on system
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For an ideal gas . . .
TncE V int
Always!!!!!
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Deriving the Adiabatic Equation
i.e. Going way beyond what you need to know for the homework and exams because you will hopefully learn something and, with luck, gain a greater appreciation of the power of differential calculus . . .
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𝑃𝑉=𝑛𝑅𝑇 Three things changing, but in a defined way suchthat if I know how one changes, I should know others.
To get rid of an unknown, I need another equation – hereit is! But I need to write it in terms of P, V, and T. And what do I do with the integral in it?
∆𝐸 𝑖𝑛𝑡=𝑄+𝑊
If I have a piston whose location is x, or a balloonwith a radius x, or a basketball being squished into the floor by an amount x, I shouldbe able to tell you any one just in terms of initialconditions and x. How do things change with x?
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They each only have one “T” thing (P and V show up twice, as P and dP, V and dV, in the left equation), so that’s the easy one to solve for and eliminate.
𝑃𝑑𝑉 +𝑉𝑑𝑃=𝑛𝑅𝑑𝑇𝑛𝑐𝑉 𝑑𝑇=− 𝑃𝑑𝑉
Take all the constants to one side and simplify. Then to keep things tidy, call it gamma.
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Get P stuff on one side, V on other, integrate.Remember, when V is equal to its initial value, P is equal to its initial value. When V is its final value, P is its final value.
−𝑃 𝛾 𝑑𝑉=𝑉𝑑𝑃
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The path shown below is isothermal (ΔT= 0). The change in internal energy of the gas is
A. PositiveB. NegativeC. zero
P
V
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The path shown below is isothermal (ΔT= 0). The heat flow is
A. Into the gasB. Out of the gasC. zero
P
V
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The path shown below is adiabatic (Q = 0). The change in internal energy of the gas is
A. PositiveB. NegativeC. zero
P
V
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The two lines below represent an isotherm and an adiabat. Which one is the isotherm?
A. The upper oneB. The lower one
P
V
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The two lines below represent an isotherm and an adiabat. Which one is the isotherm?
A. The upper oneB. The lower one
P
V