Adiabatic Processes

constantPV

V

P

cc

57

air

Pressure/Temp and Vol/Temp

Adiabatic CompressionIf 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

Blow on your hand

Malachi 2:10Have 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?

The first law of thermodynamics∆Eint = Q +

WThe 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

For an ideal gas . . .

TncE V int

Always!!!!!

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 . . .

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?

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.

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

The path shown below is isothermal (ΔT= 0). The change in internal energy of the gas is

A. PositiveB. NegativeC. zero

P

V

The path shown below is isothermal (ΔT= 0). The heat flow is

A. Into the gasB. Out of the gasC. zero

P

V

The path shown below is adiabatic (Q = 0). The change in internal energy of the gas is

A. PositiveB. NegativeC. zero

P

V

The two lines below represent an isotherm and an adiabat. Which one is the isotherm?

A. The upper oneB. The lower one

P

V

The two lines below represent an isotherm and an adiabat. Which one is the isotherm?

A. The upper oneB. The lower one

P

V