Conservation of energy in context of work, heat, and internal energy Thermodynamics Conservation of energy in context of work, heat, and internal energy

1st Law Energy is not created or destroyed. 𝑈 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 +𝐾𝐸+𝑃𝐸+…= 𝑈 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 +𝐾𝐸+𝑃𝐸+… In thermodynamics, we simplify the system and consider the relationship between work, heat, and changes to a system’s internal energy.

What is work? When physicists use the term ‘work’, we are referring to something very specific: Qualitatively: the mechanical transfer of energy to or from a system by pushing or pulling on it e.g., throwing a ball (work goes to kinetic energy) e.g., striking a match (work goes to friction, i.e., thermal) e.g., pulling a rubber band (work goes to elastic potential) Quantitatively: work = force x distance over which that force is applied 𝑊=𝐹𝑑

What is heat? When physicists use the term ‘heat’, we are referring to something very specific. Qualitatively, the transfer of thermal energy into or out of a system. Quantitatively, Q + adding thermal energy to the system - taking thermal energy from the system

What is internal energy? When physicists talk about internal energy, we are referring to something very specific: Qualitatively, the random kinetic energy of the particles that make up a substance Related to temperature Quantitatively, 𝑈 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 = 3 2 𝑛𝑅∆𝑇

1st Law of Thermodynamics A simplified version of the law of conservation of energy: A change in internal energy of a system is equal to the amount of heat added to the system from the environment plus the amount of work done on the system by the environment. ∆𝑈=𝑄+𝑊

Conceptually You can change a system’s temperature by heating it Boil water by putting it in a pot on the stove Q>0, W=0, so U>0 You can make a system do work by heating it. Shot the top off of a test tube pop-gun U>0, Q>>0, so W<0 steam pushes a piston You can change a system’s temperature by doing work to it you can boil water in a blender W>0, Q=0, so U>0 you can cool the inside of a refrigerator by expanding a gas in tubes next to the objects you want to cool W<0, Q=0, so U<0

Working with thermodynamics When gases expand, they can do work by pushing against a piston. e.g., in a car, when the spark plug fires in a cylinder in the engine, it ignites the mixture of fuel and air. The hot gas expands, pushing the piston out, and through mechanical linkages, turns the wheels of the car. Energy is transferred out of the gas as work.

Quantifying work done by a gas Imagine a gas cylinder sealed at one end by a moveable piston. Suppose the gas forces the piston out by a distance d. 𝑊=𝐹𝑑 Where the force of the gas on the piston is equal to the gas’s pressure multiplied by the area of the piston: 𝐹=𝑃𝐴 So, 𝑊=𝑃𝐴𝑑 Recall, the volume of a cylinder equals its area times height or, in this case, the distance the piston was pushed. 𝑉=𝐴𝑑 So, 𝑊=𝑃∆𝑉

Using pressure-volume diagram to track energy Assume n = 1 mol of ideal gas If you know pressure and volume, you can infer temperature

Ideal gas process: isovolumetric pressure If volume stays the same when pressure increases, what can you infer about the temperature? V=0, so W = 0, so U=Q Pf Pi

Ideal gas process: isobaric If pressure stays the same and volume increases, what can you infer about the temperature? U=Q+W So, if Q = 0, U=-PV volume pressure Vf Vi

Ideal gas process: isothermal If temperature is constant, then PV is constant. You have been wondering when you’d ever need to use a hyperbola. You’re welcome. T=0, so U=0, so Q=-W

Ideal gas process: adiabatic System does work as A goes to C without additional heat. Q=0, so U=-W

Quantifying work using PV diagram The area under the curve is amount of work done by the gas. 𝑊=𝑃( 𝑉 𝐵 −𝑉𝐴)

Internal combustion engine

“Suck”

“Push”

“Bang!”

“Bang” (part 2)

“Blow”