Lesson 6

First Law of Thermodynamics

Liceo Alfano I

Reminder• Last time: Specific heat, internal energy

• Q = ΔU = mcΔT

• This time: First law of thermodynamics

Energy Transfer• What kind of energy transfer have we been talking

about?

• Heat

• The other kind of energy transfer is work. What is work?

• Work = Force x Distance

• Ex: pushing a shopping cart forward, gravity pulling a ball down, etc.

• Remember: WORK IS ENERGY!

First Law• The First Law of Thermodynamics is:

Energy cannot be created or destroyed, only changed in form

• You’ve probably heard this before as the principle of energy conservation, but the first law adds a new part: heat

First Law

• Change in internal energy is defined as the heat transferred into the system minus the net work done by the system

• If Q is positive, heat is transferred into the system. If W is positive, there is net work done by the system

• A positive Q adds energy to the system, a positive W takes energy out

ΔU = Q - WChange in internal energy

Heat added TO the system

Work done BY the system

Concept QuestionIf I do work to a system, then W is

1)Positive

2)Negative

Concept QuestionIf I do work to a system, then W is

1)Positive

2)Negative

Concept QuestionIf I add heat to a system, then Q is

1)Positive

2)Negative

Concept QuestionIf I add heat to a system, then Q is

1)Positive

2)Negative

Concept QuestionIf a system releases heat energy, then Q is

1)Positive

2)Negative

Concept QuestionIf a system releases heat energy, then Q is

1)Positive

2)Negative

Concept QuestionIf I add 300kJ of heat to a system, then do 400kJ of work on it, then ΔU is

1)Positive

2)Negative

Concept QuestionIf I add 300kJ of heat to a system, then do 400kJ of work on it, then ΔU is

1)Positive

2)Negative

Concept QuestionIf I do 100kJ of work on a system, then the system releases 300kJ of heat, then ΔU is

1)Positive

2)Negative

Concept QuestionIf I do 100kJ of work on a system, then the system releases 300kJ of heat, then ΔU is

1)Positive

2)Negative

Human Body• Body temperature is kept constant by

transferring heat to our surroundings so Q is negative

• We generally do work on our surroundings, so W is positive

• This means ΔU is negative, so we are constantly losing energy to our surroundings

Human Body• However, when we eat, we add chemical

potential energy to our bodies

• If we eat perfectly, ΔU is zero—all that we consume is used to run our bodies

• If we eat a lot, ΔU is positive, so our bodies store the extra energy as fat

• If ΔU is negative, the body uses the fat to release heat and do work—that is how we lose weight

VideoHeat and work

Practice Problem• A system receives 1600J of heat. At the

same time 800J of work are done on the system by outside forces. Calculate the change in internal energy of the system.

Ideal Gas Work• Let’s think back to

ideal gases

• Think about the container to the right as a system.

• If I push the piston in, is W positive or negative?

Ideal Gas Work• Let’s think about how much

work I did. Remember, |W|=F*d

• What is the force?

• F=P*A

• What is the work?

• W=P*A*Δx=PΔV

x

PV Diagrams• When we talk about the first law with ideal gases,

we often use a PV diagram. How can we find work done using a PV diagram?

• A PV diagram can show some special processes:

• Isothermal

• Adiabatic

• Isobaric

• Isochoric

Isothermal• IsoTHERMal = constant

temperature

• What is our gas law when we have constant temperature?

• PV = constant

• As volume decreases, pressure increases, as volume increases, pressure decreases

Adiabatic• Adiabatic = no heat

in/out, so Q=0

• This happens with good insulation or when a process happens so quickly that heat cannot flow in or out

• If Q=0, what is ΔU?

• ΔU=-W

Isobaric

• IsoBARic = constant pressure (bar is a measure of pressure)

• Straight line on PV diagram

• Work can be calculated as W=PΔV (as we saw before)

Isochoric• Isochoric = constant

volume (chorus takes up a lot of room?)

• Vertical line on a PV diagram

• No work done because volume doesn’t change, so W=0

Cycles• Here we have a PV

diagram of a cycle

• A cycle is simply a process that ends where it starts

• Since the cycle ends where it begins, ΔU=0

• How can we find the work done on this graph?

Videos• Otto cycle

• 2-stroke engine

• Steam engine

Practice Problems• Sketch a PV diagram of this cycle: 2L of an

ideal gas at 105 Pa are cooled at constant pressure to a volume of 1L, and then expanded isothermally back to 2L, then the pressure is increased at constant volume until 105 Pa.

• 1L of air at 105 Pa is heated at constant pressure until its volume is 2L, then it is compressed isothermally back to 1L, then the pressure is decreased at constant volume back to 105 Pa. Sketch the PV diagram.

Practice Problems

The internal energy at point A is 800J. The work transfer from B to C is 300J, and from C to A is 100J. What is the total work done by the gas in this cycle? What is the final internal energy?