CH 15 Zeroth and First Law of Thermodynamics

THERMODYNAMICS

Thermodynamics – Branch

of Physics that is built upon

the fundamental laws that

heat and work obey.

Central Heating

Objectives: After finishing this unit,

you should be able to:

State and apply the Zeroth and First of Thermodynamics.

Demonstrate your understanding of adiabatic, isochoric,

isothermal, and isobaric processes.

Use a PV Diagram to solve for the work done on or by a

system.

A THERMODYNAMIC SYSTEM

• A system is a collection of objects in which attention is being focused.

• A system is a closed environment in which heat transfer can take place. (For example, the gas, walls, and container. (Example: Cylinder of an automobile engine.)

Work done on the gas or work done by the gas.

Surroundings

The surroundings mean everything else in the

environment. The system and its surroundings are separated by

some kind of wall.

Diathermal walls –Walls that permit heat to flow

through them.

Adiabatic walls – Wall that do not permit heat to flow

between the system and its

surroundings.

Four Laws of Thermodynamics.

• Of the 4 Laws, the Second Law was found first, the

First Law was found second, then the Zeroth Law, then

the Third Law was found fourth.

The Zeroth Law of Thermodynamics

• This is called the Zeroth Law because the First Law was

already in effect and those in charge did not want to change

the first law so when this law was found or stated, they

numbered the new law the Zeroth Law.

• This Law deals with thermal equilibrium and establishes

temperature as the indicator of thermal equilibrium, in the

sense that there is no net flow of heat between two systems

in thermal contact that have the same temperature.

The Zeroth Law of Thermodynamics

• Zeroth Law states:

• Two systems individually in thermal equilibrium with

a third system are in thermal equilibrium with each

other.

The Zeroth Law of Thermodynamics

(a) Boxes A and B are surrounded

by walls that do not conduct heat,

so the heat stays in the box. Both

boxes have the same temperature.

(b) When A is put into thermal

contact with B through a heat

conducting wall, grey, no net

flow of heat occurs between the

boxes.

The First Law of Thermodynamics

First Law of Thermodynamics states:

The internal energy of a system changes from an initial

value, to a final value due to heat and work.

INTERNAL ENERGY OF SYSTEM

• The internal energy, U, of a system is the total of

all kinds of energy possessed by the particles that

make up the system.

Usually the internal energy consists of

the sum of the potential and kinetic

energies of the working gas

molecules.

Work Done by a Gas

Consider the formula for work: W = Fx

Because the distance is vertical, the x

becomes Δy.

The force, in this case, is caused by the

pressure of the gas inside, using P = F/A,

so F = PA.

Substituting you have: W =PA(Δy).

Where: Δy being the displacement of the

cylinder, similar to x.

(AΔy) is simply just a change in Volume.

So we could say that work is done by the

expansion in volume of a gas. W = PΔV

PV Graph or PV Diagram

Since the above formula is similar to A = bh.

The work is determined by the area UNDER a

Pressure vs Volume diagram or graph.

According to "Conservation of Energy", we learned that WORK

can be done by a change in kinetic energy (W =ΔKE) or potential

energy or both.

We also learned that HEAT ENERGY (Q) can be done when there

is a change in temperature.

When a substance is involved in doing some form of work AND

there is a temperature change in the process the INTERNAL

ENERGY (U) can change.

TWO WAYS TO INCREASE THE

INTERNAL ENERGY, U.

HEAT PUT INTO A SYSTEM

ΔQ is positive

+U

WORK DONE BY A GAS

ΔW is positive

WORK DONE ON THE SYSTEM

ΔW is negative

-ΔU

TWO WAYS TO DECREASE THE

INTERNAL ENERGY, U.

HEAT LEAVES A SYSTEM

ΔQ is negative

Qout

hot

Wout

hot

THERMODYNAMIC STATE

The STATE of a thermodynamic system is determined by four factors:

• Absolute Pressure, P, in Pascals • Temperature,T, in Kelvins • Volume,V, in cubic meters • Number of moles, n, of working

gas

THERMODYNAMIC PROCESS

Increase in Internal Energy, U.

Initial State:

P1 V1 T1 n1

Final State:

P2 V2 T2 n2

Heat input

Qin

Wout

Work by gas

The Reverse Process

Decrease in Internal Energy, U.

Initial State:

P1 V1 T1 n1

Final State:

P2 V2 T2 n2

Work on gas

Loss of heat

Qout

Win

SIGN CONVENTIONS

FOR FIRST LAW

• • Work BY a gas is positive

• Heat Q input is positive

• Work ON a gas is negative

• Heat Q out is negative

+Qin +Wby

U

-Won

-Qout

U

WQU

Why is the formula on your Formula Sheet:

WQU

Example Problem

1. Gas in a container is at a pressure of

1.5 atm and a volume of 4 m3. What is

the work done by the gas if, at constant

pressure,

a. its volume expands by adding twice the

initial volume, and

b. it is compressed by subtracting one

quarter of its initial volume?

Example Problem

2. An ideal gas is enclosed in a cylinder. There is

a movable piston on top of the cylinder. The

piston has a mass of 8000 g and an area of 5

cm2 and is free to slide up and down, keeping

the pressure of the gas constant. How much

work is done as the temperature of 0.2 moles

of the gas is raised from 20o C to 300o C?

Problem

3. A gas is compressed at a constant

pressure of 0.8 atm from a volume of 9

L to a volume of 2 L, and in the

process, 400 J of heat energy flow out

of the gas.

a. What is the work done by the gas?

b. What is the change in internal energy of

the gas?

Problem

4. A thermodynamic system undergoes a process in which its internal energy decreases by 500 J. If at the same time, 220 J of work is done on the system, find the heat transferred to or from the system.

•

Example

5. An ideal gas is enclosed in a cylinder. There

is a movable piston on top of the cylinder.

The piston has a mass of 8000 g and an area

of 5 cm2 and is free to slide up and down,

keeping the pressure of the gas constant. How

much work is done as the temperature of 0.2

moles of the gas is raised from 20o C to 300o

C?

FOUR THERMODYNAMIC PROCESSES:

• Isobaric Process: ΔP = 0

• Isochoric Process: ΔV = 0, ΔW = 0

•Adiabatic Process: ΔQ = 0

• Isothermal Process: ΔT = 0, ΔU = 0

ΔU = ΔQ - ΔW But ΔW = P ΔV

ISOBARIC PROCESS:

CONSTANT PRESSURE, P = 0

+U -U

QIN QOUT

HEAT IN = Wout + INCREASE IN INTERNAL ENERGY

Work Out Work In

HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY

ISOBARIC WORK

400 J

Work = Area under PV curve.

Work P V

B A P

V1 V2

VA VB

TA T B

=

PA = PB

ISOCHORIC PROCESS

The substance would expand if it could, but

the rigid container keeps the volume

constant, so the pressure increases. More and

more force is exerted on the walls of the

container.

No work is done, since the volume does not change. The

area under the straight, vertical line of a pressure-volume

graph is zero.

Heat in the system increases the internal energy of the

system.

U = Q - W so that Q = U

ISOCHORIC PROCESS:

CONSTANT VOLUME, V = 0, W = 0 0

+U -U

QIN QOUT

HEAT IN = INCREASE IN INTERNAL ENERGY

HEAT OUT = DECREASE IN INTERNAL ENERGY

No Work Done

ISOCHORIC EXAMPLE:

Heat input increases P with const. V

400 J heat input increases internal energy by 400 J and zero work is done.

B

A

P2

V1= V2

P1

PA P B

TA T B

=

400 J

No Change in volume:

U = Q - W ; W = -U or U = -W

ADIABATIC PROCESS:

Occurs without the transfer of heat, Q = 0

Work done at EXPENSE of internal energy

INPUT Work INCREASES internal energy

Work Out Work In

U +U

Q = 0

W = -U U = -W

ADIABATIC EXAMPLE:

Insulated Walls: Q = 0

B

A PA

V1 V2

PB

Expanding gas does

work with zero heat

loss. Work = -U

ADIABATIC EXPANSION:

400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat

exchange is ZERO. Q = 0

Q = 0

B

A PA

VA VB

PB

PAVA PBVB

TA T B

=

A A B BPV PV

ISOTHERMAL PROCESS:

Constant Temperature, T = 0, U = 0

NET HEAT INPUT = WORK OUTPUT

U = Q - W AND Q = W

U = 0 U = 0

QOUT

Work In

Work Out

QIN

WORK INPUT = NET HEAT OUT

ISOTHERMAL EXAMPLE (Constant T):

PAVA = PBVB

Slow compression at constant temperature: ----- No change in U.

U = T = 0

B

A PA

V2 V1

PB

ISOTHERMAL EXPANSION (Constant T):

400 J of energy is absorbed by gas as 400 J of work is done on gas.

T = U = 0

U = T = 0

B

A PA

VA VB

PB

PAVA = PBVB

TA = TB

Example

6. A gas is taken through the cyclic

process described in the figure.

a. Find the net heat transferred to the

system during one complete cycle.

b. If the cycle is reversed, that is the

process goes along ACBA, what is

the net heat transferred per cycle?

Example 7. One mole of a gas initially at

pressure of 2 atm and a

volume of 0.3 L has an

internal energy equal to 91 J.

In its final state, the pressure

is 1.5 atm, the volume is 0.8

L, and the internal energy

equals 182 J. For the three

paths IAF, IBF, and IF in the

Figure, calculate

a. the work done by the gas and

b. the net heat transferred in the

process.

Assignment

•Ch 15, Pages 466 – 471,

•#1, 5, 7, 11, 13, 23, 31, 91