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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
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