Transcript 2. Thermodynamic Processes and Quantities Defined

Thermodynamic Quantities Defined
Internal Energy = U = the sum of all the energy held by the
molecules:
* the PE stored in their chemical bonds, attractions and repulsions
* the KE of their motion; translation, rotation, vibration
Heat = Q = is what flows between bodies when one loses
internal energy, and another gains it.
U1
@ T1
Q
U2
@ T2
Gases and Work
Work can be done on a
gas.
Internal energy increases
U = +
U = Won
But the side effect of the man
doing work is that the gas might
get hotter and some of the heat
leak out of the walls.
U = Won - Q lost
We can prevent this from happening by insulating
the piston and chamber so that no heat can go in or
out (adiabatic means Q = 0, no heat flows)
Gases and Work 2
Work can be done by a gas.
Internal energy is
used up (lost,
decreases) as piston
pushes up.
U = neg
U = - Wby
But the side effect is that the gas
might cool and some of the heat be
sucked into the walls.
U = - Wby + Q into
We can prevent this from happening by insulating
the piston and chamber so that no heat can go in or
out (adiabatic means Q = 0, no heat flows)
The First Law of Thermodynamics
The total change in the internal energy of a system is the sum of the work
done on the system and the heat transferred to the system.
U = Won + Q into
If Q = 0 (adiabatic) , then ΔU is positive if positive work is done on the
system.
If W = 0, (isochoric: box cant expand or contract), then ΔU is positive if
heat flows into the system
Expansions and
contractions of gases are
shown on PV diagrams
Pressure
of gas
on walls
The area under a PV
graphs has special
significance.
Volume of gas
increases as
pressure on
walls
decreases
Typical of an
expansion
Volume of gas
decreases as
pressure on
walls increases
Typical of an
compression
of gas
The area under a PV
graphs has special
significance.
x
Wgas = F x = PA x
But A x = volume compressed
Wgas = P V = area under a PV graph
Wgas = P V = area under a PV graph
Try calculating the work done by the gas in
this isobaric expansion
P = 1.01 x 105 Pa , Vi = .7 m3, Vf = 1.3 m3
W = P V = (1.01 x 105 N/m2)( .6 m3)= 60600 J
What if the arrow were switched and it was
an isobaric compression?
W = P V = (1.01 x 105 N/m2)(- .6 m3)= - 60600 J
Net work = 0
+W
-W
Name the process A to B
Isobaric expansion
How much work is done?
W= PV=Po (3Vo) = 3PoVo
Name the process B to C
Isochoric loss of pressure
How much work is done?
W= PV=Po 0 = 0
Name the process C to A
Contraction
How much work is done? W= PV=can’t be done because P is changing
W = estimate of area under curve = 4.5 boxes
4.5 boxes (1 box = ½ PoVo) = -2.25 PoV0
Net Work done in cycle = 3PoVo +0 + -2.25 PoVo= + .75 PoVo
Net Work done in cycle = 3PoVo +0 + -2.25 PoVo= + .75 PoVo
Do you see a shortcut?
Get the area of the enclosed
triangle
W= ½ bh = ½ 3Vo (½ Po) = ¾ PoVo
So for any closed cycle, the net work done
is the area enclosed .
For an open cycle
(where you don’t return to the P, V, T you started at)
the work done is the sum of the areas under the
curve
On each GRAPH below, draw lines and curves to indicate 3
processes: isobaric, isochoric and isothermal.
V
P
V
P
T
T
Closed Cycles: WHEN A SYSTEM RETURNS TO
ITS SAME P
P
and V
P
P
V
V
V
Since PV = nRT, that means it returns to the same temperature as
well. U = 0, and the first law reduces to ___________.
This means W = - Q and all work done is lost as
________. The work done is also the area of the enclosed
cycle.
Tell how the first law changes for
process that are
Adiabatic
U = W + Q
Isothermal
U = W + Q
Isobaric
U = W + Q
Isochoric
U = W + Q
Another powerful tool: if you know PV of a gas, you can
automatically calculate its temperature.
PV = nRT
# of moles
gas constant
Find temperatures at points 1 and 2:
Assume the data below applies to a 20
mole sample of ideal gas
P1V1 = nRT1
So T1 = P1V1/nR=
5e5 (1)/20(8.3)
= 7600°K
P2V2 = nRT2
So T2 = P2V2/nR=
2e5 (.5)/20(8.3)
= 60°K
Interpretation: the gas _____________
and ____________
Isothermal = at constant temperature
PV = nRT = a constant
What’s the shape of the curve xy = 1 ?
This shape is called an isotherm .
You must actually know T at all
points or calculate it to be sure;
can’t tell by shape alone.
Occurs in an ice bath (thermal resevoir)
that can exchange heat with walls if
temperature starts to change
A.k.a
isochoric
When a gas expands adiabatically, the work done in the expansion comes
at the expense of the internal energy of the gas causing the temperature
of the gas to drop. The figure below shows P-V diagrams for these two
processes.
U = Won + Q into
Which process
resulted in a
higher
temperature?
Thus the adiabat lies below the isotherm.
In the end, the internal energy of a gas depends only on its
temperature, assuming PVT changes only.
Chemical or phase changes could change PE of molecules, but we
don’t deal with that in this course.
U =  KEint +  PE int = (3/2) nkT
Example: One mole of
monatomic ideal gas is
enclosed under a
frictionless piston. A series
of processes occur, and
eventually the state of the
gas returns to its initial
state with a P-V diagram
as shown below. Answer
the following in terms of
P0, V0, and R.
A. Find the temperature at each vertex.
B. Find the change in internal energy for each
process.
C. Find the work by the gas done for each process.
Example: One mole of ideal gas is at pressure P0 and volume V0. The gas then undergoes three
processes.
1.The gas expands isothermally to 2V0 while heat Q flows into the gas.
2.The gas is compressed at constant pressure back to the original volume.
3.The pressure is increased while holding the volume constant until the gas returns to its
initial state.
A. Draw a P-V diagram that depicts this cycle. Label relevant points on the axes. In terms of
T0, the initial temperature, label each vertex with the temperature of the gas at that point.
For the remaining sections, answer in terms of T0, Q, and R.
B. Find the change in internal energy for each leg of the cycle.
C. Find the work done by the gas on each leg of the cycle.
D. Find the heat that flows into the gas on legs 2 and 3.
E. Find the efficiency of this cycle.
Example: One mole of ideal gas is at pressure P0 and volume V0. The gas then undergoes three
processes.
1.The gas expands isothermally to 2V0 while heat Q flows into the gas.
2.The gas is compressed at constant pressure back to the original volume.
3.The pressure is increased while holding the volume constant until the gas returns to its
initial state.
A. Draw a P-V diagram that depicts this cycle. Label relevant points on the axes. In terms of
T0, the initial temperature, label each vertex with the temperature of the gas at that point.
For the remaining sections, answer in terms of T0, Q, and R.
B. Find the change in internal energy for each leg of the cycle.
C. Find the work done by the gas on each leg of the cycle.
D. Find the heat that flows into the gas on legs 2 and 3.
E. Find the efficiency of this cycle.
For solution go to
http://apcentral.collegeboard.com/members/article/1,3046,151-165-0-44428,00.html
Example: Calculate the internal energy of the air in a typical
room with volume 40 m3. Treat the air as if it were a monatomic
ideal gas at 1 atm = 1.01 105 Pa.
You can use the gas law PV=nRT to express the internal energy in
terms of pressure and volume.