Download First Law of Thermodynamics 9.1 Heat and Work

Heat and Work
First Law of Thermodynamics
9.1
Heat
Heat and Work
Thermodynamic Processes
Work
Thermodynamics is the science of heat and work
Heat is a form of energy
Calorimetry
Mechanical equivalent of heat
Mechanical work done on a system produces a rise
in temperature like heat added to the system.
Work
Heat input into a system can
produce work
1calorie =4.184 J
First Law of Thermodynamics
The change in the internal energy of a system U, is equal to the heat
input Q minus the work done by the system.
∆U = Q − W
∆ W = F∆ x
∆W = PA∆x = P∆V
For the case of
thermal expansion
work is done by
the system.
Heat input
Q
Heat
Q
Internal
Energy
U
Work
W
System
The internal energy is the energy stored in the system.
For an ideal gas the internal energy is the kinetic energy of the gas
1
Thermodynamic processes in an
ideal gas
PV diagram
T1 T2
U2
U1
State 1 (P1, V1, T1)
State 2 (P2, V2, T2 )
• The state of the ideal gas is determined by the three parameters,
PVT.
• A thermodynamic process is the transition between states with input
or output of heat and work with changes in internal energy.
• The internal energy U is a property of the state. ∆U determined by
the initial and final state and is independent of path
• The heat absorbed and work done in the process depend on the
path.
P
• State 2
•
State 1
V
Each point in the diagram represents a State of the system
The temperature T is not plotted but is not an independent variable.
T can be calculated from P and V using the ideal gas law.
The points of constant temperature on the PV diagram are hyperbola
P=
Thermodynamic processes
x
Path A
P
For constant T
Work can be calculated if the
process is reversible.
x
State 2 (T2, P2, V2)
Path B
nRT
V
Path A
State 2 (T2, P2, V2)
P
x
State 1 (T1, P1, V1)
x
State 1 (T1, P1, V1)
V
The change in U going from 1 to 2 is the same for the
two paths. ∆UA =∆UB
The change in Q and W going from 1 to 2 is different
between the two paths. QA ≠ QB , WA ≠ WB
Reversible process
V
We can calculate the work done in going from 1 to 2 if
the change is a reversible process. i.e. it goes slowly
through well defined states that are in quasi-equilibrium.
Work cannot be calculated for
an irreversible process.
x
Path A
T is increased
slowly from 0 to 100o C
P
State 2 (T2, P2, V2)
?
x
State 1 (T1, P1, V1)
The temperature is changed very slowly so that
the gas sample is at a uniform temperature.
V
We cannot calculate the work done in going from 1 to 2 if
the process is irreversible. i.e. goes rapidly through states
that are not well characterized and not in quasi-equilibrium.
However, we can determine ∆U, since this is a state property.
2
Irreversible process
Work done in a process
dW = PdV
W=
∫ Pdv
V1
The gas is not in a well defined thermodynamic
state as it is heated.
The work is the area under the
PV curve.
Some thermodynamic processes
expansion of ideal gas
Question
Find the work done in going from A to B along the
line.
V2
T constant
1.
Constant Temperature processConstant T
Isothermal
2.
Constant Pressure process –
Isobaric. e.g heating a balloon
heat
bath
P constant
2P1
B
P1
3.
A
Constant Volume process. Heating
a gas thermometer
T changes at
constant V
Adiabatic process – no heat
exchange. Rapid compression or
T changes
expansion of a gas
with V
4.
2V1
V1
T2
T is constant
∆U = 0
Increasing T
Constant T
Constant P
Constant V
Adiabatic
(Q=0)
P
V constant
Q=0
No contact
with heat bath
Isothermal volume change
Thermodynamic processes
T1
T changes
at constant P
U is constant
All the heat goes
into work
W=Q
•
Work
W=
P=
V2
V2
V1
V1
∫ PdV = ∫
nRT
dV
V
Q =W
•
nRT
V
W = nRTln V2 − nRTln V1
V
W = nRTln
V2
V1
Expansion
V2
>1
V1
Compression
work is positive
work is done by the gas
V2
< 1 work is negative
V1
work is done on the gas
3
Constant Volume
At constant volume
W=0
Constant Pressure
P2 , T2
∆U = Q
P
no work is done all the heat goes
into internal energy
P1 , T1
We can use this to find ∆U
V
Q = nc V ∆T
∆U = nc V ∆T
Q
Constant Pressure
Work
W=
V2
∫ PdV
= P(V2 − V1) = P∆V
V1
Heat input
Q = ∆U + P∆V
This is true for all processes since U is a State
function and only dependent on T.
specific heat at constant volume
Q = nc V ∆T + P∆V = nc V ∆T + nR∆T
Q = nc P ∆T
Specific heat at constant pressure, cP
∆U
the specific heat of the gas cV depends
on the structure of the gas molecule.
n∆T
Units J / mole ⋅ K
cV =
cP = c V + R
Adiabatic Process
the specific heat at constant P is larger
than the specific heat at constant V.
because work is done by the gas
Adiabatic compression
Heat input Q=0
The air (γ=1.40) in an automobile engine is compressed quickly so that
appreciable heat exchange does not occur. For a compression ratio
V1/V2 =10. Find the temperature of the gas compressed from an initial
temperature of 20o C.
∆U= -W Work is done at the
expense of internal energy
Q=0
Define “Adiabatic exponent” γ
c
c +R
R
= 1+
γ = P = V
cV
cV
cV
γdepends on the nature of the gas,
γ =1.40 for air.
PV relationship
TV relationship
γ
γ
PV
1 1 = P2V2
γ −1
TV
1 1
= T2V2
PV γ = cons tan t
γ −1
TV γ −1 = cons tan t
1 1 − P2V2
W = PV
γ −1
Adiabatic compression
Santa Anna Conditions
T= 20o C
The air (γ=1.40) in an automobile engine is compressed quickly so that
appreciable heat exchange does not occur. For a compression ratio
V1/V2 =10. Find the work done in compressing a gas from 1atm if the
capacity is 2.8 l.
P= 85 kPa
P = 100 kPa
T=?
High Desert.
Coast
Santa Ana winds originate in the high desert region of California
and is heated by adiabatic compress during the rapid descent to the
coast. For the above conditions what would the temperature at the coast
be?
4
Thermodynamic processes
Work done in expansion
One mole of air (γ=1.40) at P1 = 0.25 atm, T1 = 300 K,
V1=0.1m3 expands by a factor of 2. Find the work done if
the expansion is a) isobaric,b) isothermal c) adiabatic.
Which process does the most work?
T1
Defining
Characteristic
T constant
∆U=0
V constant
W=0
P constant
No Heat input
Q=0
First Law
Q=W
∆U=Q
∆U=Q-W
∆U=-W
W=0
P(V2-V1)
PV
1 1 − P2V2
γ −1
PVγ =constant
TVγ-1=constant
Work done by nRTln(V2/V1)
the gas
Other
relations
PV=constant Q=mCV∆T
Q=ncP∆T
cP=cP +R
Temperature change in expansion
V1=0.1m3
One mole of air (γ=1.40) at P1 = 0.25 atm, T1 = 300 K,
expands by a factor of 2. Find the temperature change if the
expansion is a) isobaric, b) isothermal c) adiabatic. Which process
produces the highest temperature rise?
P
T2
V1
V2
V
Heat input on expansion
One mole of air (γ=1.40) at P1 = 0.25 atm, T1 = 300 K,
V1=0.1m3 expands by a factor of 2. Find the heat input if
the expansion is a) isobaric, b) isothermal c) adiabatic.
What process results in the largest heat input? cV (air) =5/2R
T1
T1
P
P
T2
V1
V2
V
T2
V1
V2
V
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