Today’s Lecture
Thermodynamic Processes
• Isothermal – Constant Temperature
• Isochoric – Constant Volume
• Isobaric – Constant Pressure
• Adiabatic – No Heat Transfer
First Law of Thermodynamics
ΔU = Q − W
change in the internal
energy of the system
net heat
transferred
to the system
work done
by the system
The change in the internal energy of a system depends only on the
net heat transferred to the system and the net work done by the
system, and is independent of the particular processes involved.
The equation is deceptively simple… One of the forms of the general
law of conservation of energy. BUT! Be careful about the sign
conventions. Positive Q is heat transferred to the system. Positive W
is work done by the system.
Work Done by a Gas
Work done by a gas:
Work done and heat
transferred – our major
concerns!
W = FΔx = PAΔx = PΔV
P – pressure of the gas;
Δx – displacement of the piston,
A – area of the piston
Work is positive when the gas expands!
Differential form:
dW = Fdx = PAdx = PdV
Integral form – valid for varying pressure:
W = ∫ dW = ∫ PdV
Work Done by a Gas
Work done and heat
transferred – our major
concerns!
W = ∫ dW = ∫ PdV
P – (varying) pressure of the gas;
dV – differential volume change
ΔW = P ΔV
Work equals area under curve on a P-V diagram
Multiple Ways to Get From
Initial to a Final State
Is work going to be the same
for different processes?
NO!
Is heat going to be the same?
NO!
Is internal energy going to be
the same?
YES!
Internal energy is a function of state
and will be the same independent of
path between the states.
Isothermal Process – T = const
Isotherm
nRT const
P=
=
V
V
The work integral is straightforward.
ΔU = Q − W
Since the internal energy only
depends on temperature:
ΔU = 0
⇒ Q =W
Example: Isothermal Process - T = const
A scuba diver at a depth of 25m, P = 3.5atm, exhales bubbles 8.0mm
in radius. How much work is done by each bubble as it expands while
it rises to the surface?
Solution: From the ideal gas law we know for an isothermal
process that the pressure and volume are inversely related.
Example: Isothermal Process - T = const
An ideal gas expands to 10 times its original volume, maintaining
a temperature of 440K. The gas does 3.3kJ of work.
(a) How much heat does the gas absorb?
(b) How many moles of gas are there?
Isochoric Process – ΔV = 0
ΔU = Q − W
ΔW = PΔV ≡ 0
ΔU = Q = nCv ΔT
The gas does no work for
an isochoric process.
The change in internal
energy is the heat transfer.
n – number of moles of the gas;
Cv – molar specific heat at constant volume – heat
capacity of one mole of the gas in an constant volume
process. (Compare with Q=mcΔT)
Why bother introducing a new parameter? For a gas, per mole is
more convenient.
1 ΔU
Cv =
n ΔT
Measuring Cv we learn about internal energy of
the gas as a function of temperature!
Isochoric Process – ΔV = 0
ΔU = Q − W
ΔW = PΔV ≡ 0
ΔU = Q = nCv ΔT
From the kinetic theory of gasses, we know that the (translational)
kinetic energy/molecule is:
If translational kinetic energy is the only form of internal energy then
How much heat is required to raise the temperature of 500g of He by 10oC?
Isobaric Processes – P = const.
PV diagram for an isobaric process
as the gas expands along the isobar
from state 1 to state 2.
Since the pressure of the gas remains
constant, calculation of the work it
does is particularly simple.
V2
V2
V1
V1
W = ∫ PdV = P ∫ dV = P (V2 − V1 ) = PΔV
What about internal energy and heat? Is T2< or >T1?
Isobaric Processes – P = const. W = PΔV
What about internal energy and heat?
ΔU = Q − W
and
PV = nRT
From the 1st law and equation for Cv:
Q = ΔU + W = nCv ΔT + PΔV
Ideal gas at constant pressure:
PΔV = nRΔT
hence
Q = ΔU + W = nCv ΔT + nRΔT = n(Cv + R ) ΔT
Isobaric Processes – P = const.
Molar specific heat at constant pressure
(definition):
Hence Cp is:
Why is specific heat at constant
pressure higher than at constant
volume?
C p = Cv + R
What is the specific heat at
constant pressure for He gas?
Why was there no difference in specific heats for solids?
Review of our Thermodynamic Processes
ΔU = Q − W
PV = nRT
W = ∫ PdV
Adiabatic Process – Q = 0
There is no heat transfer for an adiabatic process
Implies thermal insulation or process happens quickly.
Q=0
PV = nRT ΔU = −W
W = ∫ PdV
Adiabatic Process – Q = 0
Q=0
ΔU = −W
Positive work, W, is done by the expanding gas at expense of
reduction of its internal energy.
Since there is no heat supplied from the outside to replenish the
gas energy the temperature declines.
Ti
Tf
Vi
Vf
Work can be expressed in terms of -ΔT or -Δ(PV)
Adiabatic Process – Q = 0
Q=0
ΔU = −W
We now know the work done by an ideal gas during an adiabatic
expansion, but how do pressure and temperature behave?
From the ideal gas law:
From the 1st law:
As an exercise for the student/professor
Ti
Tf
Vi
Vf
Adiabatic Process – Q = 0
Q=0
ΔU = −W
A gas with γ = 1.4 at 105Pa occupies 5L.
It is compressed adiabatically to 2.5L.
Tf
(a) What is its final pressure?
Ti
Vf
Vi
(b) What is ratio of the final and initial temperatures?
Adiabatic Process – Q = 0
Q=0
ΔU = −W
A gas with γ = 1.4 at 105Pa occupies 5L.
It is compressed adiabatically to 2.5L.
Tf
Ti
Vf
(c) What work is required to compress the
gas?
Vi
Why is the work negative?
Adiabatic Process – Q = 0
Blue arrow – the gas is expending,
does a positive work and cooled down
(“goes to lower isotherms”)…
Ti
Tf
Vi
Vf
Red arrow – the gas is contracting,
driven by some external forces. It does
negative work and heats up (“goes to
higher isotherms”.)
An adiabatic process is described by:
γ = Cp/Cv a constant, where 1< γ < 2
It may be different for different gases!
In contrast an isothermal
process is described by:
Adiabatic and Isothermal Process for the Same Gas?
How does it happen that two equations may work for the same ideal
gas?
γ
γ
1 1
2 2
1 1
2 2
PV = P V
PV = P V
Are they both compatible
with the ideal gas law?
PV = nRT
Yes, they are both compatible!
Because the ideal gas law has 3 variables,
and P as a function of V in a particular process depends
on what happens to the temperature!
Isothermal
Adiabatic
Example: Isothermal vs Adiabatic Expansion
Consider expanding .06mol of gas with
γ = 1.4 from 2atm to 1atm. Starting at
T=300K. Find the final temperature and
volume for both an adiabatic and
isothermal process.
From the ideal gas law, the initial volume is:
For isothermal expansion the volume doubles. For an adiabatic expansion:
Fire Syringe:
A small wad of cotton bursts into
flame when the air in a narrow
tube is rapidly compressed.
Four Processes Between Two Isotherms
How do you order them in terms of heat transferred to the gas?
Things to remember:
U depends on T only!
W = ∫ PdV
Q = ΔU + W
The largest work done
by the gas will require
the most heat transfer!
What is the heat transferred to the gas in process 3?
Q = ΔU + W
W = ∫ PdV
ΔU = nCv ΔT
PV
nT =
R
P2V 2 − P1V1
nΔT =
R
Summary of Temperature Changes
Process 1 – Adiabatic
Q = 0 => ΔU = -W => T1 < Ta
Process 2 – Isochoric
W = 0 => Q = nCvΔT => T2 < Ta
Process 3 – Isobaric
P3 = Pa => Q = nCpΔT => T3 > Ta
Process 4 – Isothermal
T4 = Ta => ΔU = 0 => T4 = Ta
Summary of the First Law of Thermodynamics
First Law of Thermodynamics
• U only depends on the temperature of the gas
• Q is positive when heat is added to the gas
• W is positive when the gas does work
Thermodynamic processes,
Quasi-static - work is given by
Isothermal – constant temperature
Isochoric – constant volume
Isobaric – constant pressure
Adiabatic – no heat transfer