Lecture 2 The First Law of
Thermodynamics (Ch.1)
Outline:
1. Internal Energy, Work, Heating
2. Energy Conservation – the First Law
3. Quasi-static processes
4. Enthalpy
5. Heat Capacity
Internal Energy
The internal energy of a system of particles, U, is the sum of the kinetic
energy in the reference frame in which the center of mass is at rest and the
potential energy arising from the forces of the particles on each other.
Difference between the total energy and the internal energy?
system
boundary
system
U = kinetic + potential
“environment”
B
P
T
A
V
The internal energy is a state function – it depends only on
the values of macroparameters (the state of a system), not
on the method of preparation of this state (the “path” in the
macroparameter space is irrelevant).
In equilibrium [ f (P,V,T)=0 ] :
U = U (V, T)
U depends on the kinetic energy of particles in a system and an average
inter-particle distance (~ V-1/3) – interactions.
For an ideal gas (no interactions) :
U = U (T) - “pure” kinetic
Internal Energy of an Ideal Gas
The internal energy of an ideal gas
with f degrees of freedom:
f
U = Nk BT
2
f ⇒ 3 (monatomic), 5 (diatomic), 6 (polyatomic)
(here we consider only trans.+rotat. degrees of freedom, and neglect
the vibrational ones that can be excited at very high temperatures)
How does the internal energy of air in this (not-air-tight) room change
with T if the external P = const?
⎡
f
PV ⎤ f
U = N in room k BT = ⎢ N in room =
⎥ = PV
k BT ⎦ 2
2
⎣
- does not change at all, an increase of the kinetic energy of individual
molecules with T is compensated by a decrease of their number.
Work and Heating (“Heat”)
We are often interested in ΔU , not U. ΔU is due to:
Q - energy flow between a system and its
environment due to ΔT across a boundary and a finite
thermal conductivity of the boundary
WORK
HEATING
– heating (Q > 0) /cooling (Q < 0)
(there is no such physical quantity as “heat”; to
emphasize this fact, it is better to use the term
“heating” rather than “heat”)
W - any other kind of energy transfer across
boundary
- work
Work and Heating are both defined to describe energy transfer
across a system boundary.
Heating/cooling processes:
conduction: the energy transfer by molecular contact – fast-moving
molecules transfer energy to slow-moving molecules by collisions;
convection: by macroscopic motion of gas or liquid
radiation:
by emission/absorption of electromagnetic radiation.
The First Law
The first law of thermodynamics: the internal energy of a system can be
changed by doing work on it or by heating/cooling it.
ΔU = Q + W
conservation of energy.
Sign convention: we consider Q and W to be positive if energy
flows into the system.
For a cyclic process (Ui = Uf) ⇒ Q = If, in addition, Q = 0 then W = 0
P
W.
T
V
An equivalent formulation:
Perpetual motion machines of the first type do not exist.
Quasi-Static Processes
Quasi-static (quasi-equilibrium) processes – sufficiently
slow processes, any intermediate state can be considered
as an equilibrium state (the macroparamers are welldefined for all intermediate states).
Advantage: the state of a system that participates in a quasi-equilibrium
process can be described with the same (small) number of macro
parameters as for a system in equilibrium (e.g., for an ideal gas in quasiequilibrium processes, this could be T and P). By contrast, for nonequilibrium processes (e.g. turbulent flow of gas), we need a huge number
of macro parameters.
Examples of quasiequilibrium processes:
isochoric:
isobaric:
isothermal:
adiabatic:
V = const
P = const
T = const
Q=0
For quasi-equilibrium processes, P, V, T are
well-defined – the “path” between two states
is a continuous lines in the P, V, T space.
P
T
2
1
V
Work
A – the
The work done by an external force on a gas
enclosed within a cylinder fitted with a piston:
piston
area
W = (PA) dx = P (Adx) = - PdV
force
Δx
P
The sign: if the volume is decreased, W is positive (by
compressing gas, we increase its internal energy); if the
volume is increased, W is negative (the gas decreases
its internal energy by doing some work on the
environment).
V2
W1− 2 = − ∫ P(T , V )dV
W = - PdV - applies to any
shape of system boundary
V1
dU = Q – PdV
The work is not necessarily associated with the volume changes – e.g.,
in the Joule’s experiments on determining the “mechanical equivalent of
heat”, the system (water) was heated by stirring.
W and Q are not State Functions
- we can bring the system from state 1 to
state 2 along infinite # of paths, and for each
path P(T,V) will be different.
V2
W1− 2 = − ∫ P (T , V )dV
V1
P
T
2
Since the work done on a system depends not
only on the initial and final states, but also on the
intermediate states, it is not a state function.
V
1
ΔU = Q + W
U is a state function, W - is not ⇒
thus, Q is not a state function either.
B
Wnet = WAB + WCD = −P2 (V2 −V1 ) − P1 (V1 −V2 )
P
P2
P1
A
= −(P2 − P1 )(V2 −V1 ) < 0
D
V1
PV diagram
C
V2
V
- the work is negative for the “clockwise” cycle; if
the cyclic process were carried out in the reverse
order (counterclockwise), the net work done on
the gas would be positive.
Comment on State Functions
U, P, T, and V are the state functions, Q and W are not. Specifying an initial and final
states of a system does not fix the values of Q and W, we need to know the whole
process (the intermediate states). Analogy: in classical mechanics, if a force is not
conservative (e.g., friction), the initial and final positions do not determine the work, the
entire path must be specified.
In math terms, Q and W are not exact differentials of some functions
of macroparameters. To emphasize that W and Q are NOT the state
functions, we will use sometimes the curled symbols δ (instead of d)
for their increments (δQ and δW).
d U = T d S − P dV
y
- an exact differential
U
V
S
dz = Ax ( x, y ) dx + Ay ( x, y ) dy - it is an exact differential if it is
the difference between the values of some (state) function
z(x2,y2)
dz = z ( x + dx, y + dy ) − z ( x, y )
z(x,y) at these points:
∂Ax ( x, y ) ∂Ay ( x, y )
x
=
A necessary and sufficient condition for this:
∂x
∂y
∂z ( x, y )
∂z ( x, y )
⎛ ∂z ⎞
If this condition
⎛ ∂z ⎞
Ax ( x, y ) =
Ay ( x, y ) =
d z = ⎜ ⎟ dx + ⎜⎜ ⎟⎟ dy
holds:
∂x
∂y
⎝ ∂x ⎠ y
⎝ ∂y ⎠ x
T
⎛f
⎞
e.g., for an ideal gas: δQ = dU + PdV = Nk B ⎜ dT + dV ⎟ - cross derivatives
V
are not equal
⎝2
⎠
z(x1,y1)
Problem
Imagine that an ideal monatomic gas is taken from its initial state A to state
B by an isothermal process, from B to C by an isobaric process, and from
C back to its initial state A by an isochoric process. Fill in the signs of Q,
W, and ΔU for each step.
P,
105 Pa
2
A
T=const
1
Step
Q
W
ΔU
A→B
+
--
0
B→C
--
+
--
C→A
+
0
+
B
C
1
2
f
U = Nk BT
2
V, m3
PV = Nk B T
Quasistatic Processes in an Ideal Gas
isochoric ( V = const )
P
W1→2 = 0
2
PV= NkBT2
PV= NkBT1
1
V1,2
V
3
Q1→2 = NkB (T2 − T1 ) > 0
2
P
(see the last slide)
dU = Q1→2
isobaric
(= CV ΔT )
( P = const )
2
W1→2 = −∫ P(V , T )dV = − P(V2 −V1 ) < 0
2
1
V1
1
PV= NkBT2
PV= NkBT1
V2
V
5
Q1→2 = NkB (T2 − T1 ) > 0
2
dU = W1→2 + Q1→2
(= CPΔT )
Isothermal Process in an Ideal Gas
P
isothermal ( T = const ) :
PV= NkBT
W
V2
Wi − f
dU = 0
V2
V1
V
V
= Nk BT ln i
Vf
Wi-f > 0 if Vi >Vf (compression)
Wi-f < 0 if Vi <Vf (expansion)
V2
V
dV
= − Nk BT ln 2
V
V1
V1
W1→2 = − ∫ P (V , T )dV = − Nk BT ∫
V1
Q1→2 = −W1→2
Adiabatic Process in an Ideal Gas
Q1→2 = 0
adiabatic (thermally isolated system)
dU =W1→2
The amount of work needed to change the state of a thermally isolated system
depends only on the initial and final states and not on the intermediate states.
V2
W1→2 = − ∫ P(V , T )dV
P
V1
2
V2
PV = Nk BT
dV
V
1
PV= NkBT2
PV= NkBT1
V1
V
to calculate W1-2 , we need to know P (V,T)
for an adiabatic process
U=
∫
γ
⇒ dU =
f
Nk B dT = − PdV
2
( f – the # of “unfrozen” degrees of freedom )
⇒ PdV + VdP = Nk B dT
⎛
2 ⎞ dP
=0
⎜⎜1 + ⎟⎟ +
f
P
⎝
⎠
f
Nk BT
2
,
γ = 1+
2
PdV
÷ PV
f
V
P
dV
dP
γ∫
+∫
=0
V
P
V1
P1
PdV + VdP = −
2 Adiabatic
f exponent
⎛V ⎞
⎛P ⎞
γ
ln⎜⎜ ⎟⎟ = ln⎜ 1 ⎟ ⇒ PV γ = P1V1 = const
⎝P⎠
⎝ V1 ⎠
Adiabatic Process in an Ideal Gas (cont.)
γ
PV γ = P1V1 = const
P
2
V2
1
PV= NkBT2
PV= NkBT1
V1
V
An adiabata is “steeper” than an isotherma:
in an adiabatic process, the work flowing
out of the gas comes at the expense of its
thermal energy ⇒ its temperature will
decrease.
V2
V2
V1
V1
W1→ 2 = − ∫ P(V , T )dV = − ∫
γ
V2
PV
1
− γ +1
γ
1 1
dV
PV
V
=
−
1
1
Vγ
−γ + 1
V1
1 ⎛ 1
1 ⎞
= PV
⎜ γ −1 − γ −1 ⎟
1 1
γ − 1 ⎝ V2
V1 ⎠
γ
γ
⇒ 1+2/3≈1.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 ≈1.33 (polyatomic)
(again, neglecting the vibrational degrees of freedom)
Prove W1→ 2 =
f
f
Δ ( PV ) = Nk B ΔT = ΔU
2
2
Summary of quasi-static processes of ideal gas
ΔU ≡ U f − U i
Quasi-Static
process
ΔU
f
f
isobaric
ΔU = Nk B ΔT = P ΔV
2
2
(ΔP=0)
isochoric ΔU = f Nk ΔT = f ΔP V
( )
B
2
2
(ΔV=0)
isothermal
0
(ΔT=0)
adiabatic ΔU = f Nk ΔT = f Δ PV
( )
B
2
2
(Q=0)
Ideal gas
law
Q
W
f +2
PΔV
2
− PΔV
Vi V f
=
Ti T f
0
Pi Pf
=
Ti T f
f
( ΔP )V
2
−W
0
− Nk B T ln
ΔU
Vf
Vi
PV
i i = Pf V f
γ
γ
PV
i i = Pf V f
Problem
Imagine that we rapidly compress a sample of air whose initial pressure is
105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of
its original volume (e.g., pumping bike’s tire). What is its final temperature?
Rapid compression – approx. adiabatic, no time for the energy
exchange with the environment due to thermal conductivity
P1V1 = Nk BT1
P2V2 = Nk BT2
P2 =
P1V1γ = P2V2γ
γ
P1V1
V2γ
P1V1
P1V1
Nk
T
=
=
T2
B 2
V2γ −1
T1
For adiabatic processes:
⎛ V1 ⎞
T2 = T1 ⎜⎜ ⎟⎟
⎝ V2 ⎠
γ −1
γ
γ −1
⎛ V1 ⎞
⇒ ⎜⎜ ⎟⎟
⎝ V2 ⎠
γ −1
T1 V1
= T2 V2
also
P γ −1 / T γ = const
γ −1
=
= const
= 295 K × 4 0.4 ≈ 295 K × 1.74 ≈ 514 K
- poor approx. for a bike pump, works better for diesel engines
T2
T1
Non-equilibrium Adiabatic Processes
Free expansion
1.
2.
TV γ −1 = const
V – increases
⇒ T – decreases (cooling)
On the other hand, ΔU = Q + W = 0
U ~ T ⇒ T – unchanged
(agrees with experimental finding)
Contradiction – because approach
#1 cannot be justified – violent
expansion of gas is not a quasistatic process. T must remain the
same.
TV γ −1 = const
- applies only to quasi-equilibrium processes !!!
The Enthalpy
Isobaric processes (P = const):
dU = Q - PΔV = Q -Δ(PV) ⇒
⇒
Q = Δ U + Δ(PV)
H ≡ U + PV - the enthalpy
The enthalpy is a state function, because U, P,
and V are state functions. In isobaric processes,
the energy received by a system by heating equals
to the change in enthalpy.
isochoric:
Q=ΔU
isobaric:
Q=ΔH
in both cases, Q
does
not
depend on the
path from 1 to 2.
Consequence: the energy released (absorbed) in chemical reactions at constant
volume (pressure) depends only on the initial and final states of a system.
f
⎞
⎛f
The enthalpy of an ideal gas: H = U + PV =
Nk BT + Nk BT = ⎜ + 1⎟ Nk BT
2
⎝2 ⎠
(depends on T only)
Heat Capacity
The heat capacity of a system - the amount of energy
transfer due to heating required to produce a unit
temperature rise in that system
T
C is NOT a state function (since Q is not a
state function) – it depends on the path
between two states of a system
⇒
f1
δQ
ΔT
f2
f3
T1+dT
T1
i
V
( isothermic – C = ∞, adiabatic – C = 0 )
The specific heat capacity
C≡
C
c≡
m
CV and CP
C=
δQ
dT
=
⎛ ∂U ⎞ the heat capacity at
CV = ⎜
⎟
⎝ ∂T ⎠V constant volume
dU + PdV
dT
⎛ ∂H ⎞
CP = ⎜
⎟
⎝ ∂T ⎠ P
the heat capacity at
constant pressure
To find CP and CV, we need f (P,V,T) = 0 and U = U (V,T)
⎛ f
⎞
f
H
=
+
1
U = Nk BT
⎜
⎟ Nk BT
⎝2
⎠
2
f
f
⎛ f
⎞
CP = ⎜ + 1⎟ nR
CV = Nk B = nR
⎝2
⎠
2
2
For an ideal gas
# of moles
For one mole of a
monatomic ideal gas:
CV =
3
5
R CP = R
2
2
P
Another Problem
During the ascent of a meteorological helium-gas filled balloon,
its volume increases from Vi = 1 m3 to Vf = 1.8 m3, and the
pressure inside the balloon decreases from 1 bar (=105 N/m2) to
0.5 bar. Assume that the pressure changes linearly with volume
between Vi and Vf.
(a) If the initial T is 300K, what is the final T?
(b) How much work is done by the gas in the balloon?
(c) How much “heat” does the gas absorb, if any?
Pi
Pf
Vi
Vf
V
P(V ) = −0.625 bar/m 3 × V + 1.625 bar
(a)
PV = Nk B T
(b) δWON
PV
T=
Nk B
0.5bar × 1.8m 3
= 300K
= 270K
T f = Ti
3
1bar × 1m
Pi Vi
Pf V f
Vf
Vf
Vi
Vi
= − ∫ P (V )dV - work done on a system δWBY = P (V )dV - work done by a system
∫
Vf
δWON = −δWBY δWBY = ∫ P(V )dV = (0.5 × 0.8 bar ⋅ m3 + 0.5 × 0.4 bar ⋅ m3 ) = 0.6bar ⋅ m3 = 6 ⋅10 4 J
(c)
ΔU = δQ + δWON
Vi
⎛T
⎞
δQ = ΔU − δWON = NkB (T f − Ti ) − WON = Pi Vi ⎜⎜ f − 1⎟⎟ + δWBY = 1.5 ⋅105 J × (− 0.1) + 6 ⋅104 J = 4.5 ⋅104 J
2
2
⎝ Ti
⎠
3
3