# ΔU = Uf – Ui

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Chapter tow
Fundamentals of
thermodynamics
2.1 The zeroth law of thermodynamics:
When two systems are in thermal equilibrium, i.e. no heat flows from one system
to the other during their thermal contact, then both systems have the same
temperature. If system A has the same temperature as system B and system B has
the same temperature as system C, then system A also has the same temperature
as system C.
2.2 internal energy, U:
The total energy of a system the total kinetic and potential energy of the
molecules in the system. The change in internal energy:
ΔU = Uf – Ui
(2.1)
Uf – internal energy at a final state
Ui – internal energy at an initial state
Internal energy, heat, and work are all measured in the same units, the joule (J):
1 J = 1 kg m2 s-1
Changes in molar internal energy ΔUm are typically expressed in kilojoules per
mole (kJ mol-1).An energy of 1 cal is enough to raise the temperature of 1 g of
water by 1&deg;C:1 cal = 4.184 J
2.3 First Law of thermodynamics:
The internal energy of an isolated system is constant.
ΔU = w + q
(2,2)
w – the work done on the system
q – the energy transferred as heat to the system
If we switch to infinitesimal changes, dU, dw, and dq,
dU = dw + dq
(2.3)
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Example3:A closed system performed work of 400 J, with 1000 J of energy in
the form of heat supplied to it. How did the system’s internal energy change?
Solution: If the system performed work, then, , the work W = −400 J. Hence the
change in the internal energy is
ΔU = Q +W = 1000 + (−400) = 600 J .
2.4 work
Work = distance &times; opposing force
The work needed to raise the mass through a height h on the surface of Earth:
Work = h &times; mg = mgh
(2.4)
When energy leaves the system,
i.e., the system does work in the
surroundings, w &lt; 0.
When energy enters the system as
work, w &gt; 0.
Energy leaves the system as heat,
q &lt; 0.
Energy enters the system as heat,
q &gt;0
2.4.1 Irreversible volume work
In this case the external pressure pex is different from the system pressure p.
During expansion it is lower and during compression higher than the system
pressure. We have
dw = -F dz
(2.5)
The negative sign: when the system moves an object against an opposing
force, the internal energy of the system doing the work will decrease.
F = pex A
dw = - pex dV
dw = - pex A dz
𝑽𝒇
𝒘 − ∫𝑽𝒊
dV = A dz
𝑷𝒆𝒙 𝒅𝒗
(2.6)
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2.4.2 Free expansion:
Expansion against zero opposing force:
pex = 0 w = 0 No work is done when a
system expands freely. Expansion of
this kind occurs when a system expands
into a vacuum
Expansion against constant pressure :
𝑽𝒇
𝒘 − ∫𝑽𝒊
𝑷𝒆𝒙 𝒅𝒗
pex (V f Vi ) ΔV = Vf – Vi
w = -pex ΔV
(2.7)
Wvol = CV (T2 − T1)
(2.7*)
. 2.4.3 Reversible expansion
The reversible volume work, Wvol is connected
with a change of the system volume from the
initial value V1 to the final value V2.
dw = -pexdV
𝑽𝒇
𝒘 − ∫𝑽𝒊
dw = -pdV
𝒑𝒅𝒗
(2.8)
2.4.3.1 Isobaric process
p = const Wvol = −p(V2 − V1)
(2.9)
2.4.3.2 Isochoric process
dV = 0
Wvol = 0
(2.10)
2.4.3.3 Isothermal process, the equation of state of an ideal gas
p = nRT/V
Wvol = −nRT ln(V2/V1)
(2.11)
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2.4.3.4Isothermal process, the van der Waals equation of state
𝒏𝑹𝑻
𝑷 = 𝑽−𝒃 −
𝒂𝒏𝟐
𝑽𝟐
𝑾 == −𝒏𝑹𝑻 𝐥𝐧
𝑽𝟐 −𝒏𝒃
𝑽𝟏 −𝒏𝒃
+ 𝒏𝟐 𝒂(
𝟏
𝑽𝟏
−
𝟏
𝒗𝟐
) (2.12)
2.4.3.5 Reversible isothermal expansion work
of a perfect gas at a temperature T
For a gas to expand reversibly, the
external pressure must be adjusted to
match the internal pressure at each
stage of the expansion
2.4.3.6 Adiabatic process for which Poisson’s equations:
𝑷 = 𝒄𝒐𝒏𝒔𝒕𝑽−𝒌
𝒘=
.
𝒄𝒐𝒏𝒔
𝟏−𝒌
𝟏−𝒌
(𝑽
−
𝑽
𝟐
𝟏 ) = 𝐂𝐕 (𝐓𝟐 − 𝐓𝟏)
𝟏−𝒌
(2.13))
Note: Irreversible volume work is always higher than reversible volume work. Two cases
may occur:
• During irreversible compression, we have to supply more work to the system for
compression from the initial volume V1 to the final volume V2 than in the case of
reversible compression,
• During irreversible expansion, the work done (Wdone = −Wvol) is smaller than
during reversible expansion, and consequently the supplied work is greater.
Example4: Calculate the work done when 50 g of iron reacts with
hydrochloric acid to produce hydrogen gas in (a) a closed vessel of a fixed
volume, (b) an open beaker at 25&deg;C.
a) the volume cannot change, no expansion work is done and w = 0. In (b) the
gas drives back the atmosphere and therefore w = -pexΔV. We can neglect the
initial volume because the final volume after the gas production is so much
larger: ΔV = Vf – Vi ≈ Vf = nRT/pex
n – the amount of H2 produced
w = -pexΔV = -pex &times; (nRT/pex) = -nRT
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The reaction equation is
Fe(s) + 2 HCl(aq) → FeCl2(aq) + H2(g)
Therefore, 1 mol H2 is generated when 1 mol Fe is consumed and n can be taken
as the amount of reacting Fe atoms.
MFe = 55.85 g mol-1 nFe = 50 g / 55.85 g mol-1
w ≈ -(50 g / 55.85 g mol-1) &times; 8.31451 J K mol-1 &times; 298.15 K ≈ -2.2 kJ
2,4.3.7 Other kinds of work:
• Surface work—work connected with a change of the system surface from the
initial value A1 to the final value A2.
𝑨
𝑾𝑺𝑼𝑹𝑭 = ∫𝑨 𝟏 𝜸𝒅𝑨
where 𝜸 is the surface tension.
𝟐
(2.14)
• Electrical work—work connected with the transfer of electric charge Q across
the potential difference E.
Q
Wel = ∫0 EdQ
• Shaft
(2.15)
work : Wsh is defined by the relation
𝑷
𝐖𝐬𝐡 = ∫𝑷 𝟐 𝑽𝒅𝑷
(2.16)
𝟏
2.5 Heat capacities:
The heat capacity C for a given process is defined by the following relation 𝒄
(
𝒅𝑸
𝒅𝑻
) 𝑝𝑟𝑜𝑐𝑒𝑠𝑠
=
(2.17)
The isochoric heat capacity, CV , and the isobaric heat capacity, Cp, are defined
by the relations
CV
=
𝒅𝑸
( 𝒅𝑻 ) izochoric
Cp = (
=
𝒅𝑼
( 𝒅𝑻 ) V
𝒅𝑸
𝒅𝑻
) izobaric
=(
(218)
𝒅𝑯
𝒅𝑻
)P
(𝟐. 𝟏𝟗)
For CV and Cp we have
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CV &gt; 0
Cp &gt; 0
lim CV = 0
𝑇→0
lim Cp = 0
𝑇→0
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(2.20)
(2.21)
2.6 Enthalpy : H is a function of state defined by the relation
H = U + Pv
(2.22)
Molar enthalpy, Hm = H/n
Hm = Um + pVm
For a perfect gas, Hm = Um + RT For a general infinitesimal change of
state of the system,U changes to U + dU, p to p + dp, V to V + dV:
H + dH = (U + dU) +( p + dp)(V + dV) = U + dU + pV + pdV + Vdp +
dpdV
dpdV – the product of two infinitesimally small numbers and can be
ignored. U + pV = H
H + dH = H + dU + pdV + Vdp
dH = dU + pdV + Vdp dU =
dq + dw
dH = dq + dw + pdV + Vdp
If the system is in mechanical equilibrium with its surroundings at a pressure p
and does only expansion work, dw = -pdV
dH = dq + Vdp
The heating occurs at constant pressure:
dp = 0
dH = dq
When a system is subjected to a constant pressure, and only expansion work can
occur, the change in enthalpy is equal to the energy supplied as heat
Example5:
A system containing 5 moles of an ideal gas was heated from temperature T1 =
300 K to temperature T2 = 400 K. The internal energy of the gas increased by ΔU
= 800 J. How did the enthalpy of the system change?
Solution:
ΔH = ΔU + Δ(pV ) ,
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where Δ(pV ) = p2V2 − p1V1. For an ideal gas pV = nRT. Then
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ΔH = ΔU + nR_T = 800 + 5 &times; 8.314 &times; (400 − 300) = 4957 J
Example6:
The internal energy change when 1.0 mol CaCO3 in the form of calcite converts
to aragonite is +0.21 kJ. Calculate the difference between the enthalpy change
and the change in internal energy when the pressure is 1.0 bar given that the
densities of the solids are 2.71 and 2.93 g cm-3.
We need to relate the density and the molar mass. For 1 mol,
.
ρ = M/Vm Vm = M/ρ
The change in enthalpy when the transition occurs is
ΔH = H(aragonite) – H(calcite) = {U(a) + pV(a)} – {U(a) + pV(c)} =
ΔU + p{V(a) – V(c)} =
ΔU + pΔV
Vm(a) = 100 g / 2.93 g cm-3 = 34 cm3 Vm(c) = 37 cm3
pΔV = 1.0&times;105 &times; (34 – 37)&times;10-6 m3 = -0.3 J ΔH - ΔU = -0.3 J
The difference is only 0.1% of the value of ΔU
.Example7:Water is heated under p = 1 atm. When an electric current of
0.50 A from a 12 V supply is passed for 300 s, 0.798 g of water is
vaporized. Calculate the molar internal energy and enthalpy
changes at the boiling point (373.15 K).
ΔH = qp = IVt
ΔHm = 0.50 A &times; 12 V &times; 300 s / (0.798/18.02) mol
ΔHm = +41 kJ mol-1
H2O(l) → H2O(g)
Δng = +1 mol
ΔUm = ΔHm – RT = +38 kJ mol2.6.1The variation of the enthalpy with temperature
The enthalpy of a substance increases with temperature. The relation between the
increase in enthalpy and the increase in temperature depends on the conditions
(constant pressure or constant volume ).
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heat capacity at constant pressure, Cp :The heat capacity at constant - the
analog of the heat capacity
at constant volume, an extensive property. The molar
heat capacity at constant pressure, Cp,m, - an intensive
property.
dH = CpdT (at constant pressure)
(2.23)
ΔH = CpΔT (at constant pressure) 𝑞𝑝 = CpΔT
(2.24)
The variation of heat capacity with temperature can sometimes
be ignored if the temperature
range is small - accurate for
a monoatomic perfect gas.
Otherwise, a convenient
Approximate empirical
expression is
𝐶𝑝, 𝑚 = 𝑎 + 𝑏𝑇 +
𝑐
(2.25)
𝑇2
The empirical parameters a, b, and c are
independent of temperature.
Example8: What is the change in molar enthalpy of N2 when it is heated from
25&deg;C to 100&deg;C? For N2(g) we find in Table 2.2 a = 28.58 J K-1 mol-1, b =
3.77&times;10-3 J K-2 mol-1, and c = -0.50&times;105 J K mol-
Since dH = CpdT,
𝐶𝑝, 𝑚 = 𝑎 + 𝑏𝑇 +
𝑐
𝑇2
𝐻(𝑇2)
𝑇2
ΔH = H(T2 ) − H(T1) = ∫𝐻(𝑇1) 𝑑𝐻 = ∫𝑇1 (
T1=298K
𝑎 + 𝑏𝑇 + 𝑐2)𝑑𝑇
𝑇
T2=373K
To carry out the integration, we use
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∫ dx = x + constant
1
𝑑𝑥
1
∫ 𝑋𝑑𝑥 = 2 𝑥 2 + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∫ 𝑥 2 = 𝑥 + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
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HT2 HT1aT2 T1 𝑏(𝑇22 − 𝑇12 ) − 𝑐( − )
2
𝑇2
𝑇1
1
1
1
H(373 K) - H(298 K) = 2.20 kJ mol-1
The relation between heat capacities :
For a perfect gas,
Hm - Um = RT
ΔHm - ΔUm = RΔT
ΔHm/ΔT - ΔUm /ΔT = R
Cp,m – CV,m = R
Cp,m – CV,m = large for gases, but negligible for most solids and liquids.
We consider the change in pressure resulting from an adiabatic,
reversible expansion of a perfect gas. We will show
𝑃𝑉 𝛾 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
γ - the heat capacity ratio
γ = Cp,m/CV,m Because Cp,m &gt; CV,m, γ &gt; 1
𝛄
=
𝑪𝑽.𝒎 + 𝑹
𝑪𝒗.𝒎
𝑹
= 𝟏+
𝑪𝑽.𝒎
During an adiabatic process, the system does not exchange heat with its
surroundings,i.e. Q = 0. Work during an adiabatic process (both reversible and
irreversible) is equal tothe change of internal energy
W = ΔU .
Provided that the following conditions are fulfilled during an adiabatic process,
i.e. that
• the process is reversible,
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• the system exchanges only volume work with its surroundings,
• the system is an ideal gas,
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• the heat capacities CV and Cp do not depend on temperature,
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•the following relations, called Poisson’s equations, apply between T, p and V :
𝟏
𝜸
𝜸
𝑷𝑽 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑇𝑉
𝛾−1
𝑻𝑷
𝟏−𝜸
𝜸
𝑷 𝑽 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑽𝑻
𝟏
𝜸−𝟏
𝑷𝑻
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝜸
𝟏−𝜸
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
Example9:
An ideal gas expanded adiabatically from temperature T1 = 300 K and pressure
p1 = 1 MPa to pressure p2 = 100 kPa. Provided that Poisson’s equations hold and
that Cpm = (5/2)R, find temperature T2 after the expansion
Solution
1−γ
P1
T2 = T1(
P2
)
γ
𝛾=
Cpm
Cpm−𝑅
=
5
3
T2 = 119.3𝐾
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2.7 Second law of thermodynamics:No process is possible in which
the sole result is the absorption of heat from a reservoir and its complete
conversion to work
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2.7.1Entropy: A measure of the molecular disorder of a system
𝑑𝑞𝑟𝑒𝑣
𝑑𝑠 =
𝑇
𝐹
∆𝑆 =
∫
𝐼
𝑑𝑞𝑟𝑒𝑣
𝑇
(2.26)
dS &gt; 0 , [isolated system, irreversible process]
dS = 0 , [isolated system, reversible process]
units of entropy: J K-1; of molar entropy: J K-1 mol-1.
Example10 : Calculating the entropy change for the isothermal expansion of a
perfect gas .We need to find the heat absorbed for reversible path between the
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initial and final states .The heat absorbed during a reversible isothermal
expansion:
ΔU = q + w ΔU = 0 q = -w qrev = -wrev
𝟏 𝟐
𝒒𝒓𝒆𝒗
∆𝑺 = ∫ 𝒅𝒒𝒓𝒆𝒗 =
𝑻 𝒊
𝑻
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= −𝒘𝒓𝒆𝒗 = 𝐧𝐑𝐓 𝐥𝐧(𝐕𝟐/𝐕𝟏)
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2.7.2 Heat:
That portion of internal energy which can be exchanged between a system and
its surroundings only when there is a difference in temperature
Q = ΔU –W
(2.27)
S2
T2
Q = ∫ 𝑇𝑑𝑆 = T2S2 − T1S1 − ∫ SdT
S1
T1
Several typical examples of heat calculation are given below
Q=0
•Isochoric process, no work done
Q = ΔU = U(T2, V ) − U(T1, V )
•Isobaric process, only volume work done
Q = ΔH = H(T2, p) − H(T1, p)
•Isothermal reversible process, ideal gas
𝒒 = −𝒘 =
𝐧𝐑𝐓 𝐥𝐧(𝐕𝟐/𝐕𝟏)
•Isothermal reversible process, the van der Waals equation of state
𝒒 = 𝒏𝑹𝑻 𝐥𝐧
𝑽𝟐 − 𝒏𝒃
𝑽𝟏 − 𝒏𝒃
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•General reversible isothermal process
S2
T (S2 − S1) = TΔS𝑞 = 𝑇 ∫S1 𝑑𝑆 =
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2.7.3 Entropy of phase transitions The entropy of a substance increases when it
melts and when it boils because its molecules become more disordered as it
changes from solid to liquid and from liquid to vapor
Entropy of fusion at the melting temperature: ΔfusS = ΔfusH/T
Entropy of vaporization at the boiling temperature: ΔvapS = ΔvapH/Tb
2.8 Helmholtz energy:
The Helmholtz energy F is a function of state defined by the relation
F = U − TS .
(2.28)
The change in the Helmholtz energy _F during a reversible isothermal process is
equal to the work supplied to the system
ΔF = W ,
[T, reversible process
Example11
During a certain isothermal process, internal energy changed by ΔU and entropy
by ΔS. Derive the relation for the change in the Helmholtz energy. Is it possible
to calculate the change in the Helmholtz energy during a non-isothermal process
if in addition to ΔU and ΔS we also know the initial and final temperatures?
Solution
For an isothermal process we obtain
ΔF = ΔU − Δ(TS) = ΔU − TΔS, [T] .
For a non-isothermal process we have
ΔF = ΔU − Δ(TS) = ΔU − T2S2 + T1S1 .
Since the values of entropy in the initial and final states, S1 , S2, are not specified
and we only know that ΔS = S2 − S1, the change in the Helmholtz energy cannot
be calculated.
.
2.9 Gibbs energy
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The Gibbs energy (or the Gibbs function) G is a function of state defined by the
relation
G = H − TS .
(2.29)
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Dr.tahseen altaei
Example12:
During a certain thermodynamic process, a system passed from its initial state
defined by the values of volume V1 and pressure p1 to its final state defined by
the values p2 and V2. The change in the Helmholtz energy was ΔF. Calculate the
change in the Gibbs energy.
Solution
From the definitions for H, F, and G we obtain
G = H − TS = U + pV − TS = F + pV ,
from which it follows:
ΔG = ΔF + Δ(pV ) = ΔF + p2V2 − p1V1 .
The change in the Gibbs energy _G during a reversible isothermal and isobaric
process isequal to other than volume work, Wother, supplied to the system
ΔG = Wother = W −Wvol = W + pΔV , [T, p, reversible process]
Example13:
The change in the Gibbs energy during the oxidation of one mole of glucose,
according to the reaction
C6H12O6 + 6O2 = 6CO2 + 6H2O
is ΔG = −2870 kJ mol−1. How high can a person weighing 75 kg climb if he or
she has eaten one mole (186 g) of glucose? The biological efficiency is 25%.
Solution
In our example, Wother is equal to mechanical work needed for lifting a load to a
height h Wother = −mgh ,
where m is the mass of the load and g is the gravitational acceleration. If we
project a human body as an isothermal and isobaric system in which reversible
processes occur and which does not perform any volume work, then
ΔG = −mgh
ℎ=
∆𝐺
𝑚𝑔
=
−2870000
75&times;9.8
= 3905𝑚
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2.10 Fugacity
Fugacity is defined by the relation
𝑮𝒎 (𝑻,𝑷)−𝑮&deg;𝒎 (𝑻,𝑷𝒔𝒕 )
𝒇 = 𝑷 𝐞𝐱𝐩(
)
𝑹𝑻
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𝒔𝒕
(2.30)
Dr.tahseen altaei
where pst is the standard pressure (usually pst = 101 325 Pa), and Gm(T, p) is the
molar Gibbs energy at temperature T and pressure p, and G_(T, pst) is the molar
Gibbs energy of a system in an ideal gaseous state at temperature T and pressure
pst.
Fugacity is a thermodynamic quantity which is useful when solving phase and
chemical equilibria.
Note: For an ideal gas, fugacity equals pressure
The fugacity/pressure ratio is called the fugacity coefficient
∅=
𝑓
_
𝑝
Note: For an ideal gas = 1.
Example14:
Knowing the values of the molar Gibbs energy of carbon dioxide Gm(T, p) = −53
183 J/mol at T = 350 K and p = 10 MPa, and Gm(T, pst) = −65 675.14 J/mol,
where pst = 101.325 kPa, calculate its fugacity and the fugacity coefficient at 350
K and 10 MPa.
Solution:
f = 0.101325 &times; exp(
∅=
𝟕. 𝟒𝟏𝟓
𝟏𝟎
−𝟓𝟑𝟏𝟖𝟑 − (−𝟔𝟓𝟔𝟕𝟓.𝟏𝟒)
𝟖.𝟑𝟏𝟒 &times; 𝟑𝟓𝟎
) = 𝟕. 𝟒𝟏𝟓 𝐌𝐏𝐚 ,
= 𝟎. 𝟕𝟒𝟏𝟓.
2.11 Total differential:
Let us consider functions M(x, y) and N(x, y)
dz = M(x, y)dx + N(x, y)dy
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to be the total differential of the function z = z(x, y) is the equality of the
derivatives
𝜕𝑁
( )x = ( )𝑦
∂y
𝜕𝑥
∂M
Physicalchemistary
at all points of the
𝑀=(
𝜕𝑍
)𝑦
𝜕𝑥
,
(2.32)
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region, where
𝑁=(
Dr.tahseen altaei
𝜕𝑍
)𝑥
𝜕𝑦
(2.33)
Hence for the total differential of the function z = z(x, y) it holds
𝜕𝑍
𝜕𝑍
) 𝜕𝑥 + (
)
𝜕𝑥 𝑦
𝜕𝑦 x
∂z = (
∂𝑦
𝜕2z
𝜕2z
=
∂x ∂y ∂y ∂x
(2.34)
Example15:
Is the differential form
dz = (10xy3 + 7)dx + 15x2y2dy
the total differential of function z?
Solution
𝜕𝑍
𝜕𝑍
𝑀 = (𝜕𝑦) 𝑦 = 10xy3 + 7
𝑁 = (𝜕𝑥) 𝑥 = 15x2y2 .
It holds that
𝜕𝑀
𝜕2𝑧
( )=
= 30xy2
𝜕𝑦
𝜕𝑥𝜕𝑦
,
∂N
𝜕2𝑧
( )=
= 30xy2 .
∂x
𝜕𝑦𝜕𝑥
Sometimes we need to know the derivative of x with respect to y at a fixed z
.0
𝜕𝑍
=( )
𝜕𝑥 𝑦
𝜕𝑥 + (𝜕𝑍
)
𝜕𝑦
∂𝑦
x
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𝝏𝒙
(𝝏𝒚) =
𝜕𝑍
)
𝜕𝑦 x
𝜕𝑍
( )
𝜕𝑥 𝑦
(
2.11.1 Total differential and state functions:
All state functions (p, V, T, n, U, H, S, F, G, Cp, CV , f, . . .) have total
differentials. Heat and work are not state functions and they do not have any total
differential.
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Example16:
Prove that condition holds for the pressure of an ideal gas as a function of
temperature and volume at a constant amount of substance.
Proof
𝑛𝑅𝑇
𝑃=
𝑉
𝜕𝑃
𝜕𝑃
𝑛𝑅
𝑛𝑅𝑇
𝑑𝑃 = ( )𝑉 𝑑𝑇 + ( ) 𝑇 𝑑𝑉 =
𝑑𝑇 − 2 𝑑𝑉
𝜕𝑇
𝜕𝑉
𝑉
𝑉
1
𝜕( )
𝑛𝑅 𝜕𝑇
𝑛𝑅( 𝑉 ) 𝑇 = − 2 ( )𝑉
𝜕𝑉
𝑉 𝜕𝑇
dp is thus a total differential and pressure is a function of state. A similar proof
can be established for pressure calculated using any equation of state.
Example17:
By substituting the expression TdS for &macr;dQ into equation (3.1) we obtain
&macr;dW = dU − TdS .
Prove that work W is not a state function.
Proof
Comparison , shows that M = 1, N = −T, x = U, and y = S. Hence
𝝏𝑴
𝛛𝟏
𝛛𝐍
𝛛(−𝐓)
(
) 𝐱 = ( )𝐮 ≠ ( ) 𝐲 = (
)
𝝏𝒚
𝛛𝐬
𝛛𝐱
𝛛𝐔 𝐬
𝛛𝐓
because ( )𝐬 is generally different from zero. This shows that &macr;dW is not a
𝛛𝐔
total differential and therefore work is not a state function
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Total differential of the product and ratio of two functions
Similar rules as those governing differentiation hold for the total differential of
the product and the ratio of functions x and y.
d(xy) = xdy + ydx ,
𝒙
𝟏
𝒙
𝒅(𝒚)= 𝒚dx− 𝒚𝟐 𝒅𝒚
For example, the total differential of the product of volume and pressure
d(pV ) = pdV + V dp .
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2.11.2:Integration of the total differential
The integral of the total differential from point (x1,y1) to point (x2,y2) does not
depend on the path between these points. We can, e.g., first integrate with respect
to x at a fixed y1, and then with respect to y at a fixed x2
𝒙𝟐
𝐳(𝐱𝟐, 𝐲𝟐) = 𝐳(𝐱𝟏, 𝐲𝟏) + ∫ (
𝒙𝟏
𝒚𝟐
𝝏𝒛
𝝏𝒛
)𝒚=𝒚 𝒅𝒙 + ∫ ( )𝒙=𝒙 𝒅𝒚
𝟐
𝟏
𝝏𝒙
𝝏𝒚
𝒚
(𝟐. 𝟑𝟓)
𝟏
𝒚𝟐
𝒙
𝝏𝒛
𝝏𝒛
𝐳(𝐱𝟐, 𝐲𝟐) = 𝐳(𝐱𝟏, 𝐲𝟏) + ∫ ( )𝒙=𝒙𝟏 𝒅𝒚 + ∫ ( )𝒚=𝒚𝟐 𝒅𝒙
𝒚𝟏 𝝏𝒙
𝒙 𝝏𝒙
2.12 Combined formulations of the first and second laws of
thermodynamics:
In this section we will assume (if not otherwise stated) that every system referred
to is closed and homogeneous, that it exchanges only volume work with its
surroundings, and that all the occurring processes are reversible.
Gibbs equations:
dU = TdS − pdV ,
(2.36)
dH = TdS + V dp ,
(2.37)
dF = −SdT − pdV ,
(2.38)
dG = −SdT + V dp .
(2.39)
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internal energy is a function of variables S and V . Variables S and V will
be called the natural variables of function U. The natural variables of enthalpy
are S and p, those of the Helmholtz energy are T and V , and those of the Gibbs
energy are T and p.
2.13Derivatives of U, H, F, and G with respect to natural variables
If we consider the internal energy U as a function of S and V , its total
differential
𝝏𝑼
𝝏𝑼
𝝏𝒔
𝝏𝑽
𝒅𝑼 = ( )𝒗 𝒅𝑺 + ( )𝒔 𝒅𝑽
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By comparing () and () we obtain
𝝏𝑼
( )𝑽 = 𝑻
𝝏𝑺
𝝏𝑼
( )𝒔 = −𝑷
𝝏𝑽
,
(𝟐. 𝟒𝟎)
In a similar way we obtain for H = f(S, p), F = f(T, V ), G = f(T, p)
𝝏𝑯
( )𝒑 = 𝑻
𝝏𝑺
𝝏𝑭
(𝝏𝑻)𝒗 = −𝑺
𝝏𝑮
(𝝏𝑻)𝒑 = −𝑺
𝝏𝑯
( )𝑺 = 𝑽
𝝏𝑷
𝝏𝑭
(𝝏𝑽)𝑻 = −𝑷
𝝏𝑮
(𝝏𝑷)𝑻 = 𝑽
(𝟐. 𝟒𝟏)
(𝟐. 𝟒𝟐)
(𝟐. 𝟒𝟑)
2.13 Maxwell relations:
By applying the equalities of mixed derivatives to the total differentials of the
functions U, H, F, G, we obtain
𝝏𝑻
𝝏𝑷
( )𝑺 = −( )𝑽
𝝏𝑽
𝝏𝑺
(𝟐.𝟒𝟒)
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𝝏𝑻
𝝏𝑽
( )𝑺 = ( )𝑷
𝝏𝑷
𝝏𝑺
𝝏𝑷
𝛛𝐒
𝝏𝑻
𝛛𝐕
(𝟐.𝟒𝟓)
( )𝐕 = ( )𝐓
(𝟐. 𝟒𝟔)
𝝏𝑽
𝝏𝑺
( )𝑷 = −( )𝑻
𝝏𝑻
𝝏𝑷
(𝟐.𝟒𝟕)
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2.14 Total differential of entropy as a function of T, V and T, p:
𝝏𝑺
𝝏𝑺
𝒅𝑺 = ( )𝑽 𝒅𝑻 + ( )𝑻 𝒅𝑽
𝝏𝑻
𝝏𝑽
At a fixed volume
𝝏𝑺
𝟏 𝝏𝑸
( )𝑽 = ( )
𝝏𝑻
𝑻 𝝏𝑻
isochoric
=
𝟏 𝝏𝑼
𝑪𝑽
( )𝑽 =
𝑻 𝝏𝑻
𝑻
For the differentiation of entropy with respect to volume, Maxwell relation
applies.
𝐶𝑉
𝜕𝑃
𝑑𝑆 =
𝑑𝑇 + ( )𝑉 𝑑𝑉
𝑇
𝜕𝑇
In the same way we obtain for entropy as a function of temperature and pressure
𝜕𝑆
1 𝜕𝑄
( )𝑃 = ( )
𝜕𝑇
𝑇 𝜕𝑇
isobaric
1 𝜕𝐻
𝐶𝑃
= ( )𝑃 =
𝑇 𝜕𝑇
𝑇
and using Maxwell relation
𝐶𝑃
𝜕𝑉
𝑑𝑆 =
𝑑𝑇 − ( )𝑃 𝑑𝑝
𝑇
𝜕𝑇
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2.15 Conversion from natural variables to variables T, V or T, p
Internal energy may be converted from the function of natural variables to the
function of variables T, V
𝜕𝑃
dU = CV dT + [𝑇( )𝑉 ) − 𝑃] 𝑑𝑉
𝜕𝑇
In the same way, H = f(S, p) may be converted to H = f(T, p).
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𝜕𝑉
dH = Cp dT + [𝑉 − 𝑇( )𝑃 ] 𝑑𝑝
𝜕𝑇
the total differential of the function U = f(T, V ) is equal to the expression
𝜕𝑈
𝜕𝑈
𝑑𝑈 = ( )𝑉 𝑑𝑇 + ( ) 𝑇 𝑑𝑉
𝜕𝑇
𝜕𝑉
𝜕𝑈
( )𝑉 = 𝐶𝑉
𝜕𝑇
𝜕𝑈
𝜕𝑃
( ) 𝑇 = 𝑇( )𝑉 ) − 𝑃
𝜕𝑉
𝜕𝑇
Similarly we obtain the respective partial derivatives of the dependence H = f(T, p)
𝜕𝐻
( )𝑃 = 𝐶𝑃
𝜕𝑇
𝜕𝐻
𝜕𝑉
( ) 𝑇 = −𝑇 + 𝑉( )𝑃
𝜕𝑃
𝜕𝑇
Example18:
Prove that the internal energy of an ideal gas is only a function of temperature,
i.e. that at a fixed temperature it depends neither on volume nor on pressure
Proof
For an ideal gas we have
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𝑇(
𝜕𝑃
𝑛𝑅
)𝑉 − 𝑃 = 𝑇
−𝑃 =0
𝜕𝑇
𝑉
𝜕𝑉
𝜕𝑉
𝑑𝑉 = ( )𝑃 𝑑𝑇 + ( ) 𝑇 𝑑𝑝
𝜕𝑇
𝜕𝑃
𝜕𝑃
𝜕𝑉
𝜕𝑃
𝜕𝑉
𝑑𝑈 = {𝐶𝑉 + [𝑇( )𝑉 − 𝑃] ( )𝑃 } 𝑑𝑇 + [𝑇( )𝑉 − 𝑃] ( ) 𝑇 𝑑𝑉
𝜕𝑇
𝜕𝑇
𝜕𝑇
𝜕𝑃
From this expression it follows that
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𝜕𝑈
𝜕𝑃
𝜕𝑉
( ) 𝑇 = [𝑇( )𝑉 − 𝑃] ( ) 𝑇
𝜕𝑃
𝜕𝑇
𝜕𝑃
As was shown above, the expression in brackets for an ideal gas is zero. Hence
the internal energy of an ideal gas is independent of pressure.
2.16 Heat engines:
A heat
engine is a device that cyclically converts heat into work. According to the
Second Law of thermodynamics, this conversion cannot be complete a heat
engine means a device that cyclically receives heat Q2 (Q2 &gt; 0) from the
warmer reservoir of temperature T2. It converts part of it into work
Wdone = −W, which it supplies to the surroundings. At the same time it supplies
heat Qsupplied = −Q1 (Q1 &lt; 0) to the cooler reservoir of temperature T1, and
returns to the initial state. During this cyclic process, the change of internal
energy is zero, and consequently we have
Q1 +W + Q2 = 0 . (2.50)
Example19:
A heat engine received 600 J of heat from the warmer reservoir, supplied 500 J
of heat to the cooler reservoir, and returned to the initial state. Ascertain the work
performed by the engine.
Solution
It follows from the specification that Q1 = 600 J, Q2 = −500 J, Q = Q1 + Q2 =
100 J. Since the system returned to its original state, _U = 0. From equation
(2.50)) we thus have
W = −Q = −100 J .
The work performed by the engine is Wdone = −W = −(−100 J) = +100 J
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2.16.1The Carnot heat engine:
Let the temperature of the warmer reservoir be 𝑇ℎ and that of the cooler reservoir
be𝑇𝑐 . One cycle is comprised of four partial reversible processes:
1. Reversible isothermal expansion from A to B at Th; the entropy change
is qh/Th, qh – the heat supplied to the system from the hot source.
2. Reversible adiabatic expansion from B to C. No heat leaves the
system, ΔS = 0, the temperature falls from Th to Tc, the temperature of the
cold sink.
3. Reversible isothermal compression
from
C to D. Heat is released to theDr.tahseen altaei
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①
cold sink; the change in entropy of the system is qc/Tc, qc is negative.
4. Reversible adiabatic compression from D to A. No heat enters the
system, ΔS = 0, the temperature rises from Tc to Th.
𝐰𝐨𝐫𝐤 𝐝𝐨𝐧𝐞
−𝒘
ŋ=
=
𝐡𝐞𝐚𝐭 𝐬𝐮𝐩𝐩𝐥𝐢𝐞𝐝 𝐟𝐫𝐨𝐦 𝐭𝐡𝐞 𝐰𝐚𝐫𝐦𝐞𝐫 𝐜𝐨𝐧𝐭𝐚𝐢𝐧𝐞r 𝑸ℎ
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(2,51)
𝑇ℎ − 𝑇𝑐
ŋ=
𝑇ℎ
(2.52)
Note: A reversibly operating heat engine cannot exist in practice because, among
other reasons, the processes in such an engine would progress at an infinitesimal
rate.
ŋreversible engine &gt; ŋirreversible engine
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Physicalchemistary
course ①
Dr.tahseen altaei
Example20
The Carnot heat engine received heat Q2 = 100 J from a heat reservoir of
temperature T2 = 600 K, performed work, delivered heat −Q1 to the cooler
reservoir of temperature T1 = 300 K, and returned to the initial state. Calculate
the engine’s efficiency, the performed work and the supplied heat.
ŋ=
𝑇2 − 𝑇1 600 − 300
=
= 0.5
𝑇2
600
−𝑊 = 𝑄2 ŋ = 100 &times; 0.5 = 50𝐽
Since the initial and final state are identical, the internal energy change
is
ΔU = 0 .
0 = Q +W = Q2 + Q1 +W
Q1 = −Q2 −W = −100 − (−50) = −50 J .
The Carnot engine supplied −Q1 = +50 J of heat to the cooler reservoir.
2.16.2 Heat engine with steady flow of substance:
A heat engine with a steady flow of substance is a device that exchanges shaft
work Wsh with its surroundings at the cost of the energy of the working medium
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