First Law of Thermodynamics
State Functions- A State Function is a thermodynamic quantity whose value depends only on the state at the moment, i. e., the temperature, pressure, volume, etc…
- The value of a state function is independent of the history of the system. - Temperature is an example of a state function.
- The fact that temperature is a state function is extremely useful because it we can measure the temperature change in the system by knowing the initial temperature and the final temperature.
- In other words, we don’t need all of the nitty-gritty detail of a process to measure the change in the value of a state function.
- In contrast, we do need all of the nitty-gritty details to measure the heat or the work of a system.
Reversibility
A reversible process is a process where the effects of following a thermodynamic path can be undone be exactly reversing the path.
An easier definition is a process that is always at equilibrium even when undergoing a change.
Phase changes and chemical equilibria are examples of reversible processes. Ideally the composition throughout the system must be homogeneous.
- This requirement implies that the no gradients, currents or eddys can exist. - To eliminate all inhomogeneities, a reversible process must occur infinitely
slow!
- Thus no truly reversible processes exist. However, many systems are
approximately reversible. And assuming reversible processes will greatly aid our calculations of various thermodynamic state functions.
Reversibility during pressure changes ensures that p = pex
That is, the pressure on the inside of the container is always equal to the pressure exerted on the outside of the container.
Theorem of Maximum Work
The maximum amount of work that can be extracted from an expansion process occurs under reversible conditions.
Thus the theorem implies that during an irreversible expansion, some of the energy is lost as heat rather than work. The inhomogeneities (currents, gradients and eddys) in pressure that occur during an irreversible process are responsible for the heat.
Initial Definitions of Energy and Energy Transfer
Internal Energy- Sum of the kinetic and potential energy in a sample of matter. Microscopic modes of Internal Energy (20th century view)
- degrees of freedom for energy storage translational
rotational
librational – vibrations caused by intermolecular forces vibrational
electronic nuclear, etc…
- more specifics in second semester
- unnecessary for understanding of thermodynamics but sometimes helpful. Macroscopic view of internal energy (19th century view)
- A reservoir of energy within the sample. - The details of the energy are unknown. Internal energy is a state function.
Work
Recall the definition of work from classical mechanics. w=
∫
F d l⋅
Microscopically, work involves the concerted motion of molecules (that is, molecules moving in one direction.)
Work is a transfer of energy, not a quantity of energy. Work is not a state function.
A more thermodynamically useful rearrangement (let F = p⋅A and dV = dl⋅A) is
ex w= −
∫
p ⋅dV- the negative sign is a sign convention (discussed below) - pex is the external pressure (the pressure outside the container)
- pex is used in definition because we really do not know what kind of pressures
develop during a thermodynamic change.
- i.e., pressure currents and eddies are likely for an arbitrary change. Other forms of work are possible
w=
∫
E dq⋅ Electrical work( )
w=
∫
E d VP⋅
Polarization work (e.g., piezoelectricity) w= σ⋅
∫
dA Surface tensionw= τ⋅ θ
∫
d Twisting work Sign conventions for work-w – work done by system (expansion for pV work) +w – work done on the system (compression for pV work) Heat
- macroscopically heat is a thermal energy transfer
- energy transfer usually characterized by temperature and the zeroth law of thermodynamics.
- microscopically heat transfer comes from - inelastic collisions
- energy transfer from one molecule to another in multiple (every) direction Sign conventions for heat
-q – heat transferred away from body (heat lost) +q – heat transferred into body (heat gained)
First Law of Thermodynamics
The physicists that studied energy changes recognized that the energy of an object could be changed via heat or work.
James Joule demonstrated experimentally the law of conservation of energy, that is, that energy is not gained or lost, but merely transferred from one object to another. The law of conservation of energy is also known as the first law of
thermodynamics. Since energy changes can be expressed only as heat or work, the first law of thermodynamics has the mathematical expression
U q w
∆ = + Subtle points to make:
1) q and w are energy changes, writing ∆q and ∆w is improper.
- We can’t refer to object having an amount of heat or work. The object has internal energy, enthalpy, free energy, etc…
2) The form of the first law of thermodynamics depends on the sign convention chosen for heat and work. Older texts state that U∆ = −q w; however, the sign convention for work is different, w= +
∫
pex⋅dVDifferential Form of the First Law
The differential form of the first law is written as dU=ñq +ñw
The differential forms of heat and work are written with the slash to emphasize that they are inexact differentials.
Constant volume heat
During a constant volume process, w = 0. Therefore v
U q
∆ =
Measureable thermodynamic quantities Thermal expansivity
The thermal expansivity is the measure of how a material changes its volume as the temperature changes.
p 1 V V T ∂ α = ∂ Isothermal compressibility T 1 V V p ∂ κ = − ∂
The isothermal compressibility is the measure of how a material changes its volume as the pressure changes. Almost always, volume will decrease with increasing pressure, therefore, a negative sign is included in the definition to allow for the tabulation a positive values.
Both definitions modify an extensive change,
p V T ∂ ∂ or T V p ∂ ∂ , into an
intensive change by dividing the change by the volume.
Enthalpy
BackgroundIn order to fully explain his ideas of free energy, Josiah Willard Gibbs needed to construct an energy state function that had the definition
H = U + PV
Gibbs named the quantity the heat content because a change in the quantity
corresponded to heat gained or lost by a system at constant pressure provided no non-pV work is being done, that is.
p
H q
∆ =
Kamerlingh Onnes eventually named the function, H the enthalpy and the name stuck.
Heat capacity
- Heat is proportional to mass.
- Heat is proportional to temperature difference.
- Proportionality constant for a specific substance is the heat capacity or specific heat.
q=mC T∆
Definition of Constant Pressure Heat Capacity – Cp
Since constant pressure heat is a change in enthalpy, the constant pressure heat capacity can be rewritten as
p p H C T ∂ =∂
This quantity is one of the easily measurable quantities that is very important in thermodynamics.
Definition of Constant Volume Heat Capacity – Cv
Since constant volume heat is a change in internal energy, the constant volume heat capacity can be rewritten as
v v U C T ∂ = ∂
This quantity is another one of the easily measurable quantities that we will use a great deal in thermodynamics studies.
Relationship between Cp and CV
One way of examining the difference between internal energy and enthalpy is by examining the difference between the constant pressure heat capacity and the constant volume heat capacity.
(
)
( )
p V p V p V p p V p p V U PV PV H U U U U C C T T T T T T T U V U P T T T ∂ + ∂ ∂ ∂ ∂ ∂ ∂ − =∂ −∂ = ∂ −∂ =∂ + ∂ −∂ ∂ ∂ ∂ =∂ + ∂ −∂ At this point, we want to find another expression for p U T ∂ ∂
since it is not an easily
measurable quantity. We can find an alternative expression for
p U T ∂ ∂
based the total
differential of the internal energy.
The internal energy is a function of temperature and volume, thus the total differential is
V T U U dU dT dV T V ∂ ∂ =∂ +∂
Dividing the total differential by dT under constant pressure conditions yields
p V p T p V T p U U T U V U U V T T T V T T V T ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + = + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
Two of the partial derivatives in this expression should be familiar: V V U C T ∂ = ∂ and p V V T ∂ = α ∂
. Remember the thermal expansivity is p
1 V V T ∂ α = ∂
Let us substitute our result into the original equation for Cp – CV ,
p V p p V V T p p V T p p T p T U V U U U V V U C C P P T T T T V T T T U V V U V U P P V P V T T V T V ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ − =∂ + ∂ −∂ =∂ +∂ ∂ + ∂ −∂ ∂ ∂ ∂ ∂ ∂ ∂ =∂ ∂ + ∂ =∂ + ∂ = α ∂ +
What is the relationship between Cp – CV for an ideal gas?
p p 1 V 1 nRT 1 nR nR V T V T p V p pV ∂ ∂ α = = = = ∂ ∂ p V T T T T T U nR U nR U C C V p V p p V pV V p V nR U nR 1 U p nR 1 p V p p V ∂ ∂ ∂ − = α ∂ + = ∂ + = ∂ + ∂ ∂ = ∂ + = ∂ +
Remember that for an ideal gas T U 0 V ∂ = ∂ , therefore p V T 1 U C C nR 1 nR p V ∂ − = ∂ + =
The relationship can be rewritten in terms of molar heat capacities.
p V Cɶ −Cɶ =R
This relationship is very important and worth memorizing. Why is Cp greater than CV?
- Heat capacity is substance’s capacity to store energy.
- A substance that can store energy via work has a greater heat capacity than a substance that has its volume kept constant.
Heat Reexamined
Isothermal Heat dq=C dT ⇒ q=0 Isochoric Heat V V dq=C dT ⇒ q=∫
C dTIf CV is independent of temperature, (fair assumption for small temperature changes),
then
(
)
V V V 2 1 V q=∫
C dT=C∫
dT=C T −T =C ∆T Isobaric Heat p p dq=C dT ⇒ q=∫
C dT Adiabatic Heat By definition, q = 0Thermochemistry
Extent of ReactionTo monitor the progress, a variable known as the extant of reaction is defined as i i,0
i
n −n
ξ = ν
where ni is the number of moles of chemical species i, ni,0 is the initial number of
moles of chemical species i and νi is the stoichiometric coefficient of the chemical
species i for the specific reaction occurring.
When the extent of reaction is defined this way, the value of the extant of reaction does not depend on the chemical species used to examine the state.
Example: Consider the reaction P4 (s) + 5 O2 (g) → 2 P2O5 (s). Suppose we start
with 3 moles of phosphorus and 15 moles of oxygen.
a) What is the extant of reaction when 4 moles of P2O5 has been created?
i i,0 i n n 4 mol 0 mol 2 2 − − ξ = = = ν
Note: How much P4 has been consumed to create 4 moles of P2O5? A: 2 moles.
b) What is the extant of reaction when 2 moles of phosphorus has reacted? i i,0 i n n 1mol 3 mol 2 1 − − ξ = = = ν −
The definition includes the stoichiometric coefficient to put the changes of the all the chemical species on an equal footing.
Soon we will be dealing reaction energies and other quantities specific to reactions. The reaction energies need to be intensive quantities and yet each chemical species may have a different amount. Thus reaction energies are defined with respect to the extent of reaction so that the specific chemical species is not important, but the reaction as a whole is important.
Standard States Standard Pressure - pφ
Thermodynamic quantities are measured with respect to a pressure of 1 bar (100.000 kPa)
Standard Temperature - Tφ
Thermodynamic quantities are measured with respect to a temperature of 25 °C (298.15 K)
Standard Concentration- cφ
Thermodynamic quantities are measured with respect to a concentration of 1m (that is 1 molal).
- Molal is used as a concentration unit rather than molar because molal is independent of temperature.
Biological Standard State
The biological standard state is a pressure of 1 bar, a temperature of 37 °C and pH of 7, (that is [H+] = 1.0 × 10-7).
Hess’ Law
Since enthalpy is a state function, we can choose more than one thermodynamic path to calculate a state function.
For chemical changes, Hess’ Law states that the enthalpy of a reaction can be calculated from the enthalpies of all of the chemical step processes needed for the chemical reaction.
In other words, if reactions can be added together to form a resultant reaction, then enthalpies of the step reactions can be added to find a resultant enthalpy of reaction.
Example: Calculate the enthalpy of reaction for the following reaction: 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s), given the reactions below.
2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) ∆H° = -1049 kJ/mol
HCl(g) → HCl(aq) ∆H° = -74.8 kJ/mol
H2(g) + Cl2(g) → 2 HCl(g) ∆H° = -185 kJ/mol
AlCl3(s) → AlCl3(aq) ∆H° = -323 kJ/mol
Rearrange the equations such that their sum is the reaction of interest.
2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) ∆H° = -1049 kJ/mol
6[HCl(g) → HCl(aq)] ∆H° = 6[-74.8 kJ/mol] 3[H2(g) + Cl2(g) → 2 HCl(g)] ∆H° = 3[-185 kJ/mol]
2[AlCl3(aq) → AlCl3(s)] ∆H° = 2[+323 kJ/mol]
2 Al(s) + 6 HCl(aq) + 6 HCl(g) + 3 H2(g) + 3 Cl2(g) + 2 AlCl3(aq)
→ 2 AlCl3(aq) + 3 H2(g) + 6 HCl(aq) + 6 HCl(g) + 2 AlCl3(s)
∆H° = -1049 kJ/mol + 6[-74.8 kJ/mol] + 3[-185 kJ/mol] + 2[+323 kJ/mol]
= -1406 kJ/mol 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s) ∆H° = -1406 kJ/mol
Enthalpy of Formation
The enthalpy of formation is the enthalpy of a formation reaction for a particular substance.
A formation reaction is that where a compound is formed from elements in the naturally occurring state.
2 C(s) + 3 H2(g) + ½ O2(g) → C2H5OH (l)
½ P4 (s) + 5/2 O2(g) → P2O5(s)
7 Fe(s) + 18 C(s) + 9 N2(g) → Fe4[Fe(CN)6]3 (s)
Enthalpy of formation of the elements
By definition, the enthalpy of formation of an element in its natural state is zero. ∆fH°(B(s)) = 0 kJ/mol ∆fH°(C(graphite)) = 0 kJ/mol ∆fH°(Br2(l)) = 0 kJ/mol
∆fH°(S8(s)) = 0 kJ/mol ∆fH°(Ag(s)) = 0 kJ/mol ∆fH°(Xe(g)) = 0 kJ/mol
Reaction enthalpies can be calculated as a stoichiometric sum of enthalpies of formation. This technique is an application of Hess’ law.
Example: Calculate the enthalpy of reaction for the following reaction: CH3COOH(l) + C4H9OH(l) → C4H9OOCCH3(l) + H2O(l),
given the reactions below.
2 C(s) + 2 H2(g) + O2(g) → CH3COOH(l) ∆H° = - 483.52 kJ/mol
4 C(s) + 5 H2(g) + ½ O2(g) → C4H9OH(l) ∆H° = - 328 kJ/mol
6 C(s) + 6 H2(g) + O2(g) → C4H9OOCCH3(l) ∆H° = - 609.6 kJ/mol
H2(g) + ½ O2(g) → H2O(l) ∆H° = - 285.83 kJ/mol
Note all of the above reactions are formation reactions.
Rearrange the equations such that their sum is the reaction of interest.
CH3COOH(l) → 2 C(s) + 2 H2(g) + O2(g) ∆H° = + 483.52 kJ/mol C4H9OH(l) → 4 C(s) + 5 H2(g) + ½ O2(g) ∆H° = + 328 kJ/mol 6 C(s) + 6 H2(g) + O2(g) → C4H9OOCCH3(l) ∆H° = - 609.6 kJ/mol H2(g) + ½ O2(g) → H2O(l) ∆H° = - 285.83 kJ/mol CH3COOH(l) + C4H9OH(l) + 6 C(s) + 6 H2(g) + O2(g) + H2(g) + ½ O2(g) → 2 C(s) + 2 H2(g) + O2(g) + 4 C(s) + 5 H2(g) + ½ O2(g) + C4H9OOCCH3(l) + H2O(l)
∆H° = + 483.52 kJ/mol + 328 kJ/mol - 609.6 kJ/mol - 285.83 kJ/mol = - 84 kJ/mol CH3COOH(l) + C4H9OH(l) → C4H9OOCCH3(l) + H2O(l) ∆H° = - 84 kJ/mol
Calorimetry
Constant Pressure Calorimeter - coffee cup calorimeter - open to atmosphere
- appropriate for solution chemistry Constant Volume Calorimeter
- bomb calorimeter - sealed and isolated
- appropriate for gas phase chemistry Adiabatic Calorimeter
- heat measured by temperature needed to keep thermal energy constant. Process of measuring heat of combustion with bomb calorimeter
1. Mass wick used to start combustion. 2. Mass water used to absorb heat.
3. Mass standard (benzoic acid) to be combusted
4. Add water to combustion chamber to ensure that water from combustion will be in liquid phase.
5. Load standard and seal
6. Fill combustion chamber with oxygen (≈ 25 atm)
7. Begin combustion and record temperature change of water. 8. Calculate heat capacity of calorimeter from standard
- need to account for heat of wick, and heat capacity of water. 9. Repeat process with sample.
- with heat from wick, heat from the water and heat from the calorimeter, the heat of combustion of sample can be calculated.
Hess’ law is often used to the enthalpy of formation of a substance after its enthalpy of combustion has been calculated.
Example: 0.523 g of the military explosive, cyclotetramethylenetetranitramine (HMX), C4H8N8O8 is combusted in a bomb (!) calorimeter and an internal
energy change of – 4.620 kJ is measured. Calculate the enthalpy of formation for HMX.
First write a balanced equation for its complete combustion.
C4H8N8O8(s) + 2 O2(g) → 4 CO2(g) + 4 H2O(l) + 4 N2(g)
Next calculate the molar internal energy of reaction from the experimental data. 4.62 kJ 296.155g
2616 kJ mol 0.523g mol
To calculate the enthalpy of the reaction, we need to return the definition of enthalpy.
( )
(
)
( )
H U PV U nRT U n RT
∆ = ∆ + ∆ = ∆ + ∆ = ∆ + ∆
Reexamining the balanced chemical equation, we see that 8 moles – 2 moles = 6 moles of gas has been produced.
C4H8N8O8(s) + 2 O2(g) → 4 CO2(g) + 4 H2O(l) + 4 N2(g)
( )
0.008314kJH U n RT 2616 kJ mol 6 298.15K
mol K 2.478kJ
2616 kJ mol 6 2616 kJ mol 14.87 kJ mol 2601kJ mol mol ∆ = ∆ + ∆ = − + ⋅ = − + = − + = −
The molar enthalpy of reaction can also be written in terms of the molar enthalpies of formation.
4 ×∆fH°(CO2(g)) + 4 ×∆fH°(H2O(l)) - ∆fH°(C4H8N8O8(s)) = ∆rH°
Rearranged, the equation becomes,
∆fH°(C4H8N8O8(s)) = 4 ×∆fH°(CO2(g)) + 4 ×∆fH°(H2O(l)) - ∆rH°
From table, we find the enthalpies of formation for CO2(g) and H2O(l)
∆fH°(C4H8N8O8(s)) = 4 × (-393.5 kJ/mol) + 4 × (-285.8 kJ/mol) - (-2601 kJ/mol)
= -116 kJ/mol
Temperature Dependence of Internal Energy and Enthalpy
Recall that V V U C T ∂ =∂ and p p H C T ∂ =∂
These relationships imply that we can find the internal energy or enthalpy at a nonstandard temperature as long as we know the heat capacity.
V
dU=C dT and dH=C dTp Integrating both sides of these equations yields
V
U C dT
For large temperature changes, we can’t assume that the heat capacity is independent of temperature. Thus to calculate the internal energy or enthalpy at a nonzero
temperature, we need the temperature dependence of the heat capacity.
Example: Calculate the change in enthalpy of H2 (g) from 373 K to 1000 K, given
that the constant pressure heat capacity has the form Cp d eT fT 2
−
= + +
where d = 27.28 J/K mol, e = 0.00326 J/K2 mol and f = 0.00050 J K/mol.
(
)
(
)
(
)
(
)(
)
(
)
1000 1000 1000 2 2 1000 K 373 K p 373 373 373 2 1000 1000 1000 2 2 373 373 373 2 2 H H H C dT d eT fT dT d dT eT dT fT dT eT f e 1 1 d T | | | d 1000 373 1000 373 2f 2 T 2 1000 373 0.00326 J K mol 27.28 J K mol 627 K 860871K 2 2 0.00050 − − ∆ = − = = + + = + + = + − = − + − − − ⋅ = ⋅ + −∫
∫
∫
∫
∫
(
)
(
1)
6 J K mol 0.0016810K17105 J mol 1403 J mol 1.68 10 J mol 18508 J mol 18.51kJ mol
− − ⋅ − = + + × = = Kirchoff’s law
( )
0(
)
T r r i p,i i 298 H T H 298 K C dT ∆ = ∆ +∫
∑
νThat is by taking a stoichiometric sum of the heat capacities and integrating over the temperature range, we can find the correction to the reaction enthalpy at a non standard state temperature.
Example: 471 kJ/mol is the reaction enthalpy under standard conditions for the following reaction: 2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g). What is
the reaction enthalpy at 1000 K?
According to the NIST Webbook Internet site, the constant pressure heat capacities of the chemical species in the above smelting process can be fitted according to the Shumate equation, Cp A B T C T2 D T3 E T 2
−
= + + + +
The values for A, B, C, D and E for each species are given below.
A B C D E Fe2O3 93.43834 0.108358 -5.08645 × 10-5 2.56 × 10-8 -1610000 C 10.68 0 0 0 0 CO2 24.99735 0.055187 -3.36914 × 10-5 7.95 × 10-9 -137000 Fe 18.42868 0.024643 -8.91 × 10-6 9.66 × 10-9 -12600 Sum -70.2099 0.047418 -3.50 × 10-5 1.13 × 10-8 2762174
(
)
(
)
(
)
(
)
1000K 0 2 3 2 r r i i i i i i i 298K 0 1000 i 2 1000 i 3 1000 i 3 1000 1 1000 r i i 298 298 298 298 i 298 i H 1000 K H 298 K A B T C T D T E T dT B C D H 298 K A T | T | T | T | E T | 2 3 4 − − ∆ = ∆ + ν + + + + = ∆ + ν + + + − ∑
∫
∑
(
)
(
)(
) (
)
(
)
(
) (
) (
) (
)
2 2 r 5 8 3 3 4 4 0.047418 471kJ H 1000 K 70.2099 1000 298 1000 298 mol 2 3.50 10 1.13 10 1 1 1000 298 1000 298 2762174 3 4 1000 298 − − ∆ = + − − + − − × × + − + − − − (
)
r 471kJ 49.3kJ 21.6kJ 34.1kJ 11.2kJ 6.5kJ 413.9kJ H 1000 Kmol mol mol mol mol mol mol
∆ = − + − + − =
Kirchoff’s Law can be stated with in a differential form as well.
Include V V U C T ∂∆ = ∆ ∂ and p p H C T ∂∆ = ∆ ∂
Bond Enthalpies
A bond enthalpy is the energy needed to separate two atoms.
Tabulated bond enthalpies are average values calculated from the dissociation of many different compounds.
Thus tabulated bond enthalpies are approximate values.
*However, they can be useful for approximating enthalpies of reaction because most of the chemical energy of a compound is held in its bonds.*
Work Reexamined
Isochoric WorkpV work involves a volume change. w= −
∫
pex⋅dVSince a constant volume process has no volume change, w = 0 if there is no other work (no non-pV work).
No other assumptions have been made about the system. Isobaric Work
Since the pressure is constant, it has no dependence on the volume. Thus the pressure can pulled out of the integral. Subsequently, the integral of dV is V2 – V1 = ∆V
(
)
ex ex ex 2 1 ex
w= −
∫
p ⋅dV= −p∫
dV= −p V −V = − ∆p VThe only other assumption made is that the system does not have any non-pV work. Isothermal Work
In general we need to know the relationship between pressure and volume to perform the calculation, i.e, we need an equation of state. Let us do the calculation for an ideal gas under reversible conditions.
2 ex 1 nRT dV V w p dV p dV dV nRT nRT ln V V V = − ⋅ = − ⋅ = − ⋅ = − = −
∫
∫
∫
∫
Note restrictions on the applicability of the calculation. 1. no other non-pV work
2. ideal gas
3. reversible conditions Adiabatic Work
dU=ñq +ñw ⇒ dU=ñw ⇒ w= ∆U
Refrigeration and the Joule-Thomson coefficient
Joule coefficient - J U T V ∂ µ =∂ - internal pressure q = 0, w = 0 ⇒ ∆U = 0 isoenergetic process Joule-Thomson coefficient - JT H T p ∂ µ = ∂ - isoenthalpic, adiabatic - µ = 0 for an ideal gas Classic Experimental Apparatus- measure µJT directly. f i V 0 f i i f i i f f V 0 f i i i f f i U U U p dV p dV p V p V U p V U p V H H ∆ = − = − − = − ⇒ − = − ⇒ =
∫
∫
Modern Experimental Apparatus - measure µJT via T H p ∂ ∂
- use isothermal conditions rather adiabatic conditions.
isothermal conditions: T1 = T2 T T T p p H H T p p H H T p H T 1 T T 1 p H H p H T p C H T T ∂ ∂ − − ∂ ∂ ∂ = − ⇒ ∂ = − = ∂ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ T JT p H T C ∂ −∂ ⇒ µ =
Isoenthalps on T- p plot. - inversion temperature
- µ > 0 for substance to be used as a refrigerant
Real Gases Reexamined
Calculations on van der Waals gas - pV work ex w= −∫
p ⋅dV 2 2 nRT an p V b V = − − 2 2 ex 2 2 2 2 1 2 1 nRT an nRT an w p dV p dV dV dV dV V b V V b V V b 1 1 nRT ln an V b V V = − ⋅ = − ⋅ = − − ⋅ = − ⋅ + ⋅ − − − = − − − − ∫
∫
∫
∫
∫
- Difference in heat capacities
p v p v H U C C T T ∂ ∂ − = − ∂ ∂ T V p T V p T p V p T p V V p T U U 1 U U dU dV dT dU dV dT V T T V T U U V U T U V U T V T T T V T T U U U T T V ∂ ∂ ∂ ∂ =∂ +∂ ⇒ ∂ =∂ +∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⇒ = + = + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⇒ = − ∂ ∂ ∂ p V T ∂ ∂
( )
p p p p p pV H U U V H U pV p T T T T T ∂ ∂ ∂ ∂ ∂ = + ⇒ ∂ =∂ + ∂ =∂ + ∂ [
]
p v p p p T p p T p p T T p T T U V U U V C C p T T T V T V U V V U p p T V T T V V U U p V p V p T V V ∂ ∂ ∂ ∂ ∂ − =∂ + ∂ −∂ −∂ ∂ ∂ ∂ ∂ ∂ ∂ = ∂ +∂ ∂ =∂ +∂ ∂ ∂ ∂ = + = α + = α + π ∂ ∂ ∂ Internal Energy Reexamined
Why is ∆U = qv under isochoric conditions?
ex V
dU=ñq +ñw ⇒ dU=ñq - p dV ⇒ dU=ñq ⇒ ∆ =U q
Assuming no non-pV work.
Isobaric Internal Energy
v ex v ex v
dU=ñq +ñw ⇒ dU=C dT−p dV ⇒ ∆ =U
∫
C dT−∫
p dV=∫
C dT− ∆p VNote: no assumptions about the equation of state.
Isochoric Internal Energy
ex v v
dU=ñq +ñw ⇒ dU=ñq−p dV ⇒ dU=ñq ⇒ ∆ =U
∫
C dT=qNote: assumption of no non-pV work
Isothermal Internal Energy
v dU ñq +ñw = C dT
dT 0 dU 0
=
Enthalpy Reexamined
Why is ∆H = qp under isobaric conditions?
( )
p H U pV dH dU d pV =ñq +ñw + p dV + v dp dH =ñq + -p dV + p dV + v dp dH ñq H q = + ⇒ = + ⇒ ⇒ = ⇒ ∆ =Assuming no non-pV work and reversible conditions
Isobaric Enthalpy
( )
p p H U pV dH dU d pV =ñq +ñw + p dV + v dp dH =ñq p dV + p dV dH ñq H C dT q = + ⇒ = + ⇒ − ⇒ = ⇒ ∆ =∫
=Assuming no non-pV work and reversible conditions
Isochoric Enthalpy
( )
p H U pV dH dU d pV =ñq +ñw + p dV + v dp dH =ñq p dV + p dV + v dp dH ñq + v dp H C dT v dp = + ⇒ = + ⇒ − ⇒ = ⇒ ∆ =∫
−∫
For ideal gas under reversible conditions
p p nRT H C dT v dp C dT dp p ∆ =
∫
−∫
=∫
−∫
Why not 2 p 1 p H C dT nRT ln p ∆ =∫
− ?- temperature has a dependence on the pressure.
Isothermal Enthalpy
( )
pH= +U pV ⇒ dH=dU d pV = C dT+ ⇒ ∆ =H 0
Important: All of the above enthalpies assume that the system has no non-pV work and reversible conditions (so that P = Pex)
Summary
Isothermal ⇒∆T = 0 Isobaric ⇒∆p = 0 Isochoric ⇒∆V = 0 Adiabatic ⇒ q = 0 Definition of work: w= −∫
pex⋅dVDefinitions for heat: qp = ∆H qV = ∆U
Sign conventions for heat and work.
-w – work done by system (expansion for pV work) +w – work done on the system (compression for pV work) -q – heat transferred away from body (heat lost)
+q – heat transferred into body (heat gained)
Reversibility p = pex Extent of Reaction i i,0 i n −n ξ = ν
Definition of Heat Capacities
p p p q H C T T ∂ ∂ =∂ =∂ v v v q U C T T ∂ ∂ =∂ =∂ First Law: dU=ñq+ñw ⇒ ∆ = +U q w ⇒ ∆ =U C dT−p dVex
Partial Derivatives to be acquainted with: α, κ, Cp, Cv, πT, µJ,µJT
p 1 V V T ∂ α = ∂ T 1 V V p ∂ κ = − ∂ p p H C T ∂ =∂ v v U C T ∂ =∂ J U T V ∂ µ =∂ JT H T p ∂ µ =∂
and internal pressure T
T U V ∂ π =∂
Natural variables of U and H and total differentials
Internal energy can be expressed as a function of V and T U(V,T) Enthalpy can be expressed as a function of p and T H(p,T)
- both of these functional dependences are not unique - will consider other natural variables in the next chapter. Total differential can be expressed as
V T U U dU dT dV T V ∂ ∂ =∂ +∂ p T H H dH dT dp T p ∂ ∂ =∂ + ∂ Hess’ law
Temperature Dependence of Enthalpy and Internal Energy
V U C dT ∆ =
∫
and ∆ =H∫
C dTp Kirchoff’s law( )
0(
)
T r r i p,i i 298 H T H 298 K C dT ∆ = ∆ +∫
∑
ν V V U C T ∂∆ = ∆ ∂ and p p H C T ∂∆ = ∆ ∂ Miscellaneous Applications of Heat Capacities
p V Cɶ −Cɶ =R p v C C γ =
