Chem201 2 First Law

2. The First Law - 1 C. Rose-Petruck, Brown University, 1999

2. The First Law

© C. Rose-Petruck, Brown University, 2000


A brief philosophical excursion: Why can we just write down a postulate, let alone call it a law, and assume that it is somehow related to natural processes? This postulate like any scientific law is the essence of what has always been observed by many observers for any systems to which the law applies. But unlike mathematical laws that can often be proven without double by applying logic, laws in natural sciences are proven only to the extend that none has found a counterexample yet. Could this postulate, or other laws, be wrong? Could it be that it doesn't describe nature's mechanisms correctly? Yes, it could be wrong or incomplete. But, again, none has found it to be wrong yet.

We, in this class, might not yet have the experience and intuition yet to fully accept this postulate, or 1st law of thermodynamics, in this general form. Therefore, we shall derive it phenomenologically for systems that can do mechanical work.

2.1 Heat and Work, an Overview Heat and work, both, are energy. The difference is just the amount of ordered motion during the energy transfer.

Both heat and work are forms of energy. However, ordered motion, i.e. work, is technically useful.

2.2 Heat Heat is closely related to, but distinct from temperature. The temperature of two objects that are brought


into contact tends to even out, which is accomplished by energy flowing from one object to the other in the form of heat. The relationship between temperature and heat is defined as:

VV C

dTQd = (2.1)

PP C

dTQd = (2.2)

In these equations we have introduced the heat capacity of substances. Clearly, these equations are only appropriate for temperature changes that don't induce phase transitions, or chemical reactions.

There are two equations for heat capacities, one with a subscript of V, the other with a subscript of P. The subscripts symbolize the parameter that is held constant: the left equation is for a process at constant system volume, while the right one is for processes at constant pressure. Most everyday processes occur at constant pressure, namely the atmospheric pressure, and in such cases (2.2) should be used. This brief introduction is sufficient for now; we shall later derive how heat capacities from changes of state functions such as the internal energy.

2.3 Work As discussed above, work is a general phenomenon and refers to ordered motion of particles of systems. Many different forms of work exist; a few are listed below:

Type of work dw Comment Units

Expansion -Pex dV Pex = external pressure V = volume

N/m2 m3

Surface expansion γ dσ γ = surface tension dσ = surface area

N/m m2

Extension f dl f = tension dl = length

N m

Electrical potential Φ dq Φ = electrical potential dq = charge

V C

Electrical dipole E dp E = electrical field dp = electric dipole moment

V/m mAs

Magnetic dipole B dm B = magnetic field dm = magnetic dipole moment

kg/s2A mAs

Chemical µ dN µ = chemical potential dN = number of moles

J/mol mol

In the following we shall illustrate the first law focusing on a specific example, the compressions of a gas in a cylinder.


Illustration: Expansion of gas

Work is done when and object, e.g. a system's wall, moves against an opposing force. This is equivalent to an ordered motion done by the system on the surroundings or vice versa.


FdsWd −= (2.3)

The infinitesimal motion ds of the wall transfers the amount dW of energy in the form of work.

Sign convention: Consider the system to have an energy-account. A negative sign means: Withdraw of energy (e.g., work done by the system). A positive sign means: Deposit of energy (e.g., work done on the system).

dVPAdsPWd exex −=−= (2.4)

Imperfect differentials: In thermodynamics differentials like 2.2 are called imperfect differentials and are symbolized by a bar through the d. While they represent infinitesimal changes, in this case of work, they cannot readily be computed without additional information. Integration can be carried out but the result of the integration is dependent on the integration path. For more detail see: CJA, pp.14

Let's turn our attention to the internal energy of the system.

2.4 Internal Energy The internal energy U of a system can be defined as the capacity to do work.


QdWddU += (2.5)

This is the 1st law of thermodynamics for diabatic processes.

WddU = (2.6)

This is the 1st law of thermodynamics for adiabatic processes. Such a process allows to measure the change of the internal energy by the mechanical work that is done.

2.5 Quasistatic, Reversible and Irreversible Processes

Example: Expansion of gas We assume that the outside pressure is always nearly equal to the inside pressure, a situation called quasistatic. In such a case, the motion of the movable wall separating the system from the surroundings is infinitely slow. Compression can easily be reversed by expansion; i.e. the process is reversible. If the pressure difference is larger the motion of this wall will not be slow any more and complicated effects, such as shock waves or turbulence, can occur. Such processes are certainly irreversible. Therefore, quasistatic processes are reversible.

Then:


∫∫ −=−=path

inpath

ex dVPdVPW (2.7)

Reversible heat transfer, e.g., refers to thermal contact between systems with infinitesimally small temperature difference.

In general, systems in equilibrium will never undergo changes of state. However, since we want to study interactions and the resulting changes of state, systems must be allowed to pass though non-equilibrium states. Unfortunately, non-equilibrium states are not well defined in thermodynamics. We, therefore, divide any process into infinitesimal steps with infinitesimally small changes, which then can be considered to be quasistatic. No real process is truly quasistatic, but we can design processes to be approximately quasistatic to almost any desired accuracy. We can then write

exinpath

PPPdVPW ==−= ∫ with (2.8)

Keep in mind, that "slow" and "quasistatic" is always judged relative to the relaxation time of the system. For instance, solids with good thermal conductivity will equilibrate heat faster then solids with low thermal conductivity.

The Enthalpy We just introduced the first law, which essentially states the conservation of energy:

QdWddU += (2.5)

We already have discussed the forms of heat, Q, and work, W. Let us focus on the kinds of systems relevant to chemical reactions, namely those where the only work source is volume work (that is, we don't include other forms of work like surface work, tension work, and for now even, the chemical work). Furthermore, let's consider reversible processes. The expression for the work is then:

To measure a change in the internal energy of a process one might now be tempted to perform the process at constant volume, and measure the heat exchanged between the system and the surroundings under those conditions. For processes at constant volume, we have dV = 0, and therefore the volume work term is zero. One would obtain:

VQddU = (2.9)

For the heat we had expressions relating the heat at constant volume or pressure to the heat capacities. Those expressions are valid only if there are no chemical reactions, and no phase transitions. (Of coarse, a reaction will change the heat capacity). In a general expression we cannot substitute those equations into our formula for the energy; and we thus leave the term dQ.

Unfortunately, in many situations of Chemistry, Biology, and Physics, the proposition to do a process at constant volume makes life difficult. For instance, in many chemical reactions the density changes, and therefore the system changes its volume unless very large pressures are applied. Reaction containers would need to have very thick walls made of stainless steel. Clearly, our theory so far does not relate well to real life chemistry: one typically does not measure the heat of a process by keeping the volume constant. Rather, most processes are at constant pressure, under isobaric conditions. Since under such conditions dP=0 it follows:


( )�

PdVVdPPVdPdV +==0

(2.10)

Now we can rewrite (2.5):

( )�

( )PPPPVd

PVdQdQdWddU −=+=−

(2.11)

We rearrange to:

( ) ( ) PPPPP dHPVUdPVddUQd =+=+= (2.12)

where the quantity

PVUH += (2.13)

is called the Enthalpy. It is a measure of the energy, just like U, but it is the appropriate one for processes at constant pressure. It turns out that because much of Chemistry is at constant pressure, the enthalpy is a function with very much utility to us.

The Heat Capacities We can now rewrite the expressions (2.1) and (2.2) for the heat capacities under conditions of constant volume and constant pressure, respectively:

VV T

UC

∂∂= (2.14)

PP T

HC

∂∂= (2.15)

Inserting (2.15) into (2.5) gives:

dVPVUdTCQ

TV

+

∂∂+=∂ (2.16)

We can use this equation to consider an isobaric process:

PTV

PP T

VPVUC

TQC

∂∂⋅

+

∂∂+=

∂∂= (2.17)

Often we can use an equation of state to simplify (2.17) by calculating PT

V

∂∂

.

Heat capacities for ideal gases and gases at low pressure

Keep in mind, that an ideal gas consists of atoms and molecules that don't interact with each other (except for hard-sphere collisions). Therefore, low pressure, real gases, that where the molecules and atoms are far apart, tend to exhibit ideal gas behavior.


At low pressures gases behave approximately ideal. We will show in a later chapter that the molar heat capacity cV= CV/NA at constant volume is a function of temperature only. This result will follow directly from the equation of state for the ideal gas. It can be shown that for a gas each degree of freedom contributes 1/2R to cV. R is the universal gas constant defined as kNA=8.3143 J/mol K. Consequently, for monatomic gases, such as Noble gases, cV is 3/2R over a large range of temperatures well above the boiling point. Diatomic gases exhibit cV of 5/2R near room temperature that falls down to 3/2R for very low temperatures and increases to 7/2R for high temperatures. Gases with more complex molecules exhibit an even more complicated behavior. Nevertheless, cV is still an increasing function of temperature.

The internal energy of a monatomic ideal gas is:

nRTU23= (2.18)

Therefore:

VV T

UC

∂∂= (2.19)

0=

∂∂

TVU

. (2.20)

Using the equation of state pV=nRT

PnR

TV

P

=

∂∂⋅ (2.21)

Equation (2.17) can now be rewritten into:

nRCC VP += (2.22)

Therefore,

RRcc VP 25=+= (2.23)

The ration of the heat capacities is thus

35==

V

p

CC

γ (2.24)

Here are a couple of heat capacities at constant pressure of common substances:


Notice that gases and solids have similar heat capacities. Just the liquids, especially those with many atoms per molecule, have particularly high heat capacity values. We can see in the table below that the heat capacities relate to microscopic energy level structures in a very direct way: the more low lying states there are, the higher the heat capacity.

Classical molar heat capacities of gases:

Utrans Urot Uvib Utotal CV CP

V

P

CC

γ =

Monatomic 3/2 RT 3/2 RT 3/2 R 5/2R 5/3

Diatomic 3/2 RT RT RT 7/2 RT 9/2 R 9/2R 9/7

Polyatomic 3/2 RT 3/2 RT xRT (3+x)RT (3+x)R (4+x)R 1

34 →

++ ∞→x

xx

with x=3N-6 for non-linear polyatomic molecules with N atoms and x=3N-5 for linear polyatomic molecules with N atoms.

The Classical heat capacities, while being based on reasonable models, overestimate the heat capacities because the quantum mechanical nature of vibrations and rotations is ignored. Linear motions of particles can always be described classically, i.e. the distance between energy states for the linear motion is essentially zero. In contrast, distance between energy levels of rotation and vibration are noticeable, i.e. they will not be populated unless a molecule receives certain minimum amounts of rotational or vibration energy. This can be see from the next table. In this table the distance between rotational and vibrational energy levels are shown. To put these quantities in perspective they are compared to the temperatures at which the energy of the 1st excited level is equal to kT. According to equation (1.11) this corresponds to a temperature at which 1/e = 36.8% of the molecules are in the 1st excited state.


∆Erot [J/mol] T1st_rot [K] ∆Evib [kJ/mol] T1st_vib [K]

H2 1420 171 51.0 6140

HCl 254 30.0 35.8 4300

O2 34.6 4.2 18.8 2260

N2 48.1 5.8 28.1 3380

Cl2 5.8 0.7 6.8 810

Br2 1.9 0.23 4.0 470

I2 0.9 0.11 2.6 310

CO 46.0 5.5 26.0 3120

NO 41.0 4.9 22.8 2740

The Standard State Before we go on we have to establish some standard condition at which enthalpies are measured. One could choose any set of parameters, but it is important that all chemists refer to the same conditions. The standard conditions that have been chosen are:

Pressure: 1 bar = 105 N/m2

Temperature: depends on process; 298 K if possible.

2.6 The First Law: Examples

Example 1: Work by an electric field Let's assume an electric capacitor with parallel plates as depicted in the picture.


A electrical potential U is provided by the voltage source, which is connected to the capacitor. This potential creates an electric field

dU /nE = . (2.25)

This field generates and electrical displacement

PEED +== 00 εεε r . (2.26)

The total charge q on the plates of the capacitor is given by

Aq nD ⋅= . (2.27)

If the charge q on the capacitor is changed by a small amount, the work done is:

DEDE dVdAdUdqWd ⋅=⋅== . (2.28)

Using (2.26) we obtain:

( )VddWd PEEE ⋅+⋅= 0ε . (2.29)

This is the total work done on the volume inside the electric field. The first term represents the change of energy without any material inside the capacitor. The work done on the dielectric material is thus


VdWd PE ⋅= . (2.30)

This can be rewritten as

pE dWd ⋅= . (2.31)

Example 2: Enthalpies of Bond formation With modern techniques it is possible to study the fragmentation of molecules, and measure the energy required to break individual chemical bonds. Such studies have to be conducted in the gas phase, so as to avoid solvent effects. The energy required to break a bond is called bond enthalpy. Example:

molkJ416)()()( 0

34 =∆+→ HgHgCHgCH (2.32)

The enthalpy of the reaction is a direct measure of the energy of the bond between the carbon and the hydrogen.

It turns out that many chemically similar bonds have similar bond enthalpies. One conveniently takes the average over a number of representative molecules to assemble tables. Bond enthalpies are of coarse directly related to quantum mechanical bond energy of molecules. Compare the bond enthalpies to the table of binding energies.

There is a slight difference between the bond enthalpies and the bond energies; the difference arises from the fact that the macroscopic measurement contains other small energy terms relating to the


pressure of the gas; as apparent, these terms are small compared to most bond enthalpies. We can estimate these differences the following way:

If a gas reaction involves a change of ∆n moles of gases in the system, then we can write for an ideal gas

nRTPV ∆=∆ . (2.33)

On the other hand, equation (2.13) implies

PVUH ∆+=∆ . (2.13)

We then get:

RTnUH ∆+=∆ . (2.34)

RT at 298K 2.5kJ/mol. Therefore, the bond enthalpies should be approximately the bond energies plus 2.5kJ/mol.

Example 3: Enthalpies of phase transitions Phase transitions are processes in which a substance changes from one state to another. Examples include the melting of a solid, the vaporization of a liquid, and the sublimation of a solid. There are also examples of phase transitions between solid states with different forms. More details about phase transitions later in the semester.

Phase transitions are always connected with an enthalpy change. For example, to melt ice to liquid water one must provide an amount of energy equal to the enthalpy of melting:

molkJ0.6)K273()()( 0

22 =∆→ mHlOHsOH (2.35)

The enthalpy is the energy that is required to melt one mole of ice at 273 K to liquid water, also at 273 K. (This is an example for a process that we cannot measure at 298 K!)

The following table shows some enthalpies of phase transitions, at their respective transition temperature:


Notice how helium has a particularly small enthalpy of vaporization, and also a very low boiling point. This is a reflection of the very weak interatomic bonds between helium atoms. While the bond enthalpies were macroscopic measures of intramolecular bond strengths, the enthalpy of vaporization is the macroscopic manifestation of the intermolecular energy! Helium has very weak interatomic potentials, and thus a low boiling temperature and a small enthalpy of vaporization. Among the molecules the hydrogen has a very small enthalpy of vaporization. This is consistent with the small intermolecular attraction. The larger molecules have much stronger intermolecular attractions, which is reflected in the higher boiling points and higher enthalpies of vaporization. Water has very strong intermolecular bonds, and therefore a fairly high boiling point and a large enthalpy of vaporization. The water-water bond (hydrogen bond) is very strong only if the hydrogen points toward the oxygen of the neighboring molecule.

Example 4: Enthalpies of reaction We can now discuss the enthalpies of chemical reactions. Most chemical reactions are associated with some heat exchange between the molecules and the surroundings. Reactions that release heat to the surroundings are called exothermic, while reactions that take up heat from the surroundings are called endothermic. Example:

molkJ566)K273()()(2)(2 0

22 =∆+→ HgOgCOgCO (2.36)

In this reaction heat is consumed; it is endothermic. Since heat must go from the surroundings to the reaction vessel, the enthalpy of reaction is positive. The reverse reaction, the combustion of CO to form CO2 is exothermic and would have a negative enthalpy of reaction.


Clearly we can understand the enthalpy of the reaction as a combination of the bond enthalpy of the cleavage of one of the C-O bonds, and simultaneous formation of a new O-O bond. The implicit assumption is that the contributions to the enthalpy are additive. This almost obvious assumption is known as Hess's law, and it is a direct consequence of the fact that the enthalpy is a state function. This law comes in very handy in situations where reaction enthalpies are not easily measured. An example is the partial combustion of graphite to carbon monoxide:

)()(21),( 2 gCOgOgrsC →+ (2.37)

Under usual lab conditions the combustion proceeds completely to carbon dioxide, so that one cannot measure the combustion to carbon monoxide directly. Using Hess's law one can add reaction equations to form new equations:

molkJ110)()(

21),(

_______________________________________________________mol

kJ283)(21)()(

molkJ393)()(),(

02

022

022

−=∆→+

=∆+→

−=∆→+

HgCOgOgrsC

HgOgCOgCO

HgCOgOgrsC

(2.38)

Example 5: Standard-State Enthalpies Using many chemical reactions it is possible to write down the relative enthalpies of different molecules. If one synthesizes a new molecule one can get it’s enthalpy by measuring the heat of the reaction, and looking up the enthalpies of the reactants. This scheme is very useful to assess the feasibility of new reactions, and estimate equilibrium constants. However, it is necessary to define some standard state to which all other molecules can be referenced. By convention, one uses as a standard state the state of the element in its most stable form at 298 K: that enthalpy is set to zero.

It is then possible to assign to all molecular substances an energy that can be obtained from the reactions of their generation from the elements. Example:

molkJ285)(

21)( 0

222 −=∆→+ HOHgOgH (2.39)

The enthalpy of the reaction from the elements (both of which are defined to have ∆H0= 0)

is called the standard enthalpy of formation. It is important to reference the state of the molecule; water in the gas phase, for example, has a different enthalpy of formation than water in the liquid state.

Example 6: Electrical Motor We look at a battery and a motor. The battery provides electrical energy to the motor, which in turn generates mechanical energy and heat. Suppose that E(el) is 17 kJ, of which 15 kJ is used to provide mechanical work to some surroundings, and 2 kJ is emitted as heat.


1. Define the system as battery plus motor:

W = -15 kJ

Q = -2 kJ

So, the system has a change in internal energy of :

U=-15 kJ - 2 kJ = - 17 kJ

2. Define the system as motor only:

W = -15 kJ

Q = -2 kJ

E(el) = 17 kJ

So, the system has a change in internal energy of :

U=-15 kJ - 2 kJ + 17 kJ= 0

Unless the motor keeps some of the energy to heat itself up, it will not experience a change in the internal energy.

Note: in either case, this is an irreversible process, as the motor is faster than 'infinitely slow'. Let's


discuss that in more detail in the next section.

2.7 Problems Problem 2:

1. Derive the equation of state for an reversible, adiabatic expansion of an ideal gas. This equation is γpV =const or γγ

finalfinalinitialinitial VpVp = . Please resist the urge to look up the derivation in some thermodynamics book!! Hint: Use equation (2.5) for a reversible, adiabatic process, the equation of

state for the ideal gas, and (2.24), i.e., V

PC

C=γ . Remember that reversible processes are always

progressing though quasi-equilibrium sates. This is, the initial pressure and the final pressure can be assumed to be equal. Begin your derivation by finding an expression that relates the adiabatic work to the temperature change of the system.

Solution to Problem 2

Chem201 2 First Law

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