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Faculty of Engineering

Cairo University

Physical Chemistry

Second year Metallurgy

CHAPTER TWO

The First Law of Thermodynamics

Contents:

- 2.1 Introduction
- 2.2 The Relationship Between Heat and Work
- 2.3 Internal Energy and The First Law of Thermodynamics
- 2.4 Constant – Volume Processes
- 2.5 Constant – Pressure Processes & Definition of Enthalpy (H)
- 2.6 Heat Capacity
- 2.7 Reversible adiabatic processes for a real gas
- 2.8 reversible isothermal processes

2.1 Introduction

In a frictionless kinetic system of interacting elastic bodies, kinetic energy is conserved; i.e., a collision between two of the bodies of this system results in a transfer of kinetic energy from one to the other; thus, the work done by one equals the work done by the other, and the total kinetic energy of the system is unchanged as a result of the collision

If the kinetic system is in the influence of the gravitational filed , thus the sum of the kinetic and potential energies of the bodies is constant ; however, as the result of possible interactions kinetic energy may be converted to potential energy and vice versa but the sum of the two dynamitic energies (kinetic and potential) remains constant

If, however, frictional energies are operative in the system then with continues collision and interaction among the bodies, the total dynamic energy of the system decreases and heat is produced. It is thus reasonable to expect that there is a relationship between the dynamic energy lost and heat produced

The establishment of this relationship laid the foundations for the development of the thermodynamic subject

2.2 The Relationship Between Heat and Work

In 1798, Count Rumford suggested the first relation between heat and work; he noticed that during the boring of cannon at the Munich arsenal, the heat produced was roughly proportional to the work performed during the boring process. He suggested that heat is an invisible fluid (gas), called caloric, which resided between the particles of the substance. In the caloric theory, it had been assumed that the temperature of a substance is determined by the quantity of the caloric gas which it contains; it was also assumed that the amount of caloric per unit mass is less for smaller particles than larger particles. These two assumptions explain the flow of the heat from the bodies with higher temperature to cold bodies; and the sensible heat produced during poring larger pieces to form smaller pieces, metal turnings

In 1799, the caloric theory was discredited when Humphrey Davy melted two blocks of ice by rubbing them together. Based on this experiment, it had been proven the latent heat necessary to melt the ice was provided the mechanical work performed rubbing the blocks together

In 1840, Joule conducted experiments in which work was performed in a certain quantity adiabatically, i.e. the bath is defined contained water and measured the resultant temperature rise of the water. He observed that a direct proportionality exists between the work done and the temperature rise. He observed further that the same proportionality exist, no matter what means were employed in the work production.

▪Methods for work production used are:

1. Rotating paddle wheel immersed in water.

2. Electric motor driving current through coil immersed in water.

3. Compressing a cylinder of gas immersed in water.

4. Rubbing together two metal blocks immersed in water.

These experiments placed on a firm quantitative basis for the thermodynamics object. By defining the calorie, as a heat unit, as the amount of heat required to raise the temperature of one gram of water from 14.5°c to 15.5°c it was found that:

1calorie = 4.184 joule

This constant, 4.184, is known as the mechanical equivalent of heat

2.3 Internal Energy and The First Law of Thermodynamics

The development of thermodynamics from its early beginnings to its present state was achieved as a result of the definition of convenient thermodynamic functions of state. In these section the first of these functions, the internal energy U, is introduced.

Joule’s experiments resulted in a statement that ” the change of a state of a body inside an adiabatic enclosure from a given initial state to a given final state involve the same amount of work by whatever means the process is carried out”

This statement is a preliminary formulation of the

“First Law of Thermodynamics”

Similar to the following relation:

- W=potential energy of body m at position 2 (mgh2)-potential energy of body m at position 1 (mgh1)

For a body of mass m when lifted in a gravitational field, g, from height h, to height H (where W is the work done on the body); and also when particle with charge q, is moved in a electric field from point at potential Q1, to point a potential Q2, the work done, W, on the charged particle is given as,

- W=electric energy at point 2(qQ2) - electric energy at point 1(qQ1)

It is possible to write, for the joule adiabatic process in which W is done on a body,as a result of which it’s state moved from state A to state BW=UB-UA

In describing the sign of work in this case, W will be positive by convention, if the work is done on the system, UB > UAand thus W will be negative if the work is done by the system UB

In Joule’s experiment, the change in the state of the adiabatically contained water was measured as a temperature of the water. The same temperature rise, and hence the same change of state, could have been produced by placing the water in a thermal contact with a source of heat and allowing heat q to flow into the water. In describing the sign of q, it is by convention to assign a positive value of q if heat flows into the body (endothermic process) and negative value for q if heat flows out of the body (exothermic process); hence q = (UB –UA).

Now for a body which absorbs heat q and work W is performed on it, it is possible to show that:

∆U = q + w → (1)

Let us first consider that a thermodynamic system changes its state A to state B` as a result of heat flow q from a heat reservoir to the system, then:

Q = UB` - UA

If the body is adiabatically isolated, i.e. the path is defined, the work W done on the system causes a change from state UB`to state UB, thus:

W = UB - UB`

Therefore,

q + w = (UB` - UA) + (UB - UB`)

= UB – UA

Or ∆U = q + w

For infinitesimal changes, we can write:

dU = δq + δw → (2)

Equations (1) and (2) are the mathematical statement of the First Law of Thermodynamics.

From figure 1 (ref. figure 2.1 p. 20) it can be noticed that:

∆U = U2 – U1,

I.e. U is a state function,

w = ∫12PdV; thus dependant on the path; i.e. W is not a state function because it is path dependent; and accordingly, q is a path dependent and accordingly it is not a state function.

Fig. 2.1Three process paths taken by a fixed quantity of gas in moving from state 1 to state 2

Since U is a state function; thus,

U = U (P, V) = U (P, T)

= U (V, T)

Thus; the complete differentials of U in terms of its partial derivatives are given by:

dU = (∂U/∂P)V dP + (∂U/∂V)P dV

= (∂U/∂P)T dP + (∂U/∂T)P dT

= (∂U/∂V)T dV + (∂U/∂T)V dT

Also we have:

∫12dU = U2 – U1, and

∫ dU = 0, i.e. the cyclic integral equals zero

2.4 Constant – Volume Processes

Since δw = Pdv, then δw = 0 for constant – volume process; and from the first law

of thermodynamics, we have :

dU = δqv

And by integration we have:

ΔU = qv

2.5 Constant–Pressure Processes & Definition of Enthalpy (H)

If the pressure of the system maintained constant during a process which takes the

system from state 1, to state 2, and if the process is reversible ,i.e. Pexternal = Pinternal

, then the work done by the system is given as :

W = -∫v1v2 Pdv = -P (v2 –v1)

ΔU = qp + [-p(v2 – v1)]

Then:

qp = (u2 + pv2) – (u1 + pv1)

Then:

As the expression (u + pv) contains only state function, then the expression itself

is a state function; this expression is termed the enthalpy H, thus :

H = U + PV

2.6 Heat Capacity

The heat capacity C, of a system is the ratio of the heat added or withdrawn from the

system to the resultant change in the temperature of the system, thus;

C = q / ΔT

Or, for infinitesimal change:

C = δq / ∂T

The concept of heat capacity is only used when the addition of heat to or withdrawal of

Heat from the system produces a temperature change; thus, the concept is not used

when the process is isothermal or when a phase change occurs.

The change of temperature of a system from T1 to T2 due to the admission of a certain

quantity of heat to the system is not complete thermodynamic statement since the state

of a system is defined by two independent variables; thus, this second independent

variable could be varied in a specified manner or cold be maintained constant during

the change.

Thus, we define the heat capacity term either at constant volume, Cv, or and constant

pressure, Cp as follows:

Cv = (δq/∂T)v

Cp = (δq/∂T)p

The heat capacity as defined is an extensive function; however, by defining the specific

heat capacity, heat capacity of the system per gram, or the molar heat capacity, heat

capacity per mole; thus, the heat capacities per unit quantity of the system are intensive

functions.

Since:

Cv = (δq/∂T)v

Cv = (∂U/∂T)v

Thus:

dU = Cv dT

Or

Cp = (δq/∂T)p

Also, as:

Cp = (∂H/∂T)p

Thus:

dH = Cp dT

Or

It is expected that, for any substance, Cp will be of greater magnitude than Cv since

the heat required the increase the body temperature by one degree involved the

amount of heat required to cause expansion of the body against constant pressure

per degree of temperature increase.

The relation between Cp and Cv(Cp – Cv) can be calculated as follows:

Cp – Cv = (∂H/∂T)p – (∂U/∂T)v

= (∂U/∂T)p + P (∂V/∂T)p - (∂U/∂T)v

As:

U= f (V, T)

dU= (∂U / ∂v)T.dV + (∂U / ∂T)V.dT

Then:

(∂U/ ∂T)p= (∂U/ ∂V)T * (∂V / ∂T)p + (∂U / ∂T)v

Thus:

Cp – Cv = (∂U/ ∂ V)T * (∂V/ ∂T)p +(∂U/ ∂T)v

+P *(∂V/ ∂T)p – (∂U/ ∂T)v

So:

Cp –Cv = (∂V/ ∂T)p *(P + (∂U/ ∂V)T)

Joule's proved through experimentation that (∂U/ ∂V)T = 0 for ideal gas; thus

- Cp – Cv =P * (∂V/ ∂T)p =P * ∂/ ∂T (RT/P)p =R
- For ideal gases : U=F (t) since (∂U/ ∂V)T = 0

Joule's experiment involved two copper vessels; one is filled with a gas at same

pressure connected to a similar evacuated vessel via a stopcock.

The two-vessel system was immersed in a quantity of adiabatically contained water;

the stopcock was then opened, thus allowing free expansion of the gas into the

evacuated vessel.

After this expansion, Joule could not detect any change in the temperature of the

system. As the system was adiabatically contained and no work was performed;

Then, from the first law:

∆U = 0

dU=(∂U /∂V)T .dV + (∂U/∂T)V.dT = 0

and hence:

Since ∂T = 0, and ∂V≠0, thus (∂U /∂v)T = 0,

i.e. The Internal energy of a perfect gas is not function of volume (and hence pressure);

it is function only of temperature.

The reason for Joule's not observing a temperature rise in the original experiment was

that the heat capacity of the copper vessels and the water was considerably greater

then the heat capacity of the gas; thus, the small heat changes which actually occurred

in the gas were absorbed in the copper vessels and the water.

The Cp/Cv relation is given by :

Cp- Cv = P * (dV/dT)p +(dU/dV)T *(dV/dT)p

The term P * (dV/ dT)p represents the work done by the system per degree rise in

temperature due to expending against the constant pressure P acting on the system.

The term (∂U/ ∂V)T *(∂V/ ∂T)p represents the work done per degree rise in temperature

in expanding against the internal cohesive forces acting between the particles of the

substance.

Thus, for perfect gases the value of (∂U/ ∂V)T *(∂V/ ∂T)p =0,for real gases the

magnitude of P *(∂V/ ∂T)p is much greater then the value of (∂U / ∂V)T * (∂V/ ∂T)p which

approaches the zero value, and for condensed phase, the magnitude of

(∂U/ ∂V)T *(∂V/ ∂T)p is much greater than the value of P(∂V/ ∂T)p which approaches

the zero value.

2.7 Reversible adiabatic processes for a real gas

In a reversible adiabatic process for a real gas the value of q=0 and the external

pressure is almost equal the gas internal pressure. Thus from the first law of

thermodynamics and by using the equation of state of a real gas, we can show that:

PVγ = constant

TVγ-1 =constant

TP(1-γ)/γ =constant

2.8 reversible isothermal processes

As dT= 0, then dU= 0; accordingly:

δq= -δw = p dV

δq= -δw = p dV

Thus:

q= -w = RT ln( V2/V1)= RT ln(p1/p2)

A reversible isothermal process and a reversible adiabatic process are shown on the

P-V diagram of figure 2 in which it seen that, for a given pressure decrease; the work

done by the reversible isothermal process exceeds that done by the reversible

adiabatic process.

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