Chapter 3. The First Law of Thermodynamics

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Chapter 3. The First Law of Thermodynamics. In this chapter we will introduce the concepts of work and energy, which will lead to the 1 st Law of Thermodynamics. Configuration work : We consider some reversible process đ W= (intensive variable ) (change in extensive variable)

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### Chapter 3. The First Law of Thermodynamics

In this chapter we will introduce the concepts of work and energy, which will lead to the 1st Law of Thermodynamics.

Configuration work: We consider some reversible process

đ W= (intensive variable ) (change in extensive variable)

Let y=intensive variable X= extensive variable

đ W is not an exact differential.

The variables determine the configuration of the system.

surroundings

system

Consider a system and its surroundings.

When the system as a whole does work on the surroundings or the surroundings do work on the system as a whole, we call this external work.

When one part of the system does work on another part of the system, it is called internal work. Internal work cannot be discussed within the framework of macroscopic thermodynamics.

When we use the term work, we mean externalwork.

Convention:

 Work done on system by surroundings is negative

 Work done by system on surroundings is positive

[NOTE: Be careful. Often the opposite convention is used.]

We will be considering quasi-static processes.

This is an ideal situation in which an infinitesimal unbalanced force results in an infinitesimally slow transition from an initial state to a final state through a succession of states near thermodynamic equilibrium.

At each instant an equation of state is valid for the whole system. At present we will ignore dissipative processes such as friction.

Configuration work in changing a volume

The gas expands by exerting a pressure infinitesimally greater than the pressure on the gas exerted by the force acting on the piston.

F=PA (good approx.)

The work done by the gas is

A

F

P

đW because this is an inexact differential.

W is not a function of the system variables W  W(P,V,T).

We cannot write

Volume increases and Adx = dV

đW = PdV

The symbol đW indicates that a small amount of work is

done and the bar indicates that it is an inexact differential.

(The work done will depend on the path.)

If dV is positive then đW is positive and work is done by the system.

is positive

P

i

f

• PV Diagrams (Indicator Diagrams)

Work done by system.

V

i

is negative

P

Work done on system.

f

V

A cycle

(not zero!)

i

P

f

Net work is positive.

Net work done by system

V

Work is not a state function.

Work is an activity or process that results in a change of the state of a system.

In going from an initial thermodynamic state to a final state through intermediate states, the work done will depend on the path taken.

i

I

P

W is area under the curve and obviously depends upon the path taken.

f

II

V

Calculation of work for quasi-static processes

Isothermal expansion of an ideal gas:

i

P

f

V

T constant so

W >0 work done by gas

Isothermal increase of the pressure on a solid

Volume expansivity

We defined:

Isothermal compressibility

đ

(since dT=0)

đ

For a solid Vconstant and over a limited range constant.

Dissipative work

Dissipative work is work done in an irreversible process. It

is always done on a system. (We will discuss irreversible processes later in the course.) Dissipative effects include friction, viscosity, inelasticity, electrical resistance and magnetic hysteresis.

If there are no dissipative effects, all the work done by the system in one direction can be returned to the system during the reverse process.

Reversible processes: The dissipative work must be zero.

An example of dissipative work is work done on a liquid by stirring (see textbook). Regardless of the direction of rotation of the stirrer shaft, the external torque is always in the same direction as the angular displacement of the shaft and the work done due to the external torque is always negative, regardless of the direction of rotation. (Work is done on the system.)

Free expansion

vacuum

diaphragm

gas

If the diaphragm is punctured, the gas expands and fills all of the container. In this process, no external work is done.

In a free expansion no work is done!

Consider a system going from some initial state i to a final

state f. Many different processes are possible and the work done

will depend on the path taken.

Now we limit ourselves to adiabatic paths. (System surrounded

by an adiabatic boundary and so the temperature of the system is

independent of that of the surroundings. No energy transfer.)

Consider the processes shown on the PV diagram below:

expansion

i

free expansions

a

b

P

f

work

V

First, we have a free expansion (no work) from i to a, then an adiabatic expansion from a to f. The work done by the system is the area under the af curve.

Now we consider another path. First, an adiabatic expansion is made from i to b, such that a free expansion results in the same final state f. The work done by the system is the area under the ib curve.

Notice that the free expansions are denoted by dotted lines because they do not proceed quasi-statically through a series of equilibrium states. The area under these dotted lines is not the work done. (The work done is, of course, zero.)

It is an experimental fact that the work done during these two different processes is the same.

The total work is the same in all adiabatic processes between any two equilibrium states having the same kinetic and potential energy!!! This is called the restricted statement of the first law of thermodynamics.

there exists a function of the thermodynamic coordinates of the

system such that

U is the internal energy function

represent state variables such

as T, P, …..

In differential form, dU= -đ W If the system performs work, it

comes at the expense of the internal energy.

The internal energy function:

increase in the internal energy

then expresses:

(a) conservation of energy (b) state function U exists

For example, if we consider U=U(T,V)

Suppose a system undergoes a process from an initial state i to a final state f, by a diathermic process. In this case

By conservation of energy, energy must have entered (or left) the system by some non-mechanical process.

The energy transferred between the system and surroundings

by a non-mechanical process or electrical process is called heat (Q).

We define or

Mathematical formulation of the

first law of thermodynamics

Units: U, Q, W are all energies and must all have units of joules.

This law includes the concepts:

(a) a state function U exists

(b) energy is conserved

(c) energy is transferred due to a temperature difference.

Heat is that energy which is transferred due to a temperature difference

between a system and its surroundings. Heat is transferred from a

higher temperature region to a lower temperature region.

Heat and work are not state functions.

Q, W depend on the path. They are

path functions, not state functions.

However (Q-W) is independent of

path and is a state function.

f

P

i

V

Heat is a process, not a quantity of anything. It is not correct to say “heat in a body”. After a process is completed, there is no further

use for Q.

The differential form of the first law is :

dU=đQ-đW

We have described how to calculate a reversibleđW in a hydrostatic (i.e. P,V,T) system: đWrev=PdV and so dU= đQ-PdV.

However, as yet, we have no way of writing đQ in terms of system coordinates. Later we will see how to do this.

[Great care must be taken in the use of “heat” or “to heat”. Some

authors insist that heat should never be used as a noun or even a

verb! However avoiding such usage results in awkward

circumlocutions and so most of us are somewhat careless in this

regard. In teaching an elementary course one should be careful!]

Now that we have an equation relating energy to work, how do we use it to make calculations for particular systems?

We know how to calculate mechanical work. How can we calculate either Q or U? If we could calculate one of them, the equation would permit us to obtain the other. Clearly we have more work to do, but first let us do some examples.

EXAMPLE: Chapter 3

Compute the work done against atmospheric pressure when 10kg of water is converted to steam occupying

For water

The system, initially water, does work in pushing aside the atmosphere.

EXAMPLE: Chapter 3

Steam at a constant pressure of 30atm is admitted to the cylinder of a steam engine. The length of the stroke is 0.5m and the radius of

the cylinder is 0.2m. How much work is done by the steam per stroke?

The system (steam) does work on the piston.

EXAMPLE: Chapter 3

• A volume of contains 8 kg of oxygen at 300K. Find the
• work necessary to decrease the volume to
• At constant pressure
• At a constant temperature
• What is the temperature at the end of process (a)?
• What is the pressure at the end of process (b)?
• Show both processes on an indicator (PV) diagram.

We assume an ideal gas law equation of state.

(a) Constant P

(b) Constant T

(c) For constant P

(d) Constant T

OXYGEN

12.5

300K

150K

6.24

0 5 10

EXAMPLE: Chapter 3

• An ideal gas originally at temperature and pressure is
• compressed reversibly against a piston to a volume equal to
• one-half its original volume. The temperature is varied during the
• compression so that, at each instant, the relation is satisfied.
• Draw an indicator diagram for the process.
• Find the work done in terms of n,R and the initial temperature.

isotherms

P=KV

P

V

(b)

But

Work is done on the gas.

EXAMPLE: Chapter 3

• An ideal gas, and a block of Cu, have equal volumes of
• at 300K and 1 atmosphere. The pressure of each is increased
• reversibly and isothermally to 5 atmospheres.
• Explain, with the aid of an indicator diagram, why the work is not
• the same in the two processes.
• (b) In which process is the work greater?
• (c) Find the work done on each. For Cu,
• (d) Find the change in volume for each case.

gas Cu

5

1

0

isotherms

W=area under isotherm

Cu compressed much less

P(atm)

(d) Ideal gas

For Cu

More elegantly,

EXAMPLE: Chapter 3

• Derive the general expression for the work per kilomole of a
• van der Waals gas in expanding reversibly and at a constant T from
• a specific volume to a specific volume
• (b) Find the work done when 2 kmoles of steam expand from a volume
• of to a volume of at a temperature of

For steam

(c) Find the work of an ideal gas in the same expansion.

(b) 2 kmoles of steam

(c) ideal gas

Not a great deal of difference. What is initial pressure?

Now we have added new concepts and ideas:

Work has been introduced:

đW=PdV. It is not an exact differential

W is not a state function, but rather a process!

We also introduced the internal energy of a system. If the

system is not isolated then, since energy must be conserved,

energy must enter or leave the system by non-mechanical

means and this energy is represented by Q.

đQ=?. It is not an exact differential

Q is not a state function, but rather a process!

1st Law of Thermodynamics:

dU=đQ-đW