Atmospheric Thermodynamics –IIThe first Law of Thermodynamics and

applications

Prof. Leila M. V. CarvalhoDept. Geography, UCSB

The first Law of Thermodynamics

• Is the law that describes the relationships between heat, work and internal energy.

• It establishes the physical and mathematical framework to understand heating processes in our atmosphere, the formation of clouds, the thermodynamical modifications in parcels in movement, etc…

• Let’s explore these relationships and applications…

Motivation: Ex: formation of cloudsPressure

Volume expands and the parcel’s temperature decreases. As it cools the air becomes saturated When that begins: Lifting Condensation Level (clouds are formed) LCL: Cloud base :

Releases Latent heat:

The first Law of Thermodynamics provides the physical concepts to understand cloud formation:

Heat

• Internal Energy u: measure of the total kinetic and potential energy of a gas

Kinetic energy: depend on molecular motions -> relationship with temperature

Potential energy: changes in the relative position of the molecules due to internal forces that act between molecules (small changes)

Closed System definition:

• Is the one in which the total amount of matter, which may be in the form of gas, liquid, solid or a mixture of these phases, is kept constant

Suppose a closed system with one unity of mass

Suppose that this volume receives certain quantity of thermal energy q (joules) by ‘conduction’ and/or radiation. This system may do a certain amount of external work w (also measured in Joules) .

12 uuwq

Differences will cause changes in the internal energy

Where 1 is before and 2 after the change

This is the First Law of

Thermodynamics

In the differential form

dq is the differential increment of heat added to the system, dw is the differential element of work done by the systemdu is the differential increase in internal energy of the system

Changes in du depend only on

the final and initial state: functions of

state

dudwdq (34)

VisualizationVolume is proportional to the distance x

Frictionless

Every state of the substance, corresponding to a given position of the piston, can be represented in this diagram below

AREA A If the Piston moves outward with the same pressure p, then the work done by the substance in pushing the external force F through a distance dx is:

F=pA

FdxdW pAF

pdVpAdxdW If pressure is constant then:

(35)p

VdV

pdV

If the substance passes from state A (V1) to B (V2) during with its pressure changes than the work will be :

2

1

V

VpdVW (36)

IF V2>V1, W is positive, indicating that the SUBSTANCE DOES WORK ON ITS ENVIRONMENTIF V2<V1, W is negative, indicating that the ENVIRONMENT DOES WORK ON THE SUBSTANCE

A few useful remarks

• If we are dealing with a unity mass of a substance, the volume V is replaced by the SPECIFIC VOLUME α. Therefore, the work dw that is done when the specific volume increases by dα is• dw = pdα (37)We can then rewrite the First Law of thermodynamics as:dq = du + pdα (38)

Some considerationsF=pA Suppose that the force in the wall is

supplied by the impact of the balls. If we do not supply kinetic energy to the gas (or balls) the work required to move the walls outward comes from a decrease in the kinetic energy of the balls that rebound from the walls with lower velocities than they struck them. This decrease in kinetic energy is in accordance with the first law of thermodynamics under ADIABIATIC CONDITIONS (NO HEAT SUPPLIED).In this case, the work done by the system by pushing the walls outward is equal to the decrease in the internal energy of the system.dq = du + pdα If dq = 0pdα = - du

The Joule’s Experiment and the Joule’s LawSuppose that the gas is expanding within a Vacuum chamber. In this case, there is no F against the environment (since it is vacuum)

In this case, dw=0 (F=0)The gas also does not take or gives out heat (dq=0)As a consequence du =0 (First Law of the Thermodynamics)Under these conditions the temperature of the gas does not change, which implies that the kinetic energy of the molecules remains constant (and potential energy too) Big conclusion: The internal energy of a gas is independent of its volume if the temperature is kept constant

Specific Heats

Specific Heat at constant volumeSuppose that this volume with a unity of mass receives certain small quantity of heat dq (joules). The temperature T increases to T + dT without any changes in phase occurring within the material. If the volume is kept constant,

We define the specific heat at constant volume as:

constantvv dT

dqc

(39)

If the volume of the material is constant, the first law of the thermodynamics states that:dq = du + pdα If dα =0dq = du

constantvv dT

duc

For an ideal gas the Jaule’s law applies and therefore u depends only on temperature. Therefore, regardless of whether the volume of a gas changes, we may write

dT

ducv (40)

dq = du + pdα (38)

dT

ducv (40)

FROM:First Law of the Thermodynamics:

and

Therefore: dudTcv

pddTcdq v First Law of the Thermodynamics:

(41)

Because u is a function of state, no matter how the material changes from state 1 to state 2, the change in its internal energy from (40) is:

dTcuuT

T v 2

112

Suppose that now pressure is kept constant

The material is allowed to expand as heat is added to it and its temperature rises, as pressure remains constant. In this case, a certain amount of heat added to the material will have to be expended to DO WORK as the system expands against constant pressure of its environment We can also define a specific

heat at constant pressure cp

constantpp dT

dqc

Implications…A larger quantity of heat must be added to the material to raise its temperature by a given amount than if the volume of the material was kept constant (it is because heat will be transformed into work and also into kinetic energy to increase the temperature of the gas). Mathematically speaking…

dppddTcdq v )(

pddTcdq v First Law (Eq. 41)

If we assume that p and α are changing, then

We saw before from the ideal gas equation (3) that:RdTpdorRTp )( Therefore:

(43)

dpdTRcdq v )( (44)

At constant pressure…

dpdTRcdq v )( Last term vanishes -> dp=0

constantpp dT

dqc

Therefore: )( Rc

dT

dqv

or: )( Rcc vp (45)

The specific heats at constant volume and at constant pressure for dry air are 717 and 1004 JK-1kg-1,respectively. The difference between them is 287 J K-1 kg-1, which is the gas constant for dry air.

dpcpdTdq (46)

Curiosities…

Are there processes in the atmosphere that occur at constant pressure (isobaric process)?

All physical/dynamical processes that are considered at a given pressure level. For example air expanding at pressure p=800hPa)

Are there processes in the atmosphere that occur at constant volume?

Some horizontal movements may occur such that temperature varies, pressure varies but volume is constant.

Enthalpy

Etymology

• The term enthalpy comes from the prefix ἐν-, en-, meaning "to put into", and the Classical Greek verb θάλπειν, thalpein, meaning "to heat". The original definition is thought to have stemmed from the adjective "enthalpos" (ἔνθαλπος).

• enthalpy (denoted as H, or specific enthalpy denoted as h) is a thermodynamic property of a thermodynamic system. It can be used to calculate the heat transfer during a quasi-static process taking place in a closed thermodynamic system under constant pressure (isobaric process).

• Enthalpy is commonly referred to as ‘sensible heat’

Mathematical framework

• If heat is added to a material at constant pressure so that the specific volume of the material increases from α1 to α2 the work done by a unity of mass of the material is

p(α 2 - α1)From dq = du + pdα (38) 11221212 pupupuuq

Initial and final internal energy for a unity of mass

Therefore, at a constant pressure

1122 pupuq

h2 h1

puh h=Enthalpy of a unity mass

u and p are functions of state, h is a function of state too.

Differentiating …• The objective of computing differentials in these cases is to

provide equations that can describe small variations (differentiations) and can be used later to integrate a given process over some large changes of a given variable (could be temperature, volume, pressure, etc)

puh )( pddudh

dT

ducv

dpdhdq

USING AND dppddTcdq v )(

(48)

Some considerations

• Compare (46) and (48)

dpdTcdq p dpdhdq (48)

dTcdh p

(46)

Tch p(49) (50)

• Where h is taken as zero when T=0. Eq (50) indicates that h corresponds to the heat required to raise the temperature of a material from 0 to T K at constant pressure

Examples:

• When a layer of air that is at rest and in hydrostatic balance is heated, for example, by radiative transfer, the weight of the overlying air pressing down on it remains constant (constant pressure).

• The energy added to the air is realized in the form of an increase in enthalpy (or sensible heat) and dTcdhdq p

Enthalpy (sensible Heat)

dTcdhdq p1)internal energy increases at constant VOLUME

dTcdu v

RdTpd 2) Work is done against the overlaying environment

Because the Earth’s atmosphere is made up mainly of the diatomic gases N2 and O2, the energy added by dq is partitioned between the increase in internal energy du and expansion work in the ratio 5:2

General Expression: dry atmosphere (no clouds) Pressure

Heat

We can write a more general expression that is applicable to a moving parcel that changes pressure as it rises or sinks. Remember the definition of Geopotential:

dpd dpdhdq

dTcdh p

)()( Tcdhddq pDry Static Energy: constant provided that the parcel neither gain or loses heat (dq=0)

(This also means : no latent heat released or absorbed)

Next class we will explore Adiabatic processes