The First Law of Thermodynamics

U = Q + W

Internal Energy

The internal energy of a system, U, is the sum of kinetic energy in the referenceframe in which the center of mass is at the rest and potential energy arising fromthe forces of particles on each other.

systemsystem

boundary

“environment”U = Kinetic + Potential

The internal energy is a state function – it depends only onthe values of macroparameters (the state of a system), noton the method of preparation of this state (the “path” in themacroparameter space is irrelevant).

U depends on the kinetic energy of particles in a system and an average inter-particle distance (~ V-1/3) – interactions.

For an ideal gas (no interactions) : U = U (T) - “pure” kinetic

iii

Work and Heating (“Heat”)

We are often interested in U , not U. U is due to:Q– energy flow between a system and its environment

due to T across a boundary and a finite thermalconductivity of the boundary

– heating (Q > 0) /cooling (Q < 0)(there is no such physical quantity as “heat”; toemphasize this fact, it is better to use the term “heating”rather than “heat”)

W - any other kind of energy transfer across boundary– work

Heating/cooling processes:Conduction: the energy transfer by molecular contact – fast-moving

molecules transfer energy to slow-moving molecules bycollisions.

Convection: the energy transfer by macroscopic motion of gas or liquid.Radiation: the energy transfer by emission/absorption of electromagnetic

radiation.

The First Law of Thermodynamics

For a cyclic process (Ui = Uf) Q = - W.If, in addition, Q = 0 then W = 0.

The first law of thermodynamics: the internal energy of a system can bechanged by doing work on it or by heating/cooling it.

U = Q + W conservation of energy

Perpetual motion machines of the first type do not exist

Sign convention: we consider Q and W to be positive if energy flows intothe system.

Quasi-static Processes

Quasi-static (quasi-equilibrium) processes – sufficientlyslow processes, any intermediate state can be consideredas an equilibrium state (the macroparameters are well-defined for all intermediate states).

Examples of quasi-equilibrium processes:

isochoric: V = const isobaric: P = constisothermal: T = constadiabatic: Q = 0

For quasi-equilibrium processes, P, V, Tare well-defined – the “path” between twostates is a continuous lines in the P, V, Tspace.

Advantage: the state of a system that participates in a quasi-equilibriumprocess can be described with the same (small) number of macroparametersas for a system in equilibrium (e.g., for an ideal gas in quasi-equilibriumprocesses, this could be T and P). By contrast, for non-equilibrium processes(e.g. turbulent flow of gas), we need a huge number of macroparameters.

Work

The sign: if the volume is decreased, W is positive (bycompressing gas, we increase its internal energy); if thevolume is increased, W is negative (the gas decreasesits internal energy by doing some work on theenvironment).

2

1

),(21

V

VdVVTPW

The work done by an external force on a gas enclosed within a cylinder fitted with a piston:

W = force x displacement .= (PA) dx = P (Adx) = - PdV

dU = Q – PdV

W and Q are not State Functions

- the work is negative for the “clockwise” cycle; ifthe cyclic process were carried out in the reverseorder (counterclockwise), the net work done onthe gas would be positive.

01212

211122

VVPPVVPVVPWWW CDABnet

2

1

),(21

V

VdVVTPW

- we can bring the system from state 1 to state2 along infinite # of paths, and for each pathP(T,V) will be different.

U is a state function, W is not thus, Q is not a state function either.

Since the work done on a system depends notonly on the initial and final states, but also on theintermediate states, it is not a state function.

U = Q + W

Problem

Imagine that an ideal gas is taken from its initial state A to state B by anisothermal process, from B to C by an isobaric process, and from C backto its initial state A by an isochoric process. Fill in the signs of Q, W, andU for each step.

Step Q W U

A B

B C

C A

TUU nRTPV

0-

++-

0

-+

+

dTnCdU v

Quasi-static Processes in Ideal Gas

isochoric (V = const )

isobaric (P = const)

021 W

0),( 12

2

121 VVPdVTVPW

21212121 WUQQWU

UQQWU 212121

12 TTnCUdTnCdU vv

12 TTnCUdTnCdU vv

Isothermal Process in Ideal Gas

2

1

),(21

V

V

dVTVPW

f

ifi V

VnRTW lnWi-f > 0 if Vi >Vf (compression) Wi-f < 0 if Vi <Vf (expansion)

isothermal (T = const)

21212121 0 WQWQdU

0 dUTUU

nRTPV

2

121 ln

VVnRTW

Adiabatic Process in Ideal Gas

adiabatic (thermally isolated system)

PdVWQdTnCdU v

nRdTVdPPdVnRTPV

PdVCRVdPPdV

v

01

PdP

CR

VdV

v

The amount of work needed to change the state of a thermally isolated systemdepends only on the initial and final states and not on the intermediate states.

021 Q 21WdU

to calculate W1-2 , we need to know P (V,T) for an adiabatic process

2

1

),(21

V

V

dVTVPW

constPVPP

VV

1

1

1

2 lnlnv

v

CRC

Product Rule in Differentiation

d[f(x) • g(x)] = f '(x) • g(x) + f(x) • g'(x)

d(x • y)= x • dy+ y• dx

PV=nRT

d(PV)=d(nRT)

d(PV)=PdV+VdP

d(nRT)=nRdT

Adiabatic Process in Ideal Gas

adiabatic (thermally isolated system)

PdVWQdTnCdU v

nRdTVdPPdVnRTPV

PdVCRVdPPdV

v

01

PdP

CR

VdV

v

The amount of work needed to change the state of a thermally isolated systemdepends only on the initial and final states and not on the intermediate states.

021 Q 21WdU

to calculate W1-2 , we need to know P (V,T) for an adiabatic process

2

1

),(21

V

V

dVTVPW

constPVPP

VV

1

1

1

2 lnlnv

v

CRC

Summary of Quasi-static Processes of Ideal Gas

Quasi-Static process U Q W Ideal gas

lawisobaric

(P=0)isochoric

(V=0) 0

isothermal (T=0) 0

adiabatic (Q=0) 0

WQUUU if

P V fi

i f

VVT T

TnCU v VPTnCv

TnCvTnCU v fi

i f

PPT T

Wi

f

VV

nRT ln i i f fPV P V

U

11

1 ifff VV

VPU

ffii VPVP

Another ProblemDuring the ascent of a meteorological helium-gas filled balloon, itsvolume increases from Vi = 1 m3 to Vf = 1.8 m3, and the pressureinside the balloon decreases from 1 bar (=105 N/m2) to 0.5 bar.Assume that the pressure changes linearly with volume between Viand Vf.(a) If the initial T is 300K, what is the final T?(b) How much work is done by the gas in the balloon?(c) How much “heat” does the gas absorb, if any?

K2701mbar11.8mbar5.0K300 3

3

ii

ffif

BB VP

VPTT

NkPVTTNkPV(a)

WQU

J106mbar6.0mbar4.05.0mbar8.05.0)( 4333 f

i

V

V

dVVPW

WTTCRTVPWTTnCWUQ ifv

i

iiifv

(b)

(c)

bar625.1bar/m625.0 3 VVP

RCv 5.1 for ideal gas

J105.4 4

Summary

• U, W and Q

• First Law of Thermodynamics∆U = Q + W

• Quasi-static Processes of Ideal Gasa) Isobaric (∆P = 0)b) Isochoric (∆V = 0)c) Isothermal (∆T = 0)d) Adiabatic (Q = 0)

• Some Applications