P = RT

P = RT

EQUATION OF STATEFOR IDEAL GAS

p = RT (11.1)

low density - if average distance between molecules is 10 diameters or more, then very weak attractive forces

approximately point non-interacting particles

= unique constant for each gas[units of Kelvin or Rankine]

p = RTGood to 1% for air at 1 atm and temperatures > 140 K (-130 oC) or for room temperature and < 30 atm

At large pressures, great departure from ideal gas equation of state.

cp/cv = k = 1.4 for perfect gas

= k

cp/cv = k = 1.4 for perfect gas

= k

IDEAL GAS ~ SOME HISTORYPV = constant

IDEAL GAS: p = RT (eq. 1.11)R = Runiv/mmole

1662: Boyle (and Hooke) experimentally showed that: PV = const for const T;

New Experiments1662 ~ Robert Boyle

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..

.

.

p1, n1, m1, vx1(T1) = v , L1

p2, n1 m1, vx1(T1) = v, L2=2L1

L1

2L2

Daniel Bernoulli ~ PV = const Hydrodynamics, 1738

(early theoretical approach)

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.

.

.

.

.

If L doubled (system 2) but same v, then

(# of collisions/sec)1 = v x (1 sec)/L (# of collisions/sec)2 = v x (1 sec)/2L

(# of collisions/sec)2= ½ (# of collisions/sec)1

Daniel Bernoulli ~ Hydrodynamics, 1738

(system 1)

PV = const ( v = constant)

Daniel Bernoulli p = F/A

F {# collisions / sec}

p1 (# of collisions/sec)1/(L)2

p2 (# of collisions/sec)2/(2L)2

p2 ½ (# of collisions/sec1)/(2L)2

p2 = 1/8 p1

Vol2= 8Vol1

p2Vol2 = p1Vol1 QED

“ The elasticity of air is not only increased by compression but by heat supplied to it, and since it is admitted thatheat may be considered as an increasing internal motion of the particles, it follows that … this indicates a more intense motion of the particles of air.”

Daniel Bernoulli

Here was the recipe for quantifying the idea that heat is motion

– two generations before Count Rumford, but it came too early.

IDEAL GAS ~ SOME HISTORYPV <K.E.>

Assume perfect elastic reflections (ideal gas)

so:

- 2mvx is change of x-momentum per collision.

Initially assume vx is same for all particles.

t = 2L/vx

= (mvx)/t =2mvx/(2L/vx) = mvx2/L

Time between collisions, t, of particle with samewall is equal to:

L

Force of one particle impact = Magnitude of momentum change per second due to one particle:

nmvx2/L

Magnitude of momentum change per second due to n molecules:

<vx2> = <vy

2> = <vz2>;

<vx2> + <vy

2> + <vz2> = <v2>

<vx2> = 1/3

1/3 nm<v2>/L

Pressure = F/A = [1/3 nm<v2>/L]/L2

P = 1/3 nm<v2>/L3

PV = 1/3 nm<v2> = 2/3 n (1/2 m<v2>)

average kinetic energy per particle

PV = 1/3 nm<v2> = 2/3 n (1/2 m<v2>)

ExperimentalT(Ko) = [2/(3kB)] [avg K.E.]

PV = 2/3 n (3/2 kB T)

n = # of particles; kB = Runiv/NAvag kB=1.38x10-23 J/K

PV = (nkB) T(Ko) PV = NmolesRunivT(Ko)

n = # of particles; kB = Runiv/NAvag

kB=1.38x10-23 J/K

IDEAL GAS LAW ~ Different Representations

pV= NmRunivT

p =(1/V)Nmmmole{Runiv/mmole}T

p =(m/V){Runiv/mmole}T

p = {Runiv/mmole}T

p = RT (11.1)

pV= NmRunivT pV = nkBTp = RTpv = RT

pV= MRT

Runiv

R = Runiv/Mol. Mass

( M = total mass )

Note: It is assumed that system always in equilibrium.

Assume all gases obey ideal gas law:

p = RT

Not gauge pressureKelvin (or Rankine)

AIRR = Runiv/MolMass = 287.03 m2/(s2-K)

R = 1716.4 ft2/(s2-R)

1ST LAW OF THERMODYNAMICS

Energy Is Conserved

1st Law of Thermodynamics Q - W* = dE

q - w = de e = u + V2/2 +gz

(1st law came after 2nd Law and there actually is a 0th lawthat came after the 1st Law. All laws are based on experience.)

heat is positive when added to system; work is positive when work is done by the system*

1st Law of Thermodynamics Q - W = dE

q - w = de e = u + V2/2 +gz

dE = is an exact differential, a state variable which depends only on initial and final states

q & w depend on the process going from initial to final states; also q & w must cross boundaries

dQ - dW = dE (1st Law)

dq – pdv = du(neglect avg. K.E. and P.E.)

dW on gas = F(-dx) = -pAdx = -pdV

2ND LAW OF THERMODYNAMICS

“not knowing the Second Law or thermodynamicsis analogous to never having read a work of Shakespeare”

~ C.P. Snow

Imagine an ice cube on a hot griddle.First law says nothing about the flow of heat.First law allows heat to flow from ice cube to hot griddle, as long as the total energy is conserved.

The Second Law tells us in which direction a process can take place.

ds = qrev/TProcess always in equilibrium, applied infinitely slowly, no gradients, qrev

ds = dq/Treversible* (Eq. 11.8)

ds = dq/T + dsirrev

Change of entropy during any incremental process is equal to the actual heat divided

by the temperature plus a contribution from the irreversible dissipative phenomena of viscosity, thermal conductivity, and mass diffusion [all associated with gradients]

occurring within the system. These dissipative phenomena always increase the entropy.

A Change in Entropy Meter

Heater trickles in dq quasistatically in

such a way that T remains approximately constant

over small time.

“I propose to name the magnitude S the entropy of the body, from The Greek word for transformation. I have intentionally formed the word entropy so as to

be as similar as possible to the word energy, sinceboth these quantities…. are so nearly related”

CLAUSIUS

For heat to flow from cold to hot object is forbidden

by the second law e.g. would result in decrease in entropy,

therefore is not allowed.

T2

T1

Q1

Q1=Q2

T2 > T1

T2 > T1

1/T2 < 1/T1

Q/T2 < Q/T1

Entropy lost = Q/T1

greaterEntropy gained = Q/T2

“Boltzman’s tomb is the bridge between the world of appearance and its underworld of atoms.”

Example of reversible process that illustrates

dq/T over cycle = 0

isothermal

isothermal

adiabatic

adiabatic

isothermal

adiabatic

dq/T over cycle = 0

ideal and calorically perfect gas – reversible process (no gradients)

ideal and calorically perfect gas – reversible process (no gradients)

Step 1 ~ Isothermal: T=0, T=T1

pv = RT1

If perfect gas andadding q1 to va

and keep at T1

what is vb?ideal and calorically perfect gas – reversible process (no gradients)

For ideal gas u = f (T); p = RT pv = constant

So for step (1-2), du = 0du = q – w (cons. of energy)

q = wq1 = ba pdv = baRTH(dv/v)

= RTHln(vb/va)

ideal and calorically perfect gas – reversible process (no gradients)

Step 1 ~ Isothermal: T=0, T=T1; pv = RT1

Step 1 ~ Isothermal: T=0, T=T1

pv = RT1 c

For ideal gas u = f (T)p = RT ; pv = constantSo for step (1-2), du = 0du = q – w; q = w

q1 = ba pdv = baRTH(dv/v) = RTHln(vb/va)

ideal and calorically perfect gas – reversible process (no gradients)

Step 2 ~ Adiabatic: Q=0; T1 to T2

If perfect gas andexpanding adiabatically

from T1 to T2 and vb to vc - what is vb?

du = q – w

du = – w cvdT = -pdv

cvdT = -pdv = -(RT/v)dv

cbcvdT = -cbpdv = bcRT(dv/v)

(cv/R) cb (dT/T) = bc(dv/v)

ln(Tc/Tb)(Cv/R) = ln(vb/vc)

Step 2 ~ Adiabatic: Q=0; T1 to T2

ideal and calorically perfect gas – reversible process (no gradients)

Step 2 ~ Adiabatic: Q=0; T1 to T2

q = 0 ~ adiabaticdu = q – w; du = – w

cvdT = -pdv = -(RT/v)dv

cbcvdT = -cbpdv = bcRT(dv/v) (cv/R) cb (dT/T) = bc(dv/v) ln(Tc/Tb)(Cv/R) = ln(vb/vc)

ln(Tc/Tb)(Cv/R) = ln(vb/vc)

(Tc/Tb)(Cv/R) = vb/vc

[pv/R]c/[pv/R]b = [vb/vc](R/Cv)

pc/pb = [vb/vc](R/Cv)+1

= [vb/vc](Cp-Cv)/Cv+1 = [vb/vc]k

pcvck = pbvb

k= constant

Step 2 ~ Adiabatic: Q=0; T1 to T2

ideal and calorically perfect gas – reversible process (no gradients)

Step 3 ~ Isothermal: T=0; T=T2

pv=RT2 = RTC

For perfect gas u = f (T)P = RT So for steps (3 - 4), u = 0u = q – w q = w

q1 = dc pdv = dcRTC(dv/v) = RTCln(vd/vc)

Step 4~ Adiabatic: q=0; T2 to T1

Q = 0

du = q – w; du = – w

cvdT = -pdv = -(RT/v)dv

adcvdT = -adpdv = daRT(dv/v) (cv/R) ad (dT/T) = da(dv/v) (cv/R) ln(Ta/Td)(cv/R) = ln(vd/va)

Q = 0 ~ adiabatic

Step 4~ Adiabatic: q=0; T2 to T1

isothermal

isothermal

adiabatic

adiabatic

isothermal

adiabatic

Step#2: (Tc/Tb)(cv/R) = vb/vc

(Tcold/Thot)(cv/R) = vb/vc

Step#4:(Ta/Td)(cv/R) = vd/va

(Thot/Tcold)(cv/R) = vd/va

(Tcold/Thot)(cv/R) = va/vd

a

b

c

d

vb/vc = va/vd

Ideal gas: p = RT q1 = ba pdv = ba RT1(dv/v)

= RT1 ln(vb/va)

vb/vc = va/vd

vb/va = vc/vd

q2 = dc pdv = dc RT2(dv/v) = RT2 ln(vd/vc)

= -RT2 ln(vc/vd)

q1/T1 = -RT2 ln(vb/va) = - q2/T2

Ideal gas: p = RT q1 = ba pdv = ba RT1(dv/v) = RT1 ln(vb/va)

q2 = dc pdv = dc RT2(dV/V) = RT2 ln(vd/vc) = -RT2 ln(vc/vd) = -RT2 ln(vb/va)

vb/vc = va/vd

vb/va = vc/vd

q1 / T1 - q2 / T2 = Rln(vb/va) - Rln(vb/va)

q1 / T1 = q2 / T2

Although proved for a reversible ideal gas, true for any reversible engine

Step#2: (Tc/Tb)(cv/R) = vb/vc

Step:#4 (Ta/Td)(cv/R) = vd/va

QH / T1 = QC/T2 INDEPENDENT of cv and R!!!!

Can be shown true for any substance for reversible process

ENTROPY DEFINED AS: dS = rev Q/T (11.8)

dq/T over cycle = 0

• The concept of absolute zero extends to a great many phenomena:– volume of a gas

(Charles law - 1800)

– electrical noise in a resistor

– wavelength of radiation emitted by a body

In the early 1800’s Lord Kelvin developed a universal thermodynamic scale based on the coefficient of expansion of an

ideal gas.

Kelvin Temperature0oK (& 0oR) is related to something meaningful

cp and cvIdeal gas and calorically perfect