p = rt
DESCRIPTION
P = RT. EQUATION OF STATE FOR IDEAL GAS p = RT (11.1). [units of Kelvin or Rankine]. = unique constant for each gas. low density - if average distance between molecules is 10 diameters or more, then very weak attractive forces approximately point non-interacting particles. - PowerPoint PPT PresentationTRANSCRIPT
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