slide 1 Physics 1401 - L 23 Frank Sciulli Thermodynamics and Gases Last time l Kinetic Theory of Gases for simple (monatomic) gases l Atomic nature of matter l Demonstrate ideal gas law l Atomic kinetic energy = internal energy l Mean free path and velocity distributions l From formula for E int , can get specific heats l Specific Heats of Simplest Gases l Constant Volume l Constant Pressure Today l Specific Heats for more complex gases l Adiabatic Expansion l Entropy
Transcript
slide 1Physics 1401 - L 23 Frank Sciulli
Thermodynamics and Gases
Last timel Kinetic Theory of Gases for simple (monatomic)
gasesl Atomic nature of matterl Demonstrate ideal gas lawl Atomic kinetic energy = internal energyl Mean free path and velocity distributions
l From formula for Eint, can get specific heatsl Specific Heats of Simplest Gases
l Constant Volume l Constant Pressure
Todayl Specific Heats for more complex gasesl Adiabatic Expansionl Entropy
slide 2Physics 1401 - L 23 Frank Sciulli
Internal Energy from Atomic Nature
l pV=nRT understood from atomic nature of matteru pV=NkT is equivalent formu Both are generally applicable (up to small van der Waals
corrections) for all gases … pV ∝ kinetic energy of atomsl Internal energy of the gas is a sum of all the energy
forms (including kinetic energy) of the moleculesu simplest is monatomic gas (one atom in the molecule, rotationally
symmetric) -> energy all translational u real world: coefficient, 3/2, only applies to “noble gases”
32
32
32
atom
atom
pV NkT nRTK kT
E N K NkTE nRT
int
int
monatomic gas= =
=
= =
=
Vatom
Vatom
V
pV NkT nRTCE kTR
CE N E N kTR
E nC T
int
int
ANY gas = =
=
= =
=
review
slide 3Physics 1401 - L 23 Frank Sciulli
Specific Heat at Constant Volume (isochoric)
l No change in volume implies no work done: u dW = 0
l Heat introduced proportional to temperature change when no worku Q ≡ n CV ∆T
l Since dW=0, then the heat added must equal the change in internal energyu ∆Eint = Q = n CV ∆T
32E nRTint
monatomic gas=dE dQ dWint
1st Law of Therm.= −
V
anyE nC Tint
gas=
And we predict: Monatomic (billiard ball) gases have CV=3R/2
review
slide 4Physics 1401 - L 23 Frank Sciulli
Specific Heat at Constant Pressure (isobaric process)
l For process shown (n fixed)u Q ≡ n CP ∆T
l Here, as expansion occurs, u work is done andu internal energy increases
( )
V
V
P
V
V
Q E WnC T p V
C C R
nC T nR TQ n C R T
int= ∆ +
= ∆ + ∆
= ∆
+ ∆
=
∆
=
+
+
review
slide 5Physics 1401 - L 23 Frank Sciulli
How about molecules that are not “monatomic”?
l We found that “sphere-like” molecules (single inert atoms) had CV=3R/2 … what about the rest?u Came from considerations of kinetic energy
(Eint=nCVT) deriving gas lawl Our single atoms had only one kind of
available energy →u Kinetic energy from translation in 3 dimensions
l More generally molecules can have other “degrees of freedom”u rotationsu vibrationsu ,,,
l Maxwell:
12E f kT
fint
molecule ( )
# deg of freedom
=
=
slide 6Physics 1401 - L 23 Frank Sciulli
Molecular Specific Heats12E f kT
fint
molecule ( )
# deg of freedom
=
= int
int
gas( )
( )
V
Vf
P
anyE nf RT
E n C TC Rf
C R
=
=
=
= +
12
12
2 1
l f=different independent ways that molecule can contain and exchange energy
l Need enough thermal energy to “excite” the modes to be seen (Quantum Mechanics)u fig - H2 gas CV/R
slide 7Physics 1401 - L 23 Frank Sciulli
Real Gas at low pressures
l Empirical check
f Cv Cp γ = Cp/ Cv Translational Only 3 3R/2 5R/2 5
3 1.67= + Rotational
diatomic 2 5R/2 7R/2 7
5 1.40=
+ Rotational polyatomic
1 3R 4R 43 1.33=
12
2 1V
fP
anyf
C RfC R
gas# active d.f.
( )
=
=
= +21p
V
CC f
γ ≡ = +
slide 8Physics 1401 - L 23 Frank Sciulli
Aside: How about Solids?
l Generally expect Cv=fR/2l While gases are unbound, solids have
atoms bound to adjacent atomsu no net translations in 3D space butu there does exist
o Vibrational kinetic energy -- 3 deg of freedomo potential energy (of binding) -- 3 deg of freedom
l Recall table 19-3: many solids tend to have molar heat capacity C ~ 25 J/mol-K
l Naively expect f=6 and Cv=3R ~3(8.31 J/mol-K )
l This value approached in most solids as temperatures rise
slide 9Physics 1401 - L 23 Frank Sciulli
Adiabatic Expansion of Gas
l Adiabatic = no heat enters or leaves system (gas) … Q=0
l But temperature changes as gas does work
int
int
V
V
dE dQ dWdE dW
nC dT pdVpdVndTC
= −
= −
= −
= −
( )P V
pV nRT pdV Vdp nRdTpdV Vdp pdV VdpndT
R C C
= + =
+ += =
−
P V V
pdV Vdp pdVc c c
+= −
−
slide 10Physics 1401 - L 23 Frank Sciulli
Adiabatic Expansion - algebra
Differential equation states that fractional change in pressure plus γ times fractional change in volume must equal zero in any adiabatic expansion
( )
P V V
P V V P V
P
V P V P V
P
P
V
V
pdV Vdp pdVc c c
VdppdVc c c c c
c VdppdVc c c c ccV
dp CdVp V C
dp pdVc
γ γ
1 1
0
0
+= −
−
+ = − − −
= −
− −
+
+ = =
=
slide 11Physics 1401 - L 23 Frank Sciulli
Adiabatic Expansion - calculus
l Conclude that adiabatic change (no heat enter or leave) requires new relations among parametersu ratio (γ ) spec. heats importantu Ideal gas law also still true
1 1 2 2
0
ln ln const.ln( ) const.
P
V
dp dVp Vp Vp
V
V
CC
pV p
=+ =
+ =
=
=γ γ
γ
γ
γ
γ
2
2
1
1 21
p
V Vf
f
C RC C
RR f
γ
γ( )
= = +
+= = +
1 11 1 2 2
pV constnRT V co
T
nst
V TVV
γ
γ
γ γ− −
=
=
=
slide 12Physics 1401 - L 23 Frank Sciulli
Gases: Isothermal Expansion
l Unitsu R = 8.31 J/(mol-K) = kNA
l Work (constant temperature) done obtained from integral in p-V (see sample prob 20-1)
pV nRT=
f f
i i
V Vf
iV V
VnRTW p dV dV nRTV V
ln
= = =
∫ ∫
IsothermspV =nRT= const
review
slide 13Physics 1401 - L 23 Frank Sciulli
Reversibilityl All these (isothermal,
isobaric, isochoric, adiabatic) are reversible processesu each point on P,V diagram
is a possible state … state can be changed in any direction along a curve
l Contrast free expansion of gas u not reversibleu Tf=Ti
u pfVf=piVi
u but f state cannot return to i state
slide 14Physics 1401 - L 23 Frank Sciulli
Sample Problem 20-9al Recall sample problem 20-2,
isothermal expansion of 1 mole O2 at 310K & pi=2.0 atm from 12liters to 19 liters gave Pf = 1.26 atm & W=Q=1180 J.
l Now adiabatic expansion of same sample.u Final pressure?u How much work done?
.
K. K
VT TV
T
TT
γ −
=
= =
∆ = −
11
2 12
0 40
2
2
1231019
25852 0
.
.
. atm
Vp pV
p
p
γ
=
=
=
12 1
2
1 40
2
2
122 019
1 05
V
QW E
nC T
W
int
0
(1)(20.85)( 52.0)1085 Joules
=
= −∆
= − ∆
= − −
=
slide 15Physics 1401 - L 23 Frank Sciulli
Contrast Gas Expansions (sample 20-2 & 20-9)
l Check you get the answers for thesel Big difference for irreversible: ENTROPY
Variable Isothermal Adiabatic Free
P1 2.0 atm 2.0 atm 2.0 atm
V1 12 L 12 L 12 L
T1 310 K 310 K 310 K
P2 1.26 atm 1.05 atm 1.26 atm
V2 19 L 19 L 19 L
T2 310 K 258 K 310 K
W 1180 J 1085 J 0 J
Free Expansion
reversible irreversible
All involve expansion of gas at 2 atm, 310K, and 12liters to a volume 19liters in different ways
slide 16Physics 1401 - L 23 Frank Sciulli
Entropy Definition
l precise quantitative definition measuring heat flow energy at specific temperature (S=Q/T, units J/K)u Sign convention: heat added (∆Q>0), then entropy increases (∆S>0)
l simplest underlying concept underlying thermodynamics:u Closed system … total entropy change zero or positive
l Reversible process: an entropy increase is compensated by an entropy decrease in another part of system … entropy recoverable
l Irreversible: entropy increases and cannot be recoveredl can calculate entropy change for irreversible process by connecting
initial & final states with (series of) reversible processes
for fixed
f
ifi
dQS S ST
Q TT
∆ = − ≡
∆=
∫
slide 17Physics 1401 - L 23 Frank Sciulli
Second Law of Thermodynamics
l Simplest (and simplistic) expression: “Heat energy always flows from hotter bodies to cooler bodies”u intuitively obvious
l Take a more complete statement (though more abstract) with concept of entropyu “In any closed system, the entropy always increases for
irreversible processes and is constant for reversible processes.” Entropy never decreases in closed systems, and typically increases in real processes!
u Heat flow from hot to cold is a consequence.u Found to be provide experimentally correct predictions in
wide array of circumstances u Has a physical interpretation in terms of the system’s orderu Provides an understanding of where our sense of the “arrow
of time” arises.
slide 18Physics 1401 - L 23 Frank Sciulli
Reversibility: isothermal expansion of gas
l Entropy increase of the gas during expansion occurs as a consequence of heat (Q) into gas at temp T
l An equal and opposite decrease in the entropy of the reservoir occurs at the same time
l If the gas is isothermally compressed, the gas entropy decreases and the reservoir entropy increases
reversibleisothermalexpansion
gasexpansion ln
f
iV
f
i
VpdVQ W nRT dVST T T T
V
VVS nR∆ = +
∆ = = = =∫ ∫
gascompression ln f
i
VS nRV
∆ = −
slide 19Physics 1401 - L 23 Frank Sciulli
Isochoric temperature change of gas
l Add heat with doing work by increasing reservoir temp
l Temperature change of gas in contact with temp reservoir is also reversible
i
i
f
i
i
f
f f
fT T V
i
iT T T TV
T
f
VT
TS n
Q E dTS n
CT
TS nC ST
CT T T
int
ln
ln
→
→ →
∆ = +
∆ = + =
= =
−∆
∆ ∆∆ = ∫ ∫
slide 20Physics 1401 - L 23 Frank Sciulli
Entropy (more)l Entropy is a physical measurable property of a
system (like T,V, p,…) that is calculable in terms of the othersu “state function”
l How does one calculate entropy change when the net entropy increases? (in an irreversible process like free expansion of a gas)
l Answer: Calculate entropy change for irreversible process by connecting initial & final states with (imagined series of) reversible processes. Examplesu Simple heat flow between two solids.u Free expansion.
slide 21Physics 1401 - L 23 Frank Sciulli
Entropy Change in Irreversible Process
0 00 0
350 650t t
tQ QS = =
=
∆ ∆∆ = − >
l Note that entropy will increase so long as heat flows from hot body to cold body. The reverse (cold to hot) is not possible from 2nd Law.
l Simple heat flow from hot body to cold is irreversible
l Calculate entropy change in irreversible process: connect initial & final states with series of pretend reversible processes
slide 22Physics 1401 - L 23 Frank Sciulli
Entropy change: solidsl Net entropy change of any process of a closed system is either zero
(reversible) or greater than zero (irreversible)u Example sample problem 21-2: 2 equal mass (m) solids at different temps …
equilibrium when both at final temperature Tf … what is entropy chg?
ln
ln ln
f
i
Tff
ii T
f frev L R
iL iR
TdQ dTS mc mcT T T
T TS S S mcT T
∆ = = =
∆ = ∆ + ∆ = +
∫ ∫ 2f
reviL iR
rev irrev
TS mc
T TS S
ln∆ =
∆ = ∆
slide 23Physics 1401 - L 23 Frank Sciulli
Sample Problem 21-2
l Same net entropy increase for the two massesu Lost in irreversible processu Stored in reservoir for reversible process
ln ln
L R
f f
iL iR
S S S
T TmcT T
∆ = ∆ + ∆
= +
[ ]
( ) ( )( . kg)( J/kg K) ln
. J/
Equal masses:
K
ln
iL iR
frev
iL iR
rev
f
irrev
TS mcT T
S
T
S
S
T T C
S
∆ =
∆
=
∆
∆
+
=
=
∆
=
=2
2
012
3131 5 386293 333
2 37
40
i
600 C 200 C 400 C l m= 1.5 kgl c= 386 J/kg-K
slide 24Physics 1401 - L 23 Frank Sciulli
Entropy increase of free expansion of gas
l Derived previously for isothermal expansion of the gas
l Since,in the two processes (reversible and irreversible), the gases have same initial states and same final states u the entropy change of the gas is the
same for both processesl Note that free expansion process is
irreversiblel Note that net entropy of closed
system in isothermal expansion is zero
irreversible in a closed system
reversibleisothermalequivalent
f
f
i
i
VQ W nRTST T T V
VS nRV
ln
ln∆ =
=
=
∆
=
slide 25Physics 1401 - L 23 Frank Sciulli
2nd Law of Thermodynamics: gases
l Net entropy change of any process of a closed system is either zero (reversible) or greater than zero (irreversible)u discussion section 21-2
l Entropy is a state function; depends only on parameters of the system: calculate general change for gas
V
ii f ff
i
rev irrev
dQ dE dWdQ nC dT pdV
S S S
dQT
S S
int
→
= +
= +
∆ = −
=
∆ = ∆
∫
0S∆ ≥
f f
i i
T Vf
Vi T V
f fV
i i
dQ dT dVS nC nRT T V
T VS nC nRT V
entropy change for ideal gas
ln ln
∆ = = +
∆ = +
∫ ∫ ∫
slide 26Physics 1401 - L 23 Frank Sciulli
Thermodynamics and Gases
Next l More on Entropyl Uses in engines, heat pumps, refrigeratorsl Origin of Entropy
Todayl Specific Heats of Simplest Gases
l Constant Volume (isochoric) l Constant Pressure (isobaric)
l Specific Heats more generallyl Adiabatic Expansion l Reversible and Irreversible Processesl Entropy
