Chap 5. Simple applications of thermodynamics

single external parameter V. Then a macrostate can be specified by two parameters, V and E.

[eqn: (E, V ) ∀ atop {determines} (T, p )]. So, sometimes, it is more convenient to choose (E, p) or (T, V) or (T, p) for two independent parameters.

And thermodynamic laws [eqn: □Q = T dS = dE + □W~~]

[eqn: □Q = TdS = dE + pdV]

∀

5-1. Ideal gas

(1) Equation of state and internal energy

equation of state ; [eqn: pV = N k_B T = nu R T ~~[R ; gas ~c o n stant]]

[eqn: pile {―――――-> # derive }] [eqn: E = E (T)]independent of V.

<Actually it is obvious because there are no interactions between particles so interparticle distance is not relevant.>

<derivation>

Consider [eqn: E=E(T,V)]and [eqn: S = S(T,V)]

[eqn: left( pile { dE& = & left( ∂E over ∂T right) _V dT & + &left ( ∂E over ∂V right) _T dV # dS &=& left( ∂S over ∂T right) _V dT &+& left ( ∂S over ∂V right ) _T dV }]

thermodynamic law : [eqn: TdS = dE + pdV]

∀ [eqn: TdS = left ( ∂E over ∂T right)_V dT + left ( ∂E over ∂V right) _T dV + nuRT over V dV]

[eqn: ∴ left [ lpile { left ( ∂S over ∂T right) _V &=& 1 over T left ( ∂E over ∂T right) _V ~&―①# left ( ∂S over ∂V right ) _T &=& 1 over T left ( ∂E over ∂V right) _T + nuR over V ~&―② }]

[eqn: {∂^2 S } over ∂V∂T = left ( ∂ over ∂V right ) _T ① = left ( ∂ over ∂T right ) _V ②]

[eqn: ∀~~ 1 over T {∂^2 E } over ∂V∂T = 1 over T {∂^2 E } over ∂T∂V - 1 over T^2 left ( ∂E over ∂V right ) _T]

[eqn: ∴ ~~ left ( ∂E over ∂V right) _T = 0 ~~~~~~∀ E = E(T)]

* Actually, for monatomic Ideal gas,

[eqn: Omega (E ) = B V^N E ^ { 3N over 2 }]

[eqn: 1 over T = ∂S over ∂E = k_B {∂lnOmega} over ∂E = k_B 3 over 2 N 1 over E]

[eqn: ∴ E = 3 over 2 N k_B T = 3 over 2 nu R T]

(2) specific heat

ⓐ [eqn: c_V]= ?	[eqn: □Q = dE]

[eqn: ∴ ~~c_V = 1 over nu left ( ∂E over ∂T right ) _V = c_V (T)]independent of V.

and

[eqn: dE = left ( ∂E over ∂T right)_T dT = nu c_V dT](only correct for Ideal gas i. e. [eqn: left ( ∂E over ∂V right ) _T = phi])

[eqn: dE = nu C_V dT]

ⓑ [eqn: c_p]=?	[eqn: □Q = dE + p dV = nu c_V dT + pdV]

When p is fixed,	[eqn: d(pV) = nu R dT]

[eqn: pdV = nu R dT]

[eqn: ∴ ~~□Q = nu ( c_V +R ) dT]

[eqn: ∀ c_p = 1 over nu left ( □Q over dT right ) _p = c_V + R]

[eqn: c_p = c_V +R]

∴

ratio  [eqn: gamma == c_p over c_V = 1 + R over c_V > 1]

* Actually, for monatomic Ideal gas,

[eqn: E = 3 over 2 nu R T][eqn: ∴ c_V = 1 over nu left ( ∂E over ∂T right ) _V = 3 over 2 R]

[eqn: ∴ left { pile { c_V = 3 over 2 R # c_p = 5 over2 R } and~ gamma = 5 over 3]

(3) Adiabatic process

· isothermal process ∀ [eqn: PV = nu RT ~=~c o n stant]

· Adiabatic process (no heat change)

[eqn: □Q = phi ~~~~∀p~~↔ V~??]

[eqn: ∴ dE + p dV = 0]

[eqn: dT = - p over nuc_V dV]

[eqn: ∀ nu c_V dT + p dV = 0 ~~∀~~]

[eqn: pV = nu R T]

[eqn: pdV + V에 &= nu R dT = nu(c_p -c_V )dT#& =(1 - gamma) p dV]

[eqn: ∀ V dp + gamma p dV = 0]

[eqn: ∀ dp over p + gamma dV over V = phi ∀ d(ln p + gamma ln V) = phi]

[eqn: pV ^gamma = c o n s t a nt]

∀

(4) entropy

[eqn: Tds = dE + pdV]

[eqn: ds = nuc_V over T dT + nuR over V dV]

[eqn: pile{∴ S(T,V) - S(T_0 , V_0 ) &= nu int from T_0 to T {c_V (T') dT'} over T' + nu R int from V_0 to V dV' over V'# {pile{(Ideal ~gas ~; # c_V = co n s ta n t )} }& = nu c_V ln(T/T_0 ) + nu R ln ( V/V_0 ) }}]

[eqn: pile { ∴ S(T,V) &= nu [ c_V ln T + R ln V + const. ]# & = nu R [ 3 over 2 ln T + ln V + const.] }]

* Actually, for monatomic Ideal gas,

[eqn: lpile{S = k_B ln Omega & = k_B [ ln B + N ln V + 3 over 2 N ln E]#&= k_B [ ln B + N ln V + 3 over 2 N ln ( 3 over 2 N k_B T )]}]

[eqn: ∴ S(T, V ) = nu R [ 3 over 2 ln T + ln V + const.]]

5-2. Thermodynamic relations

(1) Maxwell relations

consider [eqn: T dS = dE + p dV]

For convenience, we rewrite this as

[eqn: dE = T dS - pdV]

∀ shows that [eqn: E = E(S, V)]such that

[eqn: left ( ∂E over ∂S right) _ V = T ~~~ a n d ~~~left ( ∂E over ∂V right ) _S = -p]

[compare [eqn: left ( ∂S over ∂E right ) _ V = 1 over T]and  [eqn: left ( ∂s over V right ) = p over T ~:~c o m plex]]

∀ So, instead of [eqn: (E, T ; P, V )]variables, one can use (S, T ; S, V) variables.

∀ Four possible pairs : (S, V), (S, P), (T, V), (T, P)

simplest thermodynamic functions with variables (S, V), (T, V), (T, P) ??

· internal energy : E

[eqn: under {dE = TdS - pdV }―――->][eqn: E = E(S, V)]

· enthalpy : [eqn: H == E + pV]

[eqn: under {dH} = dE + pdV + V dp =under{ TdS + V dp} ~~~∴ H = H(S,``p)]

· Helmholtz free energy : [eqn: F == E - TS]

[eqn: under {dF} = dE - TdS -SdT = under {- S dT - pdV} ~∴F = F(T, ~V)]

· Gibbs free energy : [eqn: G = E - TS + pV]

[eqn: under {dG} = dE - TdS + p dV + V dp =under { - S dT + VdP} ~∴G = G(T, p)]

·E ; [eqn: {∂^2 E} over ∂S∂V = left ( ∂ over ∂S right ) _V left ( ∂E over ∂V right ) _S = left ( ∂over ∂V right ) _S left ( ∂E over ∂S right ) _V]

[eqn: left ( ∂V over ∂S right ) _p = left ( ∂T over ∂p right ) _S]

∴

·H ; [eqn: {∂^2 H} over ∂S∂p = left ( ∂ over ∂S right ) _p left ( ∂H over ∂p right ) _S = left ( ∂over ∂p right ) _S left ( ∂E over ∂S right ) _p]

[eqn: left ( ∂V over ∂S right ) _p = left ( ∂T over ∂p right ) _S]

∴

[eqn: left ( ∂V over ∂S right ) _p = left ( ∂T over ∂p right ) _S]

[eqn: left ( ∂V over ∂S right ) _p = left ( ∂T over ∂p right ) _S]

·F ; [eqn: {∂^2 F} over ∂T∂V = left ( ∂ over ∂T right ) _V left ( ∂F over ∂V right ) _T = left ( ∂over ∂V right ) _T left ( ∂F over ∂T right ) _V]

[eqn: +left ( ∂p over ∂T right ) _V = +left ( ∂S over ∂V right ) _T]

∴

·G ; [eqn: {∂^2 G} over ∂T∂p = left ( ∂ over ∂T right ) _p left ( ∂G over ∂p right ) _T = left ( ∂over ∂p right ) _T left ( ∂G over ∂T right ) _p]

[eqn: left ( ∂V over ∂T right ) _p = -left ( ∂S over ∂p right ) _T]

∴

There 4 relations between S, T, p, V are called

"Maxwell relations"

(2) Specific heats

[eqn: □Q = TdS]

[eqn: ∴ atop □ left { lpile { C_V & = & left ( □Q over dT right )_V& = &T left ( ∂S over ∂T right )_V # C_p & = & left ( □Q over dT right )_p& = &T left ( ∂S over ∂T right )_p } ~right } pile { r e lation ~ between # C_V ~ and C_p~ ? }]

Consider S = S(T, V). Then,

[eqn: lpile { TdS & = T left { left ( ∂S over ∂T right )_V dT + left ( ∂S over ∂V right )_T dV right } # & = C_V dT + T left ( ∂S over ∂V right )_T dV }]

Consider V = V(T, p)  (remember [eqn: C_p]use T, p variables).

Then,

[eqn: TdS = C_V dT + T left ( ∂S over ∂V right )_T left [ left ( ∂V over ∂T right )_p dT + left ( ∂V over ∂p right )_T dp right ]]

Fix p (dp=0) and divide by dT.

[eqn: ∀ T left ( ∂S over ∂T right )_p = C_V dT + T under { left ( ∂S over ∂V right )_ T } left ( ∂V over ∂T right )_p]

[eqn: = left ( ∂p over ∂T right )_V]by Maxwell relations.

[eqn: C_p = C_V + T under { left ( ∂p over ∂T right )_V left ( ∂V over ∂T right )_p }]

∴

can be calculated from eq. of state.

· Define

[eqn: left{ lpile { alpha &==& 1 over V left ( ∂V over ∂T right )_p &;& volume~ coeff.~ of~ thermal~ e x pansion#& & & & at~ c onstant~ pressure # kappa &==& - 1 over V left ( ∂V over ∂p right )_T &:& i so thermal ~ compressibility }]

* Consider V = V(T, p) again

[eqn: dV = left ( ∂V over ∂T right )_p dT + left ( ∂V over ∂p right )_T dp]

Fix V(dV = 0) and divide by dT.

[eqn: ∀ atop □lpile { 0 &=left ( ∂V over ∂T right )_p + left ( ∂V over ∂p right )_T left ( ∂p over ∂T right )_V # & = alpha V - kappa V left ( ∂p over ∂T right )_V }]

[eqn: ∴ left ( ∂p over ∂T right )_V = alpha over kappa]

[eqn: ∴C_p = C_V + T alpha over kappa cdot alpha V = V_V + VT alpha^2 over kappa]

[eqn: C_p = C_V + VT alpha^2 over kappa]

∴

molar specific heats;

[eqn: c_p = c_V + v T alpha^2 over kappa ~~( v = V over nu ~;~molar~volume)]

* Ideal gas

[eqn: pV = nu R T ~~∀~~pv = RT]

[eqn: left { lpile { alpha &=& 1 over v left ( ∂V over ∂T right )_p &=& R over vp &=& 1 over T # kappa &=& - 1 over v left ( ∂v over ∂p right )_T &=& RT over vp^2 &=& 1 over p }]

[eqn: c_p - c_V = v T p over T^2 = pv over T = R]

(3) entropy and internal energy

If one knows S = S(E, V), then other thermodynamic quantities can be easily obtained like [eqn: left ( ∂S over ∂E right )_V = 1 over T], [eqn: left ( ∂S over ∂V right )_E = p over T], ... .

However, in experimental situations, T is much easily handled than E by putting the system in contact with heat reservoir at T.

So, let's look at S as function of (T, V) and E as function of (T, V). Then, convert T by E in S later.

S = S (T, V)

[eqn: lpile { dS &= left ( ∂S over ∂T right )_V dT + left ( ∂S over ∂V right )_T dV # & = C_V over T dT + under { left ( ∂p over ∂T right )_V } dV }]

can be calculated from eq. of state.

So, if we know eq. of state and [eqn: C_V = C_V (T, V )], then S can be calculated as follows.

[eqn: pile{S(T,V) - S(T_) ,V_0 ) &= under {S(T,V) - S(T_0 , V) }#& dV = 0 ~ process# &+under { S(T_0 ,V) - S(T_0 ,V_0 ) } # & dT=0 ~process }]

[eqn: lpile {∴ S(T,V) = S(T_0 ,V_0 ) &+ int from T_0 to T {(C_V T', V) } over T' dT' #& + int from V_0 to V left ( ∂p over ∂T (T_0 , V' ) right )_V' dV' }]

E = E(T, V)

[eqn: lpile { dE &= TdS - pdV = C_V dT + T left ( ∂p over ∂T right )_V dV = p dV # & = left ( ∂E over ∂T right )_V dT + left ( ∂E over ∂V right )_T dV }]

[eqn: ∴ atop□ left [ lpile { left ( ∂E over ∂T right )_V &=& C_V # left ( ∂E over ∂V right )_T &=& T left ( ∂p over ∂T right )_V -p {(calculable ~ f r om~ }atop {eq.~ of ~ state ) } }]

[eqn: lpile {∴~ E(T,V) &= E(T_0 , V_0 ) + int from T_0 to T C_V (T', V ) dT' #& + int from V_0 to V left ( T_0 left ( ∂p over ∂T ( T_0 , V' ) right )_V' - p(T_0 , V' ) right ) dV' }]

Actually, volume dependence of [eqn: C_V]can be obtained from eq. of state.

[eqn: lpile { left ( ∂C_V over ∂V right )_T & = left ( ∂ over ∂V right )_T left [ T left ( ∂S over ∂T right )_V right ] = T {∂^2 S } over ∂V∂T # & = T left ( ∂ over ∂T right )_V left ( ∂S over ∂V right )_T = T left ( ∂ over ∂T right )_V left ( ∂p over ∂T right )_V # &= T left ( {∂^2 p } over ∂T^2 right )_V # & (can ~ be~ calculated ~f r om~eq.~of~ state.)# }]

So, with the knowledge of eq. of state and dependence of [eqn: C_V]on T at one fixed volume [eqn: V_1], S and E can be obtained.

[eqn: left [ C_V (T, V) = C_V (T, V_1 ) + int from V_1 to V left ( ∂C_V over ∂V' (T, V') right )_T dV' right ]]

<Example> Van-der Waals gas

eq. of state ; [eqn: left ( p + a over v^2 right ) left ( v-b right ) = RT](v : molar volume)

① volume dep. of molar internal energy □.

[eqn: lpile{ left ( ∂epsilon over ∂v right )_T &= T left ( ∂p over ∂T right )_v -p = T left [ R over v-b right ] - left [ RT over v-b - a over v^2 right ] #& = a over v^2}]

② volume dep. of molar

[eqn: lpile{ left ( ∂c_V over ∂v right )_T &= T left ( {∂^2 p} over ∂T^2 right )_v = T left ( ∂ over ∂T right )_v left [ R over v-b right ] =phi #& ∴~c_v = c_v (T)~~i n dependent~ of~volume}]

③ molar internal energy □(T, v)

[eqn: lpile{ epsilon (T, v) &= epsilon (T_0 , v_0 ) & + int from T_0 to T c_v ( T', v)dT' &+ int from v_0 to v a over v'^2 dv' # &= epsilon (T_0 ,v_0 ) & + int from T_0 to T c_v (T')dT' &- a left ( 1 over v - 1 over v_0 right ) # }]

∴ [eqn: epsilon (T, v ) = int from T_0 to T c_v (T' ) dT' - a over v + const.]

[v → large ∀ r → large (interparticle distance)

∀ negative value of [eqn: epsilon]  → smaller,

∀ [eqn: epsilon] → increase ]

④ molar entropy s(T, v)

[eqn: lpile { s(T,v) & = s(T_0 ,v_0 ) + int from T_0 to T {c_v (T') } over T' dT' + int from v_0 to v R over V'-b dv' #&= int from T_0 to T {c_v (T' ) } over T' dT' + R ln(v-b) + c o nstant }]

5-3. Free expansion & Throttling process

(1) Free expansion

Insulated

[eqn: pile { i n i tial &;& V_1 , T_1 # final & ; & V_2 , T_2 } right } T_2 = T_2 ( V_1 , V_2 , T_1 ) ~??]

total system ;	Q = 0 ( adiabatic but irreversible)

W = 0 ( no work)

∀ △E = 0

∴ [eqn: E(T_1 , V_1 ) = E(T_2 , V_2 )] (energy-conserving process)

ideal gas ; E = E(T)      ―→ T2 = T1 (no temp. change)

real  gas ; E = E(T, V)

(at fixed T, E increases as V increases like in V.d.W. gas.)

(ex) V.d.W gas

[eqn: epsilon (T_1 , v_1 ) = epsilon (T_2 , v_2 )]

[eqn: ∴ int from T_0 to T_1 c_V (T' ) dT' - a over v_1 = int from T_0 to T_2 c_V (T') dT' - a over v_2]

[eqn: ∴ int from T_2 to T_1 c_V (T') dT' = a left ( 1 over v_1 - 1 over v_2 right ) > 0 ~~(because v_1 < v_2 )]

[eqn: c_v > 0 ~~∀∴ T_2 < T_1 ~~(C oo l i n g)]

·If [eqn: c_V]is constant, [eqn: epsilon(T,v) = (T-T_0 ) c_V - a over v + const.]

∀ □-constant lines ∀ [eqn: Tc_V = a over v ~∀~T = a over c_V cdot 1 over v]

And

[eqn: T_1 - T_2 = a over c_V left ( 1 over v_1 - 1 over v_2 right )]

In reality, the container has an appreciable heat capacity.

So [eqn: c_V ―→ v_V + c_{container}]

∴ Cooling rate is much reduced especially

at low temperature

(2) Throttling (or Joule-Thomson) process

; Better cooling process but needs precooling.

Instead of single free expansion, use "continuous" flow in steady state.

Consider particle of mass M as a system

① Q = 0 (adiabatic)

② M does work.

Total work done by the system from initial to final state:

[eqn: W = P_2 V_2 - P_1 V_1]

③[eqn: △E + W = phi]

[eqn: ∴ E( T_2 , P_2 ) - E(T_1 ,P_1 )][eqn: + P_2 V_2 - P_1 V_1 = phi]

∀ [eqn: E_2 + P_2 V_2 = E_1 + P_1 V_1 ~∀~H_2 = H_1]

∀ [eqn: UNDER { H(T_1 , P_1 ) = H(T_2 , P_2 ) }](enthalpy-conserving process)

idal gas ; h = h(T, P)

[eqn: mu == left ( ∂T over ∂P right )_H]: Joule-Thomson coefficient ( ∼ cooling rate)

·calculation of [eqn: mu]

[eqn: dH = T dS + V dP]

H : fixed (dH=0) and needs relation between T and P.

So use [eqn: S = S(T, P)]

[eqn: lpile{ 0 & = T left [ left ( ∂S over ∂T right )_P dT + left ( ∂S over ∂P right )_T dP right ] + V dP# & = C_P dT - T left ( ∂V over ∂T right )_P dP + V dP # & = C_P dT + V ( 1 - alpha T ) dP }]

[eqn: ∴ mu = left ( ∂T over ∂P right )_H = V over C_P ( alpha T - 1 ) = v over C_P ( alpha T -1 )]

·[eqn: mu]in terms of Virial coeff.

5-4. Heat engines and Refrigerators

(1) Heat engines

It is easy to transform mechanic energy into thermal energy.

Important question : how about the "reverse" procedure?

Is it possible to devise the engine such that all absorbed heat can be transformed into the mechanical work with the final macroscopic state of the engine being the same asthe initial state for continuous work? ("perfect" engine)

Answer : NO

Why ? : total system = heat reservoir + engine

so [eqn: △S = under {△ S _T } + under { △S_M }]

[eqn: - q over T]    [eqn: = phi]

[eqn: ∴ △S = - q over T < 0] (violate the 2nd law)

So, find a way to increase △S in total system!!

[eqn: △S = - q_1 over T_1 + q_2 over T_2 ge phi]

efficiency of engine : [eqn: eta == omega over q_1]

[eqn: eta = {q _1 - q_2 } over q_1 = 1 - q_2 over q_1 le 1 - T_2 over T_1]

[eqn: eta_max = 1 - T_2 over T_1]When [eqn: △S = phi]

[Ideal engine]  ∀ quasi-static process

(2) Refrigerators

[eqn: (T_1 > T_2 )]

[eqn: q_1 = q_2 +omega]

(when [eqn: omega = phi], we call it "perfect" refrigerator)

[eqn: △S = q_1 over T_1 - q_2 over T_2 ge phi]

[eqn: {q_2 + omega } over T_1 ge q_2 over T_2]

∴	[eqn: omega ge q_2 left ( T_1 over T_2 -1 right )]

[eqn: omega _min = q_2 left ( T_1 over T_2 - 1 right )]: minimum work needed for cooling

(quasi-static process)

[Ideal refrigerator]

(H. W.) 1, 2, 3, 11, 14, 17, 26