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Fis 2 > Termodinámica > Flashcards

Termodinámica Flashcards

1
Q

Los diferenciales de calor y trabajo son de tipo…

A

INEXACTAS (quita puntos)

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2
Q

F normal form

name

A

Energía de Helmholtz

F=U-TS

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3
Q

dvpF/dV

dvpF/dT

A

On the square, go up to the denominator and to the other corner.
dºF/dV = -p
dºF/dT = -S

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4
Q

(dvpF/dT)^2

A 
dºF/dT = -S
(dºF/dT)^2 = -(dºS/dT) = -Cv/T  (V cte)
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5
Q

(dvpF/dV)^2

A 
dºF/dV = -P
(dºF/dV)^2 = -(dºP/dV) = 1/(kt V)  (T cte)
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6
Q

kt

name

A

Compresibilidad
-1/V (dºV/ dP)(T)
where º means partial derivative and (T) means T=cte

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7
Q

α

name

A

Dilatación térmica

α = 1/V (dºV/ dT)(P) = Pβκt

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8
Q

dV/V ito coeffs

A

αdT-ktdP

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9
Q

β

name

A

Piezotérmico

1/P dºP/dT

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10
Q

ks

A

Es como kt pero con P y a S cte

= -1/V (dºV/dP)(S)

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11
Q

H normal form

A

H=U+PV

Hitler got a paid visit

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12
Q

(dºH/dS)^2

What is cte?

A

P=cte
Since the first derivative wrt S is T, and TdS=CpdT
so
dºT/dS=T/Cp

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13
Q

(dºH/dP)^2

What is cte?

A

S is cte
Since the first dº is V,
dºV/dP = -ksV

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14
Q

Potencial químico

A

μ = (dºF/dn)

where n is moles

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15
Q

G normal form

A

G=F+PV=U-TS+PV=H-TS

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16
Q

(dºG/dP)^2

What is cte?

A

The first dº is V

then the second’s = -ktV

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17
Q

(dºG/dT)^2

What is cte?

A

The first dº is S

then the second’s = -Cp/T

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18
Q

When finding diffs using Born, always…

A

add +μdN at the end

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19
Q

U normal form

A

U=TS-PV

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20
Q

When do you use μ?

A

Sistemas abiertos

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21
Q

Negative signs for condiciones de Schwartz

A

The only things with negative OUTCOMES are
TV
and
StriPpers

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22
Q

Trabajos ito in(ex)tensivas

A

intensiva . diff(extensiva)

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23
Q

1a ecuación de Gibbs-Helmholtz

A

ΔU/T^2 = -dº/dT(ΔF/T)

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24
Q

2a ecuación de Gibbs-Helmholtz

A

ΔH/T^2 = -dº/dT(ΔG/T)

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25
Q

Relación F y variables clásicas

A

El cambio de F es W

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26
Q

W máximo a T=cte

A

The maximum work obtainable from a system where there is not change in temperature is the change in the Helmholtz potential.

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27
Q

Calor isobárica ito potenciales

A

The enthalpy of a reversible isobaric process is the heat involved in the process. This also occurs in chemical
reactions under constant external pressure.

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28
Q

Trick for Schwartz conditions with differentials

A

Let N=normal
f=differential
(dºN/df)(f of N)

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29
Q

Trick for getting α and κt

A

V(P,T)
dV = (dºV/dP)(T)dp + (dºV/dT)(P)dT
Divide everything by V
Then the first term is -kt and the second’s α

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30
Q

Van der Waals equation

A

(p+an^2/v^2)(V-nb)=nRT

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31
Q

If given a coeff…

A

Write down the definition, might help later

32
Q

Entropía

A

S=Q/T

dS=dQ/T

33
Q

Work

A

pdV

34
Q

Relación entre incremento de energía libre y trabajo

A

El incremento de energía libre y el trabajo máximo son iguales y opuestos.

35
Q

dS para gases ideales ito C if
P cte
V cte

A

dS=nCv(dT/T)+nR(dV/V)

dS=nCp(dT/T)-nR(dP/P)

36
Q

“Ecuación fundamental” entropía

A

dS = (1/T)dU + (p/T)dV = nCv(dT/T)+nR(dV/V)

37
Q

¿Qué hacer si necesitas calcular la entropía pero no puedes conocer su valor inicial?

A

Si no conozco el origen de la entropía pero necesito calcularla, utilizar un valor referencial de origen S1. Así, la integral es S = S1 + etc

38
Q

Interpretación de un diferencial con dos términos

A

Se puede interpretar un diferencial con dos términos como un proceso realizado en dos pasos. En un paso, se mantiene cte la variable del diferencial del otro paso. P.e.:
F=-SdT-pdV se hace en un paso a V cte seguido de otro a T cte

39
Q

When given variables = numbers, what do first?

A

Always get all numerical values out of the way FIRST

40
Q

Possible uses for a given coeff. in enunciado

A

IT may be used in an integral, but

Sometimes, a coeff is given just so you can find a missing thermo var. (say V2). Then use that in an integral.

41
Q

Trick for the definition of β

A

Expand dV(T,P), divide by V. The first term is β and the second -kt

42
Q

Sign of work done on system

A

Positive

43
Q

Ecuaciones Poisson

A

PV^γ
TV^γ-1
TP^1-1/γ

44
Q

γ adiabático

A

Cp/Cv

45
Q

Relación de Mayer

A

Cp-Cv=R

46
Q

Formula with Cv ito S

A

TdS=CvdT
if V=cte!
This comes from the fundamental eq. for S

47
Q

Cv monoatómico

diat

A

3/2 and 5/2. Cp is always Cv+1

48
Q

γ mono

A

5/3

49
Q

Trabajo isócoro

W isóbaro

A

W=0

Pext=cte

50
Q

W si expansión libre

A

Es contra el vacío, Pext=0 y W=0

51
Q

Primer principio forma diff

A

dU = d!Q + d!W

where ! means inexact diff.

52
Q

Incremento de energía interna a V cte

A T cte

A

ΔU=nCvΔT

cero

53
Q

H electrica

H magnética

A

He=U-EPV

Hm=U-μoHMV

54
Q

Ecuación de Clapeyron

A

dp/dT=L/TΔv=Δh/TΔv

donde L es el calor latente molar de cambio de fase

55
Q

Aproximación de Clausius-Clapeyron

A

dlnp/dT=Δh/RT^2

solo cuando una de las fases es gaseosa, y donde p es la presión de vapor de la fase condensada correspondiente

56
Q

1 cal in SI

A

4,184J

57
Q

Other R. Units.

A

R’=0,082 [atm.L/mol K]

58
Q

Calor latente

A

E necesaria para cdf

59
Q

Incremento de entalpía libre = 0 cuando

A

Δg=0 cuando estamos en equilibrio de fases, o sea, el punto de transición

60
Q

Ecuación puente

A

(dºU/dV)(T)=T(dºp/dT)-p

61
Q

Punto triple. Deducción matemática.

A

El punto triple es donde confluyen las líneas de fusión, vap. y sublimación. Igualando las condiciones para dos de estas fases, y despejando una variable (p.e. T), se deduce que esa misma es la T para la fase restante.

62
Q

How to know whether a reaction is stable based on potentials

A

We can know whether one side of a transition is stable by calculating a potential when the reaction happens one way (say left to right) and deciding whether it’s what it should be: a min (max).

63
Q

Mins and Max for all potentials

A

At eq.: S max (dº^2)
U min
F min
G min

64
Q

Límites C(p,v) cuando T->0

A

0

65
Q

Límites 3er principio

A

limT->0dºs/dv=0
limT->0dºs/dP=0
Can be paired with maxwell relations for cash money
limT->0dºs/dP=inf means no se puede usar una ecuación de estado dada en el límite inferior de temperaturas.

66
Q

Relación Q y H

A

Q=mΔhv

that’s entalpía de vap.

67
Q

Rendimiento máquina térmica

A

η=W/Q1=1+Q2/Q1=1-T2/T1

IMPORTANT: es trabajo realizado/ calor ABSORBIDO

68
Q

Eficiencia máquina

A

E=1/η

69
Q

Rendimiento Carnot

A

η=1-TL/TH=1-QL /QH

where T1 is hot

70
Q

Relación Q y C

A

dQ=mcdT
c es el calor esp. de la sustancia
add a negative sign if it’s cooling

71
Q

If given some eq. of state, how can you get the coeffs.?

A

You must EXPAND the derivada total dV and replace the definitions based on what you get from the partial derivatives. Likewise, equations of state are obtained from coeffs by using the full dV expansion, say (two terms).

72
Q

Condiciones máquina reversible:

A

dW=dQ+dQo
dQ/T+dQo/To=0
where dQ is heat taken from substance and dQo heat released into the ENVIRONMENT dawg.
if you know dQ ito c, you can replace that in second formula (& then 1st formula) for quick CASH $$$

73
Q

Sign of heat taken from system

A

Negative

74
Q

ΔS en ciclo reversible

A

0

75
Q

β en gas ideal

A

En gas ideal, β=T^-1

76
Q

Rendimiento frigorífico

A

η=|QL/W|

77
Q

alfa can also be written as

A

Pβκt

Fis 2 (7 decks)

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