Section 1: Zeroth and First Laws

Question 1

zeroth law of thermodynamics

Answer 1

If two systems are separately in thermal equilibrium with a third, then they must also be in thermal equilibrium with each other.

Question 2

zeroth law leads to the concept of

Answer 2

temperature as a measurable quantity

legitimises the use of thermometers and calibration of temperature

Question 3

thermal equilibrium

Answer 3

bulk physical properties are uniform and invariant

in particular, the state variables are unchanged

Question 4

consider 3 systems A,B,C each with state variables P and V

AB thermal equilibrium implies

Answer 4

some fixed relationship between PA,VA, PB, VB

so PA=f(VA,PB,VB)

Question 5

similarly, AC equilibrium implies

Answer 5

so PA=f(VA,PC,VC)

Question 6

equating AB and AC equilbirum

Answer 6

f(VA,PB,VB) = f(VA,PC,VC)

can be solved for PB as PB=g(VA,VB,PC,VC)

Question 7

the zeroth law implies equilibrium between B and C so we also have

Answer 7

PB=F(VB,PC,VC)

Question 8

PB=F(VB,PC,VC) and PB=g(VA, VB,PC,VC) imply that

Answer 8

VA must cancel and there must also exist functions

θB(PB,VB) = θC(PC,VC)

Question 9

this we have a function θB(PB,VB) - an EQUATION OF STATE - whose value is

Answer 9

the same for two thermally equilibrated systems

the empirical temperature

Question 10

makes sense to define a temp scale based on

Answer 10

1. a thermometric property ‘X’
2. two arbitrary fixed points, θ1 and θ2
3. the form θ=θ1+(θ2-θ1)(X-X1/X2-X1)

Question 11

form of first law taught last year

Answer 11

ΔU=Q+W

Question 12

differential form of the first law

Answer 12

dU= δQ+δW

Question 13

dU is the

Answer 13

infinitesimal change in internal energy in J for a process

Question 14

δQ is the

Answer 14

infinitesimal heat (thermal energy in J) transferred in the process

Question 15

δW is the

Answer 15

infinitesimal work (in J) done by the surrounding on a system during the process

Question 16

dU is an exact differential so

Answer 16

U is a function of state
so ΔU is path independent

Question 17

δQ and δW are inexact differentials so

Answer 17

neither Q nor W are uniquely defined for a process and depend on the path taken between initial and final states

Question 18

inexact differentials often relate to

Answer 18

irreversible processes

Question 19

irreversible processes don’t map onto

Answer 19

a unique line on a PV diagram

Question 20

example of an irreversible process

Answer 20

free expansion of gas

Question 21

what can be calculated for an irreversible process

Answer 21

the end points

Question 22

the 1st law tells us that although the distribution between heat (thermo) and work (dynamics) can’t always be determined,

Answer 22

their combination (thermodynamics) is calculable

U is a state function

Question 23

we can choose quasistatic, reversible processes to calculate alternative paths with

Answer 23

well-defined δQ and δW and therey explore ΔU

Question 24

from the first law - need to consider calculable circumstances eg

Answer 24

δQ=0, δW=0 or where δW can be determined through an equation of state

Question 25

1st law calculations - typical situation is

Answer 25

a frictionless piston compressing a fluid

Question 26

piston - we consider

Answer 26

reversible changes from state (P1,V1) to state (P2,V2) by applying a force to the piston treat as a quasistatic process

Question 27

quasistatic process

Answer 27

a series of infinitesimal equilibrium states

Question 28

for the reversible process, the incremental work done by the surroundings on the gas is given by

Answer 28

δW=-PdV

Question 29

work done by the surrounding on the gas for the complete process is given by

Answer 29

ΔW= - ∫ Pdv

Question 30

for an ideal gas under isothermal conditions, ΔW=

Answer 30

-∫ PdV = -ΔnRT/V dV = nRTln (V1/V2)

Question 31

alternatively, the combination of isobaric process and isochoric process yields, ΔW=

Answer 31

-∫ between 1 and3 of PdV - ∫ between 3 and 2 PdV =P1∫ between 1 and 3 dV = P1(V1-V2)

Question 32

clearly, from the two versions of ΔW, it is

Answer 32

path dependent *ΔW is the area under the process curve*

Question 33

the work done by the surroundings increases the system's internal energy so

Answer 33

a reduced volume, increased pressure corresponds to a positive ΔW

Question 34

path independence implies that

Answer 34

the order of differentiation does no matter ∂/∂x)y (∂z/∂y)x = ∂/∂y)x(∂z/∂x)y

Question 35

formal requirement for dz to be exact

Answer 35

(∂x\∂y)x = (∂y/∂x)y

Question 36

reciprocal theorem

Answer 36

(∂x/∂z)y = (∂z/∂x)^-1 y

Question 37

reciprocity theorem

Answer 37

(∂x/∂y)z (∂z/∂x)y (∂y/∂z)x = -1

Question 38

chain rule of differentiation

Answer 38

(∂x/∂y)z = (∂x/∂A)z (∂A/∂y)z

Question 39

the non-natural derivative

Answer 39

(∂A/∂x)z = (∂A/∂x)y + (∂A/∂y)x (∂y/∂x)z

Question 40

we have δQ=dU+PdV for a reversible process so the heat capacity at constant volume, Cv is

Answer 40

lim as ΔT approaches 0 of (δQ/ΔT)v = (∂U/∂T)V

Question 41

for Cp, we want dP=0 so

Answer 41

rewrite the first law

Question 42

new potential, enthalpy given by

Answer 42

H=U+PV

Question 43

enthalpy in differential form

Answer 43

dH=d(U+PV) = dU+PdV+VdP = δQ + VdP

Question 44

Cp=

Answer 44

lim as ΔT approaches 0 of (δQ/ΔT)p = (∂H/∂T)p

Question 45

since U, P and V are functions of state,

Answer 45

so is H

Question 46

enthalpy is commonly used for

Answer 46

isobaric conditions including fluid flow through pipes and chemistry

Question 47

devices working on a continuous fluid flow - fluid in device is

Answer 47

pushing fluid in front and being pushed by fluid behind use energy conservation

Question 48

enthalpy - fluid flow - volume=

Answer 48

piston area x distance

Question 49

change in enthalpy =

Answer 49

H2-H1 = Q-W

Question 50

consider specific enthalpy, h, as the enthalpy per unit mass then

Answer 50

mass flow per unit time dm/dt and heat per unit time dQ/dt yields power as P=(h1-h2)dm/dt +dQ/dt

Question 51

energy conversion including kinetic and potential energy terms

Answer 51

ΔH+Δ(KE+PE)_bulk = Q-W

Question 52

for specific (ie per unit mass) terms, w=

Answer 52

h1-h2+1/2(v1^2-v2^2) + g(z1-z2) +q

Question 53

bernoulli's equation - for an incompressible fluid (constant density p), with no change in internal energy, u, zero heat q and zero work w

Answer 53

0=1/p(P1-P2) +1/2(v1^2-v2^2) + g(z1-z2) where p=density, P=pressure

Question 54

Clausius' form of first law

Answer 54

In a thermodynamic process involving a closed system, the increment in the internal energy is equal to the diference between the heat accumulated by the system and the work done by it.

Question 55

Inexact differentials often relate to

Answer 55

irreversible processes which cannot be depicted as a continuous line on a P-V diagram

Question 56

first law tells us that although the distribution between heat and work cannot always be uniquely determined

Answer 56

the combination as internal energy can be U is a state function

Question 57

how to explore the change in U

Answer 57

choose quasistatic (slow) reversible processes to calculate alternative paths with well defined δQ and δW

Question 58

Born's formulation of the first law

Answer 58

For a closed system, in any arbitrary process of interest that takes it from an initial to a final state of internal thermodynamic equilibrium, the change of internal energy is the same as that for a reference adiabatic work process that links those two states. This is so regardless of the path of the process of interest, and regardless of whether it is an adiabatic or a non-adiabatic process. The reference adiabatic work process may be chosen arbitrarily from amongst the class of all such processes.

Question 59

δW for irreversible processes

Answer 59

>-PdV before, piston was frictionless and process quasistatic. Without these provisions, infinitesimal work will be larger

Question 60

W for isobaric conditions

Answer 60

W=- integral PdV =P(V2-V1)

Question 61

W for isothermal conditions

Answer 61

integral nRT/V dV =nRT ln(V1/V2)

Question 62

W for adiabatic conditions

Answer 62

- integral C/V^gamma dV where C=P1V1^gamma = P2V2^gamma

Question 63

if δW=-PdV then first law can be rewritten as

Answer 63

δQ=dU+PdV allows us to calculate Cv and Cp