1

ME2121/E Thermodynamics Chapter 2 Definitions and Basic

Concepts, and Temperature Scales by

Professor KC Ng (EA-07-22) Email: mpengkc@nus.edu.sg

(August 2014)

As in all subjects of mechanical engineering, classical thermodynamics is defined by sets of terminologies that engineers and scientists used to describe the behaviours of matters, processes, reactions and energy interactions. As students taking thermodynamics for the first time, one of the most common questions asked is:

What is a thermodynamic system? It is simply a study of behaviour of matter (gases, liquids or solids) in an enclosed space which an engineer or scientist would like to investigate. Matters (such as gases, liquid or solids) are contained within a physical or imaginary boundary of a closed or open system. The energy generated or consumed by the matters may be transferred in and out from the boundary of system. Outside the system boundary is called the surroundings, as shown schematically in Figures 2.1.

Fig. 2.1 Boundary and surroundings of a thermodynamic system

A close system has no mass leaving its boundary (dm = 0). An example to demonstrate this concept is the piston-cylinder assembly, commonly found in engineering devices such as the shock absorber of a car or the catapult on an

aircraft carrier that ejects a jet-plane taking off from the deck of an aircraft carrier. (http://science.howstuffworks.com/aircraft-carrier3.htm). One of the terminologies used in thermodynamics is called property. It is a variable is used to describe the behaviours or characteristics of a system. The changes in the states of matter are measured by variables such as temperature (T), pressure (P), volume (V), enthalpy (h), entropy (s), etc. They are measured by using suitable sensors, calorimeters, etc., where their changes are monitored by quantifiable readings of sensors. One common example of a system is the piston-cylinder assembly, as shown in Fig. 2.2. For example, the matter contained inside the cylinder is air (a mixture of N2 and O2). They have direct relationships between its volume (V) and the temperature (T) as well as the pressure (P) that one can measure or monitor. These measurable variables are termed the properties or characteristics of a system.

Fig. 2.2 An example of the piston-cylinder assembly

Similarly, there are other variables of a thermodynamic system that we will come across in the coming lectures; For example, mass (m), energy of matters in system or internal energy (U), enthalpy (H=U+PV), specific heats (Cp), entropy (dS=dQ/T), etc.

system

Matter, e.g., gases, liquids or solids

Heat transfer (Q)

Gas matter, as measured by P, V & T

Work transfer (W)

Boundary Boundary

air

2

Thus, a thermodynamic system is described by the following key points:

It can be characterize by variables called properties,

a property measures the state of matter(s) without causing changes to it,

a property is used to describe the path of an event or process: from a starting point (1) to an ending point (2). Eg., we define the specific heat of a matter as;

2

1

1v

v TU

mC

where the subscripts 1 denotes the starting point, 2 the ending point of a process. The subscript v is a variable (outside the bracket) being held at constant during the integration along the process.

a property can be defined in terms of other established properties, e.g., the enthalpy (H=U+PV) and it is given by (in the extensive form function of mass)

PVUH

or in the intensive (per kg) form as

PvuhormVp

mU

mH .

where the lower case variables represent mass independent or intensive variables. Note that P is not dependent on the mass in the control volume.

Quantity of matters In SI units, the basic amount of substance is given by mass in kilogram (kg). It is also common for an engineer to refer to the amount of matter as mols, and 1000 mols or 1 kmol. For

example, 1 kmol of Carbon (C12) contains the number of carbon molecules (basic structure of matter) equal to a mass of 12 kg. For gases at the standard temperature and pressure (STP), this number of molecules (volumetric basis) per kmol is termed the Avogadros number, i.e.,

No = 6.023 1026 [molecules/kmol]. Hence, the mass of 1 kmol of any substance is also called the molar mass (M), i.e.,

m [kg] = nM [mol x kg/mol] or

m/M = n, also equal to its volume fractions

where n is the number of moles. This implies that the mass fraction is given by volume fractions since mols are related to the volume via the Avogadros number (No) . Example: Air is a mixture and its mass fractions in the atmosphere can be computed via the mol fractions) Consti-tuents

Volume Fraction (n/n)

Mole-cular mass [kg/kmol]

mi= ni xMi [ kg]

By mass Fraction (m/m)

Nitrogen (N2)

0.79 (79%)

28 =0.79x28 =22.12

22.12/28.82 =0.767

Oxygen (O2)

0.21 (21%)

32 =0.21x32 6.72

=6.72/28.82 =.233

1.0 100%

=28.82 1.0

State of a thermodynamic system: The state of a system is defined by the thermodynamic properties and at least two independent properties are needed to locate a state, e.g., the (P & T) or (P & v). When two points on a thermodynamic diagram have the same numerical values, it is termed as identical states. Process(es): A process is the path between two given states where there may be interactions of heat

3

(Q) and work (W) transfer in or out of a control volume. The Q or W are deemed as a useful effect arising from the process.

The area under a process path is of interest to an engineer because it represents a variable e.g., Q or W, which is a non-property as it is path dependent. Fig. 2.3 shows that when two processes are connected in such a manner that at the end of the processes, and should the path returns to its initial state point, 1 this is termed a cycle. Fig. 2.3 A cycle is demonstrated by two process

paths between 2 state points: 12 is an outgoing process path whilst the dotted line ---- is the return path of a cycle.

As the initial (1) and final (2) states are defined by thermodynamic properties, say (pi). The net change of a variable undergoing a cyclic process would be zero (naught), i.e.,

0 (1) because is the property of a system, such as P or v or T or any variable of the system. A property of a thermodynamic system must satisfy the definition of a continuous function, that is, it not an abrupt or discontinued function. Thermodynamic property relation for a cycle

A cycle, as shown below, is described by two thermodynamic variables, say X and Y. If their states are defined numerically by the values indicated (in bracket), demonstrate from these states that the principle of a property is valid. A requirement of a property is that the variable of the states (numerically) must sum to zero when the process path returns to the same starting point. That is: 0 where VorTPbecan , 0)22()42()44()24(P The numerical numbers at every state point in the cycle represent the respective properties of the matter. You can repeat with any other thermodynamic variable, say V, and it would sum to zero when taken over a cycle. This proves the point that a property in a thermodynamic system always adds to zero when it is returned to its original state. Mathematical Test of a Property We start with a temperature (T) relation for air in a piston-cylinder assembly (see Fig. 2.2), expressed mathematically as a function of two other variables of the system (such as P and v), namely;

T = T (P,v). (2)

P

v

Path A

Path B 1

(P1, v1)

2 (P2, v2)

X= V

cycle

Y=P

A (2,2)

B (2,4) C (4,4)

D (4,2)

cycle

4

However, from experiments conducted on a piston-cylinder assembly, one obtains a relation for the gas behavior that is given by; Pv = RT or T=Pv /R - an experimental fact (3) where R is a gas constant (= Ro /M), Ro is the universal gas constant (= 8314 J/kmol.K), M is the molecular mass of the gas (kg/kmol).

(1) For T to be a property of the thermodynamic system, its change must satisfy the continuous function requirement. From mathematics, the total differential of a variable T in terms of independent variables P and v, that is T = T(P,v)) :

vinChangeldifferntiaPartial

P

PinchangealdifferentiPartial

v

aldifferentiTotal

dvvTdP

PTdT

(4) where the subscripts P and v indicate that they are held constant that follows the partial gradients of T vs. P and T vs. v, respectively.

(2) Another point to note is that a property is independent of path. The change from points A to B (see fig below) can be obtained by a partial integration of any path to obtain the total change on a surface (eg., path C or the sum of dotted paths).

From equation (3)- the experimental fact, the gradients of T to P and T to v are

Rv

PT

v

and

RP

vT

P

, (5)

From mathematics, a continuous function of the form z = z(x,y), the order of partials of z with respect to x and y would be immaterial, i.e.,

yxxyyz

xxz

y

.

Thus, extending this mathematical principle to a thermodynamic system, we demonstrate that the continuous function of the property as given by T = T (P,v), would be

vPT

PvT

..

22

(6)

Differentiating the gas relation again to obtain the second order differentials, we have

RPT

vPvT 1

.

2

and

RvT

PvPT 1.

2

,

(7)

Note: we have the same result in Left Hand Side term = Right Hand Side term in equation (7): - Proven the requirement of continuous function. Hence, we conclude that the equation (3) is a property of a thermodynamic system because it satisfies the requirements of a continuous function. Furthermore, a property can be expressed in terms of other thermodynamic properties. This relation is also known as the equation of state (EOS) i.e., P v = R T.

Independency Test or Cyclic relation Independency is also a necessary condition for being a property in a thermodynamic system. To

v

p T

A

B

Path along constant V

aldifferentiPartial

vPT

Path along constant P

ldifferntiaPartial

PvT

Path C

A continuous surface

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demonstrate independency of variables in an equation of state (EOS), we need to use two mathematical relations in addition to the thermodynamic relations. The first is the chain rule:

aaa xy

yz

xz

and the second is the derivative inversion: i.e.,

y

y

xzz

x

1

Invoking Equation (4) on the total differential on the change in temperature (T= T(p,v) ):

dvvTdP

PTdT

Pv

If we now take the partial derivative of each term with respect to P but at constant T;

TPTvT Pv

vT

PP

PT

dPdT

Since T is held constant, the only possible solution for the left hand side term is

0

TdP

dT

and noting that

1

TPP

Applying the above results and invoking the derivative inversion rule on the partial derivative, gives

TPv

TPv

Pv

vT

TP

orPv

vT

PT

1

10

This proves the independency requirement of the variables (P, v, T) in the system as the LHS

term is a constant (-1). The individual partials on the RHS can have any value (independent of each other) as long as their product is summed to unity, An alternate proof of independency with thermodynamic variables is demonstrated below by assuming the variables P,v, T of a thermodynamic system. Let us choose arbitrary the relations for P and T as P = P (T,v). From mathematics, the total differential is

dvvPdT

TPdP

Tv

Should we arbitrary select T = T(P,v) as the variable of interest, then the total differential is given by

dvvTdP

PTdT

Pv

Now, substitute dT of the second equation into the first equation, and re-arrange, we get

dvvP

vT

TPdP

PT

TP

TPvvv

01

1

We note that

(i) dP can have any value, i.e., independent of the others if the product of bracketed terms is equal

to unity, ie.,

1

vv PT

TP - this

is obviously a trivial solution. Similarly,

(ii) On the right-hand side, dv can have any value, i.e., independent of the others if

TPv vP

vT

TP

0

=1 =0

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From the second condition of independency, and re-arranging, we can easily show that

TPv Pv

vT

TP

1

Hence, we prove again the independency requirement for P,v and T in a thermodynamic system. It is a necessary requirement for being a property of a thermodynamic system. Example 2.1 Consider the relationship between V, T and P of a gas contained in a vessel (closed system). If the experimental relationship between the expansion and compression coefficients of the gas are inversely proportional to T and P, respectively, then from mathematics, derive from 1st principles the ideal gas law, PV / T = constant (mR). Note the definitions of the expansion ( ) and compression ( ) coefficients for real gases are respectively given by

TTV

V P11

and

PPV

V T11

.

(note that these approximations are found from experiments). Solution: The general expression for the variables can be given by V= V (P,T) From mathematics, the total differential of V is given by

dPPVdT

TVdV

TP

Using the definitions and the experimental relation of the gas, the above can be re-written as

0

PdP

TdT

VdVor

PdPV

TdTV

VdPVdTdV

Integrating, we have,

ln(V) - ln(T) + ln(P) = ln (constant),

i.e., taking the anti-log, we arrived at the equation of state (EOS) of an ideal gas:

tconseT

PV tcons tantan

(= mR) Example 2.2 Estimate the change in the specific volume of air (assuming ideal gas) if the temperature and pressure changes are from T1 = 25oC and P1 =122 kPa to T2 =65o C and P2 =102 kPa, respectively. Compare the changes as calculated directly from the ideal gas law (as derived in example 2.1). Solution: The relationship for the specific volume is given by

dPPRTdT

PR

dPPvdT

Tvdv

T

TP

2

Here, R= Ro /M = 8314.5 (kJ/kmol.K) / 28.98 (kg/kmol) = 0.287 kJ/kg.K, Tmean= ((65+25)/2+273.15) = 318.15 K, Pmean= (122+102)/2 =112 kPa,

7

kgm

xx

xxxxdv

/248.01457.01025.0

)1020()10112(

15.31810287.0)40(1011210287.0

3

323

3

3

3

Using the ideal gas equation, (Pv=RT), the specific volume, v = V/m, can also be estimated by using a finite differencing approach, i.e.,

./2499.07010.09510.0122

298287.0102

338287.0

3

1

1

2

2

12

kgm

xx

PRT

PRT

vvdv

Obviously, the partial differential approach provides a more accurate answer than the finite difference method. The latter employs the equation of state across the initial and final states whilst the former tracks the changes of fluid properties along the paths of processes. 2.1 Zero Law of Thermodynamics When two bodies are said to be in equilibrium with each other, they are said to have the same property (eg., temperature, T). Although we termed this statement (below) as a law, it can not be proven mathematically but it is merely a statement of truth: - that is, an axiom. The Zero Law states that when two systems, are each in mutual equilibrium with a third system, they are also in equilibrium with each other in the same manner Let us illustrate this statement graphically. Say, a master thermometer, such as a mercury-in-bulb type, has an accuracy of 30 0.05o C. Now one brings but separately the thermometer A to be in equilibrium with C and thermometer B to be with C. According to the Zero Law, if B & C are said to be in thermal equilibrium, then A&C are also in the same

state of thermal equilibrium. This implies that A & B are in the same state, 30 oC.

However, many students/researchers tend to extrapolate the results to imply that thermometers A & C would have the same accuracy of 0.05o C. Accuracy of instrument does not follow in such calibration techniques as the latter is dependent only on the properties of instrument. 2.2 Units Before proceeding further, it is timely to examine the units that are likely to come across in this course and they are tabulated below: Quantity Units SI symbols Length meter m Mass kilogram kg Time seconds s Electric current Ampere A Temperature Kelvin K Amount of substance

mole mol.

Solid angle Steradian sr Plane angle Radian rad Energy Joule J Pressure Pascal; Pa Atm pressure Bar 105 Pa Specific internal energy

J/kg u

Specific enthalpy J/kg h Specific entropy J/kg.K s

Thermometer :A

Equilibrium with each other

Thermometer B

Master thermometer C

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2.3 Empirical and Thermodynamic Temperature scales

Have you ever wondered how a numerical number of a temperature scale at ice point of water takes a value of 273.15 K or 491.67 R ? Why not any other numerical numbers? One employs an instrument to read the temperature of a body. This implies that the sensor device has a thermometric property, X, (measurable changes) that responds to the changes in the temperature of a body. Some of the commonly used devices are as follows:

a liquid-in-bulb device, called thermometer (X= expansion of liquid)

resistance change of an inert wire, called resistance-temperature device or RTD, - (X= change in resistance)

the electro-motive force generated across a pair of wires of different materials, called the thermocouples, (X= change in electro-motive force or emf.).

the constant-volume bulb with pressure changes (indicated by mercury), called constant-volume bulb thermometer, (X=change in gas pressures and hence the liquid column).

For any of the above-mentioned devices, if X denotes the thermometric property, the ratio of thermometric properties across two temperatures (easily reproducible) is described by a linear empirical law,

referenceorXX

1

2

1

2 (8)

where the variable X is dependent on the type of instrument used. Constant-Volume Gas Bulb Thermometer (CVGB)

The below figure shows a schematic of a constant-volume bulb thermometer. It comprises a bulb of finite volume, a connecting tubing with a marked point, a U-tube with one end open to atmosphere and a bottom connection to a flexible tubing where the other end is attached to a reservoir containing a fluid, say mercury. The volume in the gas bulb is always held constant by adjusting the height of the reservoir in the flexible arm. When a temperature change of the surroundings of the bulb changes, it causes the pressure in the bulb to vary. From equation (8), the empirical temperature is

given by

1

2

1

2

1

2

1

2

PP

hh

XX

columnsofheight

, (9)

The state 1 of equation (9) refers to a reference temperature (3), taken to be the triple-point of water, 3 (a state where three

Gas bulb containing gas molecules

9

phases of water co-exist). In general, the empirical equation becomes

333 PP

XX

. (10)

Based on experiments reported in the literature, the thermometric properties of three types of temperature sensors are compared, using the common physically reproducible states such as the boiling point (NBP), melting point (NMP), triple point (TP), etc. For example,

Thermocouples (X = electro-motive force or emf)

RTDs (X = Resistance in Ohms) CVGB (X= Pressure of gas)

The table below compares the thermometric variables for six types of natural phenomena (such as Normal Boiling Point (NBP), Normal Sublimation Point (NSP), Normal Melting Point (NMP), Triple Point (TP), etc.) of Nitrogen, Oxygen, water (H2O), Tin (Sn) etc.: Natural points of pure substances

Thermo couples

RTD sensor

CVGB CVGB

3

3RR

3P

P 3P

P

N2 (NBP) 0.12 0.20 0.27 0.29 O2 (NBP) 0.15 0.25 0.31 0.33 CO2 (NSP)

0.56 0.68 0.71 0.72

H2O (TP) 1.0 1.0 1.0 1.0 H2O (NBP)

1.51 1.30 1.37 1.37

Sn (NMP) 2.79 1.89 1.97 1.85 NBP= normal boiling point; NSP= normal sublimation point; TP= triple-phase point; NMP= normal melting point. Let us now re-visit the question posed at the beginning of section 2.3. How are the numerical numbers of an absolute

temperature scale (K or R) for the freezing or boiling point of water assigned?

The answer lies in the celebrated experiments of Lord Kelvin where he employed the CVGB thermometer (the gas inside the bulb follows thermodynamic behavior), filled (i) with different types of gases such as O2, N2, H2, and air and (ii) at assorted gas-bulb pressures (partial vacuum) when measuring the triple point and boiling states of water, that is .

The CVGB thermometer is first filled with one of the mentioned gases at 1000 Torr (1 Torr = 1mm Hg or 101.325/741= 0.136 kPa). Experiments were carried out for two points, namely the triple point of water (where 3-phases of water could be formed when Psat = 0.006112 bar) and the boiling point of water at ambient pressure (boiling could be observed) and record the values of thermometric readings, Ps/P3.

Evacuate the gas bulb slightly, say to 750 Torr, and repeat the experiments to obtain various readings of Ps/P3

Repeat the steps with a lower gas pressure in the CVGB thermometer, until near vacuum.

N2

O2

Air

H2 1.3650

1.3660

1.3670

1.3680

3PPs

250 500 750 1000 P3 (Torr),

1.3661

The extrapolated values of Ps/P3 when pressure in bulb approaches zero.

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A plot of Ps/P3 versus the gas pressure in the bulb is then plotted, as shown above. There are two important facts needed to formulate the numerical values, namely;

The arbitrary number of divisions (A) assigned between the steam to ice point (nearly at Triple Point),i.e.,

divisionsofnosetis

orAwhere

A

,180100

,

(11)

The experimental value of the thermometric property at steam point to the triple point, Ps/P3, see the experiments, i.e.,

3661.133

PPS

i

ss

(12)

From equations (11) and (12), for an arbitrary number of A, say 100 divisions on a selected temperature scale, one obtains

15.2733661.0100

1

100

i

si

PP

(13)

This is called the Kelvin scale (or K). Since, the triple point is found to be 0.01 K higher than the ice point, the thermometric formulae for the CVGB thermometer is written with triple point as the reference point;

3

,mod

16.273PPBT

estemperaturynamictherandempirical

betweenEquivalent

(14)

where the empirical temperature () is now equivalent to absolute temperatures (T). This is because the gases (molecules) in a constant-volume gas bulb thermometer () obeys the Laws of Thermodynamics (hence. the constant

B=1). Should the number of divisions used in equations (11) or (13) is set to 180 (as in Fahrenheit scale), then the freezing point of water is 491.67, or known as the Rankine scale. In the International Temperature and Pressure Systems (ITPS) meeting (some 6 decades ago), all scientists/engineers from many countries agreed to employ only two temperature scales in honour of Celsius (100 divisions) and Rankine (180 divisions). In principle, you could select say 50 or any number of divisions but it will be unlikely to have anyone using the new scale. Example 2.3 This exercise relates to the expansion or contraction coefficients when a liquid experiences a change in temperature. The volume of mercury contained in a bulb type thermometer is Vb,o (m3) at ice point. If the capillary of thermometer has a cross sectional area As,o (m2) and the linear expansion coefficients of glass and mercury are

)/1()/1( KandK mG respectively, determine the change in height of mercury column for a small change in temperature, t (K)? Solution:

NBP of H2O

NFP of H2O

NBP of O2

Absolute Zero

373.15 100 671.67 212

K oC R oF

273.15 0 491.67 32

90 -183 162 -297

0 -273.15 0 -459.67

100 divisions

180 divisions

11

The incremental increase in the volume of mercury (that fills the bulb and stem) due to its properties being sensitive to a t change is given by;

)1(,,,1 tVtVVV mob

tofeffect

mobob

where subscript 1 refers to the state after there is a small change in the temperature. Similarly, the effect of temperature change on the cross-sectional area of capillary tube (stem) is given by

)21())(21(

)(

,

0

22,

2

,

,,1,

tAttr

trrA

Gos

small

GGos

tetemperaturofeffect

Gososs

Similarly, the effect of a temperature change on the volume of glass bulb is

)31(

))()(331(34

)(34

,

0

323,

3,,1,

tV

tttr

trrV

Gob

comparisonbytermssmall

GGGob

rradiusontofeffect

Gobobb

bo

Simultaneously, the mercury (liquid), also expands due to a temperature change, would have to fill the cavity left behind by expanded glass tube and the expanded bulb, i.e.,

volumecapillarynews

volumebulbnew

b

mercuryended

hAVV 1,1,exp

1

Note that 0

,osV =0 (the stem initially is hs,o=0)

)21()31()1)((

,

,

0

,,

thAtVtVV

Gos

Gobmosob

After re-arranging, that is

ttA

Vh

G

Gm

os

ob

21

3,

,

where h is the height of mercury in the stem in meter. Note that t = (t - to). Note that this exercise is for a simple case where hs,o=0. This expression is important when making of a master thermometer. The typical values of KandK mG /1/1 are 910-9 [1/K] and 210-4 [1/K] for the glass and mercury, respectively. A picture of the process is shown below: If the initial height in the tube is hs,o( 0), show that the change in height is given by

tV

hA

AtVhh

G

mGbo

sosoGm

so

osos

21

)2(3

1

If t = 0, then h equals to zero (naught). Also,

h

As1h

Vbo Vb,1 to =T C t= T+t

hso=0

V1=Vb,1 + As1h

12

if hso is zero (the mercury is just at the brim of bulb), then the expression reverts back to the earlier case shown in example 2.3. Home Assignment #1 (Please submit to your tutor (during tutorial session) two weeks after the lecture of Chapter 2. It carries 2.5 marks out of 10 marks).

Question:

Experiments were conducted using a constant volume gas bulb (CVGB) thermometer to measure the triple-point of water (where 3 phases of water co-exist). Four types of gas were used in the CVGB thermometer to measure this condition.

For each type of gas charged into the thermometer, the product of pressure and volume of gas of sensor, i.e., (PV), is calculated. By varying the gas pressure charged into the thermometer, say from 250 Torr to 1000 Torr, a range of PV readings of the CVGB are taken and recorded against the charged pressures.

These experiments are repeated for H2, N2, Air and O2. As the charged pressure of gas in the CVGB thermometer reduces to zero (P 0), as shown in the figure below, the product (PV)* at the triple point of water converges to a single value of 22414 cm3.atm/g.mol.

(a) Write down the empirical temperature scale for the gases at any temperature, .

(b) If the numerical number for the triple point of water is adopted internationally as * = T* = 273.16 K, then, write down the absolutely temperature scale T(K).

(c) From the figure, demonstrate that the gas

constant is 82.05 cm3.atm/mol.K or 8.314 J/mol.K. (Note: 1 atm = 1.01325 bar, 1 m3 = 106 cm3). Explain, briefly, why this is a universal value?

(d) Explain why the product PV converges to a single point in the experiments?

Tutorial #1 (Chapter 2 -ME2121/2121E)

Q1. The behaviour of an ideal gas in a thermodynamic system can be expressed by three variables in an explicit form:

vRTP

where R is a gas constant [J/kg.K]; P, T and v are the pressure (Pa), temperature in Kelvin [K] and specific volume of the gas [m3/kg], respectively. Using mathematics, show that

(i) the variable, P, is a property of the thermodynamic system,

(ii) the remaining variables, T and v, as specified in the given relation, are independent of each other.

(Hint: Read up the requirements of exact differentials and the test for independency of an exact function.) Q2. A mass of gas is trapped in a cylinder fitted with a leak-proof piston. The initial temperature and pressure of the gas are 30C and 3 bar. The pressure of the gas in the cylinder is then

0

P (Torr)

(PV)

(PV)*P0 = 22414 cm3.atm/g.mol

H2

N2

Air

O2

T*= Ttriple point, of water=273.16 K

250 500 750 1000

13

reduced to 2.5 bar and its temperature is increased to 60C. Describe qualitatively, how you would conduct the above-mentioned process in a quasi-static manner. You may make suitable assumption(s) about the boundary and the surroundings. Would friction between the wall and piston affect your answer? (There is no exact answer for this question. You should read up on the basic definitions and the concept of heat and work interactions for a simple system). Q3. Write down the thermometric property of a mercury-in-glass thermometer. The length of a mercury column in a mercury-in-glass thermometer is 5 cm when it is in contact with water at triple point. Calculate

(a) the empirical temperature when the mercury length is at 6 cm,

(b) the length of mercury column when it is at steam point,

(c) If the thermometric property X of the sensor can be read with a precision of 0.1 mm (height of the mercury column of CVGB), what is the error uncertainty of a measurement of temperature difference, i.e., dT = (T1 T2)?

Note: From mathematics, the uncertainty of a

variable (f) is given by =2

1

ki

ixi

i

exf .

You may wish to read about experimental uncertainty in a paper by RJ Moffat, entitled, Describing the Uncertainties in Experimental Results, Robert J. Moffat, Experimental Thermal and Fluid Science 1988; 1:3-17. Ans: (a) 327.8 K, (b) 6.83 cm, (c) 0.77 K. Q4. Write down the thermometric property of a constant-volume gas thermometer. The best

experimental value of vapour pressures between the ratio of the boiling to the ice point of water is found to be

3661.1i

s

PP

where the subscripts s and i refer to the steam and ice points, respectively (See attached figure). If an engineer decides to have an absolute temperature scale that divides equally in 50 divisions between the boiling (steam) and ice points of water (i), show from first principles that the numerical numbers for such an absolute temperature scale (called the S Scale for convenience) at these states of water are 136.57 S and 186.57 S, respectively.

The Y-axis is (Ps / P3) which is the ratio of the gas pressure in Constant-Volume Gas Bulb (CVGB) thermometer when the bulb is dipped separately in steam point and triple point of water.

-- End of Tutorial #1 --

P (torr)

Ps = pressure at steam (boiling) point of water, P3 = triple-point of water, which is 0.01 K higher than ice point of water (i). 1 Torr = pressure at 1 mm of mercury (1 atm = 746 mm Hg)