Systems, States, and Processes:Temperature and the Zeroth Law

  Temperature and the Zeroth Law

So far we have discussed only properties known from mechanics, namely mass m, volume V , pressure p, and velocity V. Temperature, as a measure of how hot or cold a body is, is the first thermodynamic quantity that we introduce.

Indeed, through touching objects we can distinguish between hot and cold.

However, our sense for temperature is relatively inexact, just feel the metal and the wood of your chair, which have the same temperature, but feel different. Objective measurement of temperature requires (a) a proper definition, and (b) a proper device for measurement—a thermometer.

Observation of nature and of processes towards equilibrium have established the following definition of temperature:

Two bodies in thermal equilibrium have the same temperature.

This statement is so important that it is known as the Zeroth Law of Thermodynamics. As example, consider two bodies, e.g., a cup of hot coffee and a spoon, or two stones, at different temperatures T¯A > T¯B which are brought into thermal contact, see Fig. 2.6 for a schematic representation. An equilibration process occurs, and after a while the system comprised of the two bodies reaches its equilibrium state, with a common temperature T . While we shall need the first law—the conservation of energy—to compute its actual value, we know from experience that the final temperature will lie between the initial temperatures, T¯A > T > T¯B .

The zeroth law as stated above implies that if body A is in thermal equilibrium with bodies B and C, than also bodies B and C will be in equilibrium.

All three will have the same temperature.

Thus, to measure the temperature of a body, all we have to do is to bring a calibrated thermometer into contact with the body and wait until the equ-librium state of the system (body and thermometer) is reached. When the size of the thermometer is small compared to the size of the body, the final

temperature of body and thermometer will be almost equal to the initial temperature of the body, see Sec. 3.12.

Thermometers and Temperature Scale

So what is a thermometer? Thermometers rely on the change of physical properties with temperature. The volume of most liquids grows with temperature, the volume change is employed in mercury or alcohol thermometers: liquid thermometers rely on the measurement of length. Resistance thermometers rely on the change of ohmic resistance of electric conductors with temperature. Thermocouples use thermoelectric effects—voltage caused by temperature difference—to measure temperatures.

Thermometers must be carefully calibrated, so that different thermometers, and different types of thermometers, will agree in their measurements. The calibration requires reference points that can be reproduced accurately, and a proper definition of the scale between the reference points.

The temperature scale used in daily life is the Celsius scale which measures temperature in degrees Celsius [ ◦C]. The Celsius scale was originally defined based on the boiling and freezing points of water at p = 1 atm to define the temperatures of 100 ◦C and 0 ◦C. The Fahrenheit scale, which is employed in the USA, assigns these points the temperatures 212 ◦F and 32 ◦F. National and international bureaus of standards now use a larger number of well- defined fix points for the calibration of thermometers.

Just having reference points is not enough, there must be a well-defined scale for the temperatures between the reference points. As an example we consider two liquid thermometers filled with different liquids A and B, which are build such that their liquid columns have the same heights for the reference points at 0 ◦C and 100 ◦C, see Fig. 2.7. However, the change of volume with temper- ature might follow different non-linear functions V (T ) for the two liquids, so that both thermometers show different heights for temperatures between the reference points, as example the figure shows different readings for 50 ◦C.

Gas Temperature Scale

To define a proper temperature scale between the reference points, one has to agree on a particular reference substance, and define the scale for that sub- stance. Thermometers involving other substances, or other physical effects, can then be calibrated based on the reference.

The reference substance used is the ideal gas. Any gas at sufficiently low pressures and large enough temperatures (see Sec. 6.10), behaves as an ideal gas. From experiments one observes that for an ideal gas confined to a fixed volume the pressure increases with temperature. The temperature scale is defined such that the relation between pressure and temperature is linear, that is

T ( ◦C) = a + bp (2.9)

where the two constants a and b can be found from two well-defined reference points. With this, temperature is determined through measurement of pressure, see Fig. 2.8. For the Celsius scale one finds a = −273.15 ◦C independent of the ideal gas used. The constant b depends on the volume, mass and type of the gas in the thermometer.

By shifting the temperature scale by a, one can define an alternative scale, the ideal gas temperature scale, as

Thermal Equation of State

Careful measurements on simple substances show that specific volume v (or density ρ = 1/v), pressure p and temperature T cannot be controlled independently. Indeed, they are linked through a relation of the form p = p (v, T ), or p = p (ρ, T ), known as the thermal equation of state. For most substances, this relation cannot be easily expressed as an actual equation, but is laid down in property tables, see Chapter 6.

The thermal equation of state relates measurable properties. It suffices to know the values of two properties to determine the values of others. This will still be the case when we add energy and entropy to the list of thermodynamic properties, which can be determined through measurement of any two of the measurable properties, i.e., (p, T ) or (v, T ) or (p, v).

For inhomogeneous states, where the properties are space dependent, we assume the validity of the thermal equation of state in the local volume element dV . This assumption reflects our understanding that the atoms and molecules of the considered substance are interacting frequently, and thus behave collectively, see Sec. 2.2.

To summarize: The complete knowledge of the macroscopic state of a sys- tem requires the values of two intensive properties in each location (i.e., in each infinitesimal volume element), and the local velocity. The state of a sys- tem in equilibrium, where properties are homogeneous, is described by just two intensive properties (plus the size of the system, that is either total vol- ume, or total mass). In comparison, the knowledge of the microscopic state would require the knowledge of location and velocity of each particle.

Ideal Gas Law

The ideal gas is one of the simplest substances to study, since it has simple property relations. Ideal gases are employed in many engineering applications. Arguably, the most important ideal gas is air, which is the working substance in a large number of systems, including internal combustion engines.

Careful measurements have shown that for an ideal gas pressure p, total volume V , thermodynamic temperature T , and mass m are related by anexplic  it thermal equation of state, the ideal gas law pV = mRT . (2.12)

Here, R is the gas constant that depends on the type of the gas. Alternative forms of the equation result from introducing the specific volume v = V/m or the mass density ρ = 1/v so that

pv = RT , p = ρRT . (2.13)

The ideal gas law is our first property relation. According to this equation, the properties appearing in the equation cannot be changed independently: the change of one property must necessarily lead to a change of at least one other property. When the temperature is kept constant, an increase in pressure leads to a reduction of volume; when pressure is kept constant, the increase of temperature leads to increase of volume; when volume is kept constant, reduction of temperature leads to lower pressure. Of course, there can be processes where all three, pressure, volume, and temperature, change.

This equation does not contain any quantities that depend on the type of gas, accordingly the behavior of ideal gases is universal.

A Note on Problem Solving

Before we start solving our first problems, it might be worthwhile to briefly list good practices for problem solving. Typically, any engineering problem should be tackled by the following steps:

1. Understand the problem, i.e., read the question carefully. Nothing good can come from a solution that is based on a misunderstanding.

2. Make a sketch of the relevant system, and proper diagrams. A good sketch can summarize a complicatedly worded problem in a far more accessible form.

3. Indicate all quantities that are known on the sketch, or in a list, so that they are easy to find when needed. List all processes that occur in the system

4. List the quantities that need to be determined.

5. List the relevant thermodynamic equations for their determination.

6. Simplify the equations based on what is known about the processes in the system.

7. Solve the equations for the quantities of interest. Do not insert values for quantities before all manipulation of equations is complete; in other words, solve symbolically, and insert values as late as possible. This simplifies double checking of computations, and makes it far easier to correct errors that occur due to mistyping of values or wrong unit conversions.

8. Carefully consider and simplify all units. Before you have sufficient practice, never assume the outcome of a unit conversion, or the final unit for a value in a computation. Wrong unit conversions are a rather frequent source of major problems: always double-check. Note that each property value must be accompanied by a unit, i.e., never just write a number but [number value×unit].

9. Add comments throughout your treatment of the problem, so that you have text and equations/values on your answer. Written out sentences make the solution accessible, and you can explain assumptions, simplifications etc. This makes it far easier to follow through the line of argument for any reader—including yourself at a later point in time; just equations make for an unreadable submission.

10. Finally, use experience and common sense to scrutinize the final results.

Do they make sense, e.g., are the values for temperatures, pressures, energies etc. realistic?

We shall as much as possible adhere to these steps in the examples through- out this book. However, due to space restrictions, we will, e.g., not always have a sketch, and will skip over algebraic reformulations of equations. More- over, explicit unit conversions will be shown only in few early examples. It is strongly recommended that the reader goes through the examples carefully, including making a sketch, and double checking of all calculations, including the units.

Example: Air in a Room

A room of dimension 5 m × 10 m × 3 m is filled with air at 20 ◦C, 1 atm. Compute the mass of air in the room, the number of moles and the number of particles. If the temperature in the room increases to 25 ◦C for the same pressure, what amount of air has left through doors and windows?

We use the ideal gas law (2.12) with the values V = 150 m3, p = 101.325 kPa, T = 293 K. Note that the ideal gas law requires the Kelvin temperature! We find, with R = 0.287 kJ as the gas constant for air,

Example: Air in a Refrigerator

A refrigerator of volume VR = 330 litre which maintains food at TR = 5 ◦C is located in a kitchen at T0 = 22 ◦C, p0 = 1.02 atm. When the refrigerator door is opened, warm air enters the cooling space, and when the door is closed again, this warm air is cooled to TR. We ask for the amount of air inside, and for the net force on the door after cooling is complete.

To simplify the problem, we assume that the air in the refrigerator is completely exchanged, so that in the moment of closing all air in the interior is at T0, p0. Then, the mass in the interior is, with p0 = 103.35 kPa, VR = 0.33 m3, T0 = 295 K and R = 0.287 kJ ,

This force must be overcome to open the door. Note that the calculation assumes perfect sealing, and complete replacement of the cold air with warm air. Actual kitchen refrigerators have imperfect seals, so that some air creeps through during cooling, hence the observed forces are weaker. Nevertheless, in particular not too long after closing, one can observe this effect. Try your refrigerator at home!

More on Pressure

Pressure p is defined as the force (F ) exerted by a fluid per unit area (A), p = F/A, in the limit of infinitesimal area. Pressure is isotropic, that is the force on a surface is independent of the orientation of that surface.

To proceed, we have to differentiate between water and air. First we con- sider water: Water can be assumed in good approximation to be an incompressible substance, that is the water density is constant, with the well-known value of ρH O "’ 1000 kg . Integration of (2.21) is straightforward, and gives, together with the condition

This relation is valid for all incompressible liquids, where the appropriate mass density ρ must be used. For water, depth increase by Δh = 10.33 m increases the pressure by about 1 atm. Hydrostatic pressure depends only on depth, not on the actual weight of liquid above. This implies that hydrostatic pressure is independent of the geometry of the container, see Fig. 2.11 for an illustration.

Air, on the other hand, is compressible, it obeys the ideal gas law p = ρRT .

Using this to eliminate density from the differential equation for pressure (2.21), we find

Integration is only possible when we have additional information on the tem- perature T (z) as a function of height z.

Figure 2.12 compares both pressure functions for air and T0 = 293 K. For the non-isothermal atmosphere the pressure decreases slightly faster, but at moderate heights the difference is almost not noticeable. The Canadian town of Banff is located at an altitude of 1463 m above sea level. When the sea level pressure is p0 = 1 atm we compute from (2.25) and (2.26) local pressures of 0.843 atm and 0.839 atm, respectively. Note that most weather forecasts do not present the actual local pressure p, but the normalized pressure, that is the corresponding sea level pressure p0. For instance, when the forecast gives the pressure for Banff as 990 kPa, based on (2.26) the actual pressure in town will be 831 kPa.

The example shows a marked influence of gravitation on pressure for heights on the kilometer scale. Most engineering devices are relatively small, at most on the scale of several meters, and the variation of gas pressure can be safely ignored. Therefore it is sufficient to assign just one value of pressure to a gaseous system in equilibrium.

Gas pressure results from the momentum change of those gas particles that hit the wall and bounce back, and thus exert a force. When a gas filled container is put on a scale, the scale will show the total weight of container and gas, although most of the gas particles are not in contact with the container

walls. Indeed, it is the small variation with pressure between top and bottom of a container which puts the weight of the gas on the scale.

Problems

Ideal Gas I

A 5 litre container holds helium at a temperature of 25 ◦C and a pressure of 2 atm. Determine the mass of gas in the container, the number of moles, and the number of particles.

Ideal Gas II

A cylinder with radius 5 cm contains 5 g of pure oxygen at a temperature of 200 ◦C. The cylinder is closed with a freely moving piston, which in equilib- rium rests at a height of 50 cm. Determine the mass of the piston.

Ideal Gas Thermometer

An ideal gas thermometer holds a fixed gas volume of 1000 cm3. For calibration, the thermometer is brought into contact with melting ice and boiling water, both at 1 atm, where the pressures measured are p1 = 51.6 kPa and p2 = 70.5 kPa.

1. For Celsius temperature, assume a linear relation of the form T ( ◦C) = a + bp and determine the constants a and b.

2. Determine the mole number of particles enclosed.

3. Careful measurement shows that the mass of gas enclosed is 1 g. Find the

molar mass—what gas is it most likely?

Ideal Gas and Spring

The following process is done in a room at a temperature of 20 ◦C and a pressure of 100 kPa: A container with quadratic base of 10 cm side length is closed by a piston of mass mp = 100 g. Initially, the piston is fixed at a height of H0 = 10 cm, and the cylinder is filled with 2.5 g of carbondioxide. A spring is attached to the piston from above, so that at the initial state the spring is at its rest length. When the fixing of the piston is removed, the piston moves up, and the spring is compressed. The system comes to an equilibrium state such that the piston has moved up by ΔH = 3 cm.

1. Determine the spring constant k.

2. The gas in the container is now heated to 120 ◦C. Determine the final displacement of the piston.

Climbing a Mountain

1. Compute the mass of the helium filling.

2. As the balloon ascents, the volume of the filling increases. Compute the volume of the balloon as function of height. Above which height is the balloon completely filled? (Hint: the pressures of helium filling and the surrounding air are equal as long the balloon is not filled completely).

3. Compute the buoyancy for V < Vf , and show that it is independent of height. The buoyancy force is given as FB = ρair (z)gV where V is the actual balloon volume, and ρair (z) the density of the surrounding air at height z.

4. Set up an equation for the buoyancy for the case that the balloon is completely filled. How high will the balloon rise?

An Experiment

As you accelerate in a car, you are pressed in the seat—what happens to a helium filled balloon? Think about it, or try it, then explain!

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