Concept of a Perfect Gas

The starting point of thermodynamics related to gases is the behaviour of an ideal or perfect gas. A perfect gas is a gas which follows certain physical laws or rules under a prescribed set of conditions. For this module, a perfect gas is a gas which is sufficiently removed from its condensation temperature that it will remain in a gaseous state when subjected to the changes of temperature and pressure of the operation being considered.

The characteristic equation of the state of a perfect gas is:

Where R is a constant.

The relationship between temperature, pressure and volume is consistent throughout a selected working range. The behaviour of a perfect gas can be used as a model for the behaviour of real gases.

Note that pressure of a perfect gas will normally refer to absolute pressure rather than to gauge pressure. Unless gauge pressure is specifically indicated, it is assumed that thermodynamic calculations will utilize absolute pressures, which is the convention used throughout this chapter. Similarly, thermodynamic calculations normally utilize absolute temperature, using the Kelvin scale.

Pressure, temperature and volume are used to describe the state of a gas. The General Gas Law is derived from:

• Boyle’s Law

• Charles’ Laws

• Gay-Lussac’s Laws

Boyle’s Law

Boyle’s Law describes the relationship between pressure and volume at a constant temperature. It states that the product of pressure and volume is always the same if the temperature is constant, or:

The results of this equation produce a curve such as the one shown in Figure 1. Note that the pressure is in absolute units, not gauge pressure, and temperature is in Kelvin (K).

When Boyle’s Law is used to compare two different sets of conditions, the equation becomes:

This is a reasonable assumption if the process of expansion or compression is slow enough for the temperature to remain constant. Constant temperature operations are also called isothermal operations.

Gay-Lussac’s Law

Gay-Lussac’s Law describes the relationship between pressure and temperature at a constant volume. It states that the ratio of pressure and temperature is always the same if the volume is constant, or:

An alternate expression by comparing two different sets of conditions is:

Charles’ Law

Charles’ Law describes the relationship between volume and temperature when pressure is constant:

An alternate expression is:

General Gas Law

The three laws can be combined to give the expression:

The version that compares the two sets of conditions is:

Example 1

A gas is measured to be at a pressure of 150 kPa (gauge), 50°C and a volume of 3 litres. Find the volume the gas would occupy at standardized conditions of 15°C and 101.325 kPa.

Solution 1

Standardized conditions are stated as 15°C and 101.325 kPa. Converting the pressure and temperature to absolute, using the equation $\frac{P_1{V}_1}{T_1}$ = $\frac{P_2{V}_2}{T_2}$ , gives the solution as follows:

Ideal Gas Law

The ideal gas law explains the relationship between pressure, volume and temperature that includes the mass of the substance.

where m is the mass (in kg) and R is a characteristic constant (with units of kJ/kgK), specific to the type of gas. Table 1 provides a list of some common gases and their values for R.

Example 2

A steel tank contains 2 m3 of oxygen at a pressure of 13 790 kPa (gauge) and a temperature of 21°C. What is the mass of oxygen in the tank? (R = 0.261 kJ/kgK)

Solution 2

Mass is calculated using the equation PV = mRT:

Dalton’s Law

It is often necessary to perform calculations that involve multiple gases, such as mixtures of water vapour in air, or natural gas, which consists of a mixture of hydrocarbons. Dalton’s Law states that “the total pressure of a mixture of gases is equal to the sum of the individual (or partial) pressures that each gas would exert if the other gases were not present”. In other words, the partial pressure of each gas is calculated using the total volume, and the total pressure is the sum of all of the partial pressures.

Example 3

A tank with a volume of 5 m3 is half full of water. The top half of the tank contains air (at 40°C and atmospheric pressure) and water vapour. If the partial pressure of the water vapour is 7.384 kPa at 40°C, what is the mass of the air in the tank? (R for air = 0.287 kJ/kgK)

Solution 3

The volume of gas above the water is a mixture of air and water vapour. Since the partial pressure of the water vapour is 7.384 kPa, the partial pressure of the air is:

The partial pressure always applies to the total volume, so the volume of the air to be considered is 2.5 m3 at 40°C.

The mass can now be calculated using the equation: PV = mRT

Specific Heat

The specific heat of a substance is the amount of heat that is required to raise a unit quantity of the substance through 1°C rise in temp.

Solids and liquids have only one specific heat. However, the specific heat of a perfect gas varies, depending on the conditions under which it is heated. Is the gas confined within a closed container or is it free to expand as it is heated? Is the gas only partially confined and therefore partially free to expand? If the latter, to what degree is it allowed to expand? These situations collectively demonstrate the variation in heat required to raise a gas from one temperature to another. The amount of heat varies, depending on how strictly the gas is contained.

To help understand this, let’s consider the two extreme conditions, involving two identical cylinders. Each cylinder is perfectly insulated from its surroundings, eliminating heat loss. Each contains a piston that is perfectly sealed so no gas can escape between it and the cylinder wall. Between the piston and the cylinder end, both cylinders contain the same mass of the same perfect gas and at the same temperature. The piston of cylinder #1 is firmly fixed and cannot move. The piston of cylinder #2 is free to move, with negligible friction.

Now, if the same amount of heat is applied to both cylinders, the temperature of each will increase. Ignoring any expansion of the cylinders themselves, we see that the volume in cylinder #1 remains constant, while the volume in cylinder #2 increases (ie. the piston moves). We will also observe that the pressure in cylinder #1 increases, while the pressure in cylinder #2 remains constant.

Thus, it can be seen that each cylinder represents an extreme, purely theoretical situation.

• In cylinder #1, the heat is added under constant volume. In cylinder #2, the heat is added under constant pressure.

• In cylinder #1, no work is performed by the gas, since it is not free to expand and move the piston. In cylinder #2, the piston moves, so the expanding gas performs work.

• In cylinder #1, since all the added heat goes toward increasing the internal energy of the gas, less heat must be added to increase the temperature from T1 to T2. In cylinder #2, however, some of the added heat is converted into work, so more heat must be added to achieve the same temperature increase, from T1 to T2.

The above demonstrates why a perfect gas is always assigned two specific heats, one for heat transfer under constant pressure and one for heat transfer under constant volume. As shown, the heat required under constant pressure must be greater to accommodate the work done by the gas. Therefore, the specific heat of a gas under constant pressure (Cp) is always greater than the specific heat under constant volume (Cv).

Heat Added at Constant Volume

In a cylinder where the piston is fixed so that the volume remains constant, the heat added to the gas may be calculated from:

where:

Heat Added at Constant Pressure

In a cylinder where the piston moves, causing the cylinder pressure to remain constant, the heat added to the gas may be calculated from:

where:

Characteristic Gas Constant, R, related to Specific Heats

The characteristic gas constant of a gas has a direct relationship to the specific heats of the gas, which is Cp – Cv = R. This relationship is derived as follows.

Consider the two heat transfer formulae above. For the perfectly confined gas, with constant volume, the heat added is Qv = mCv(T2 – T1). If the same volume of gas is allowed to expand and do work, with constant pressure, the heat added becomes Qp = mCp(T2 – T1). It follows that the work done can be found by subtracting Qv from Qp.

Work results when a force is applied through a distance. If there is no motion, no work can be done. If there is no force, no work can be done. An expanding gas can perform work by applying pressure against the surface area of a piston, resulting in a force, which then moves the piston through a distance (the stroke).

Consider the First Law of Thermodynamics, which states:

Since Qv = mCv(T2 – T1) is used in situations where no work is performed and therefore represents only the change in internal energy of the gas, it can replace ΔU in the First Law equation. Likewise, since the work performed by a gas at constant pressure is equal to P(V2 – V1), this term can be substituted for WD in the First Law equation. The First Law equation then becomes:

But the TOTAL “Heat Supplied” will be enough to both change the internal energy AND perform the work. So, the left side of the First Law equation can now be replaced by mCp(T2 – T1) and the entire equation becomes:

Since PV equals mRT, we can substitute “mRT” wherever we find “PV”.

And, since P(V2 – V1) can also be written as PV2 – PV1, it follows that:

This now allows us to substitute mR(T2 – T1) for P(V2 – V1) in the revised First Law equation, as follows:

Now, this can be simplified by dividing through by ‘m’ and by ‘(T2 – T1)’, which leaves:

The characteristic gas constant, R, of a gas is the difference between the specific heat at constant pressure, Cp, and the specific heat at constant volume, Cv.

Ratio of Specific Heats

Another useful relationship between specific heats is their ratio. The ratio of the specific heat at constant pressure over the specific heat at constant volume of a particular gas is shown as:

$\frac{C_p}{C_v}$ , which is assigned the Greek symbol, γ (Gamma)

Table 2 shows specific heats and ratios for some common gases.

Example 4

5 m3 of air at atmospheric pressure is heated in a closed vessel from 21°C to 45°C. If the specific heat is 718 J/kg°C (per Table 2), how much heat is required? (R for air = 0.287 kJ/kgK)

Solution 4

First, find the mass of air, using:

Then, find the heat required, using:

Concept of Heat

Heat is the energy associated with the motion of molecules within a body. Heat can be converted to or from other forms of energy. Temperature is an indication of the heat intensity in a body and, therefore, informs the engineer of the following:

• the direction of heat transfer to or from the body under consideration

• the rate of heat transfer between bodies of different temperatures, and

• the material requirements for particular services.

When two substances at different temperatures come into contact with each other, over time their temperatures equalize. Heat is transferred from the hotter object to the colder one. There is a transfer of energy but it is necessary to understand what occurs at the microscopic molecular level to fully explain what happens.

First Law of Thermodynamics

There is a very close relationship between heat and mechanical energy, or work, which is called the “mechanical equivalent” of heat. During compression, work is applied to a gas, causing an increase in its internal energy. Therefore, the work transferred to or from the gas appears as a gain or loss of heat energy, resulting in a change in the gas temperature and pressure. Heat and work are just different forms of energy. Expansion and compression of gases depend on this fact.

The First Law of Thermodynamics states that work and heat are mutually convertible. This is evidenced by the units of measure of work and heat. Work is the product of force and distance and is measured in Nm or kNm. Heat and other forms of energy are measured in Joules or kJ. One kNm is equivalent to one kJ. That is, one unit of work done is equivalent to one unit of energy which is either expended or made available.

If work is applied to a system, there is a corresponding increase in the internal energy in the system which is measured by the amount of heat contained within the system. For example, if a gas is compressed and no heat is allowed to escape from the system, this is measurable by an increase in its temperature. Expansion causes a reduction of heat energy and a decrease in temperature as work is performed.

Second Law of Thermodynamics

The Second Law of Thermodynamics states that external work must be done to transfer heat from a lower temperature system to a higher temperature system. It is impossible for heat to flow from a colder system to a hotter one without expending energy in the form of work.

This principle led to the development of theoretical thermodynamic cycles that uses heat to produce work and also ones that use work to transfer heat such as in refrigeration. For the moment, we will consider the work required to compress gases and the work that can be done by the expansion of a gas. It will be seen in the following objectives that work can be done on a gas or by a gas in several different ways, which depend essentially on how heat flows within the system due to compression or expansion.

Work at Constant Pressure

Studies of expansion and compression of gases are of interest to the power engineer to allow calculation, for example, of the work that can be done by an expanding gas in an internal combustion engine or the work that must be done to compress a quantity of air in a compressor.

Expansion and compression of gases may be carried out under many different conditions. It is customary to study certain set conditions and to use these as a guide to the actual, existing operating conditions.

An expansion or a compression taking place at constant pressure is called an isobaric process. The process can be represented on a PV diagram, as shown in Figure 3. The line 1 – 2 represents an expansion at constant pressure, P, from volume, Vl , to volume, V2. The work done during this process is represented by the area under the line. In equation form, this work (area) can be shown as:

A constant volume process (called isochoric) does not cause any work to be done because the area is zero. (V2 – V1) = 0 This can be seen in Figure 4.

Example 5

2712 kNm of work is done to compress 5 m3 of a gas to a volume of 1 m3. If the work is done at constant pressure, what is the pressure?

Solution 5

Work Done During Isothermal Process

Another possible compression or expansion process is the isothermal process, during which the temperature of the gas does not change.

An isothermal process for an ideal or perfect gas follows Boyle’s Law, which states that “PV = a constant”. Since compression normally causes an increase in temperature, heat must be removed from the system if the gas temperature is to remain constant. Normally, this is not fully achievable in a compressor because of the difficulty in removing all the heat of compression. Some isothermal compressors have cooling jackets to extract heat. Figure 5 illustrates the isothermal process graphically, where the curved line, labelled ‘T1’, represents the constant temperature as the pressure and volume change.

The work done during an isothermal process can be found by calculating the area under the curve. This is illustrated in Figure 6, where the shaded area is the work done. The text that follows the figure shows how the formula for work done is arrived at, using calculus to determine the area under the curve.

The area under the curve can be found by taking a small strip of volume, dV, and using calculus to integrate the volume from V1 to V2. In mathematical terms, this is shown as:

Since the pressure, P, varies with volume, Boyle’s Law allows the ‘P’ to be replaced with ‘constant/V’, as follows:

Integration of this equation gives the equation:

Using Boyle’s Law again, the ‘constant’ can be replaced by PV to give the equation:

If the temperature is known, the General Gas Law equation (PV = mRT) may be used to substitute ‘mRT’ for ‘PV’, giving:

We can also express this work in terms of pressure, since P1V1 = P2V2, which can be rearranged as $\frac{V_2}{V_1}$ = $\frac{P_1}{P_2}$ . Substituting this in both equations gives:

Example 6

A quantity of air with an initial volume of 20 m3 at a pressure of 700 kPa (absolute) is expanded to a pressure of 100 kPa. What work is done under isothermal conditions?

Solution 6

For isothermal work, expansion occurs at a constant temperature. Since both pressures are known, use the equation that involves pressure:

Example 7

A volume of 10 m3 of air at a pressure of 280 kPa (absolute) is expanded isothermally to a volume of 25 m3. What work is done?

Solution 7

Since both volumes are known, use the equation that involves volumes:

Adiabatic Compression and Expansion

During many expansion and compression processes, there is insufficient opportunity for the heat to flow out of or into the working fluid. This process, called adiabatic, occurs when no heat moves across the system boundary. This means that during compression, heat is retained and the temperature of the working fluid increases. During expansion, the temperature decreases.

In an adiabatic process, the relationship between pressure and volume involves the ratio of specific heats, Cp/Cv designated by γ. The pressure-volume relationship is:

Another way to express this is:

Figure 7 represents an adiabatic expansion, from point 1 (P1, V1) to point 2 (P2, V2). The curved lines, T1 and T2, represent temperature. Notice that the temperature does not remain constant, but drops from T1 to T2.

Again, the work done is represented by the shaded area under the expansion line. Also, the area under the curve can be found by taking a small strip of volume, dV, (see Figure 6) and using calculus to integrate the volume from V1 to V2.

When the calculus is completed, we end with the following derived formula for work during an adiabatic process:

Example 8

A volume of 10 m3 of air at a pressure of 280 kPa absolute is expanded adiabatically to a volume of 25 m3. What work is done? (Assume that Cp = 1.005 and Cv = 0.718)

Solution 8

First, find the value of γ, using the ratio of specific heats:

Then, calculate P2 using the equation P1V1γ = P2V2γ:

Finally, calculate the work using the adiabatic equation:

Polytropic Expansion and Compression

Most expansion or compression processes are neither isothermal nor adiabatic. Some heat is always expended during expansion and retained during compression. This condition of varying pressure, volume and heat flow is called a polytropic process. These processes look much like an adiabatic process, but the ratio of Cp to Cv (ie. γ) does not apply. Instead, a different exponent, n, is used to define the relationship between pressure and volume, as follows:

This can also be expressed as: P1V1n = P2V2n

Figure 8 shows the curve for a polytropic expansion from point 1 (P1, V1) at temperature, T1, to point 2 (P2,V2), at temperature, T2.

The work done during the polytropic process is represented by the shaded area under the curve. The reasoning and mathematics (calculus) involved in developing the work formula is identical to that for the adiabatic process, except that n is used instead of γ. Thus, for a polytropic process, the work done can be calculated from:

Example 9

10 m3 of air at 280 kPa (absolute) is expanded to a volume of 25 m3. If the expansion is polytropic, how much work is done? (Assume that n = 1.2)

Solution 9

First, calculate the final pressure, P2, using P1V1n = P2V2n:

Now, calculate the work, using the polytropic equation:

Comparison of Expansion and Compression Processes

The work done by a gas during expansion can be compared for the various processes by observing the processes imposed on a single PV diagram. This is shown in Figure 9. The premise is that each process starts at the same point 1 (P1, V1) and expands to the same final volume (V2). Since the work performed in each case is represented by the area under the curve, the comparison can easily be observed. The greatest work is done during the isobaric process, followed by isothermal, polytropic, and adiabatic.

Figure 10 shows a similar comparison for a compression from point 1 (P1, V1) to the final volume (V2). Again, comparing the areas under the curves, we see that the work done on the gas during the compression is greatest during an adiabatic process, followed by polytropic, isothermal, and isobaric.

1. A receiver contains 1.5 m3 of air at 860 kPa and 20°C. Calculate the final temperature after 2.75 kg of air is added if the final pressure is 1325 kPa. (R for air = 0.287 kJ/kgK)

2. 2 m3 of hydrogen at atmospheric pressure in a closed vessel needs to be heated from 0°C to 21°C. If the specific heat at constant volume is 10 070 J/kg°C, how much heat is required? (R for hydrogen = 4.139 kJ/kgK)

3. A tank with a volume of 5 m3 is half full of water. Top half of the tank contains 2.75 kg of air at atmospheric pressure. If the partial pressure of water is 7.3814 kPa, what is the temperature of the air in the tank? (R for air = 0.287 kJ/kgK)

4. The values of the specific heats of a gas at constant pressure and at constant volume are 0.9839 and 0.7285, respectively. Find the value of γ for this gas. If 1.8 kg of this gas is heated from 16°C to 155°C, find the heat absorbed if the heating takes place at

a) Constant pressure

b) Constant volume.

5. With respect to compression and expansion processes, define the following:

a) Isothermal process

b) Adiabatic process

c) Polytropic process

6. 100 m3 of a perfect gas at atmospheric pressure and a temperature of 21°C is compressed to 25 m3. What work is required if the gas is compressed:

a) Isothermally

b) Adiabatically (Assume γ = 1.32)

c) Polytropically (Assume n = 1.18)

Self-Assessment

Question 1

Internal energy of a perfect gas is directly related to:

temperature

pressure

pressure and volume

pressure and temperature

volume

Question 2

P₁/T₁ = P₂/T₂ represents:

first law of thermodynamics

Boyle's law

Charles' law

second law of thermodynamics

Joule’s law

Question 3

An air receiver is cylindrical with hemispherical ends. The diameter is 1.2 m and the overall length is 4.2 m. How much air (air at atmospheric pressure of 100 kPa) is to be taken from the atmosphere and pumped into the vessel at 500 kPa gauge. The air receiver initially contains air at atmospheric pressure and the compressed air is the same temperature as the atmospheric air.

25.8 m³

21.5 m³

28.25 m³

33.9 m³

8.59 m³

Question 4

The general gas law may also be written as:

P₁V₁ = P₂V₂

W = P (V₂-V₁)

PV = mRT

Delta Q = Delta U + Work

R = C_p-C_v

Question 5

An air receiver is cylindrical with hemispherical ends. The diameter is 1.1 m and the overall length is 3.1 m. How much air (air at atmospheric pressure of 100 kPa) is to be taken from the atmosphere and pumped into the vessel at 700 kPa gauge. The air receiver initially contains air at atmospheric pressure and the compressed air is the same temperature as the atmospheric air.

25.5 m³

20.78 m³

18.18 m³

29.14 m³

7.79 m³