6.4 Use the compressibility factor for real gases | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Gas Mixtures and Real Gases

MEC E 340 Applied Thermodynamics — Study topic 6.4

Consider a fixed amount of gas in a closed tank, or a gas flowing through a control volume. The working fluid is a gas at numbered state 1, or at inlet and outlet states when there is flow. For the calculations in this lesson, assume equilibrium states and use the compressibility factor supplied in each example. No phase change or cycle is being modelled. The ideal-gas equation is a useful reference, but real gases can depart from it, especially at higher pressures or near conditions where the gas is less dilute. The compressibility factor provides a simple correction to the ideal-gas equation. The first-law balances still apply; the correction changes how the gas’s pressure, temperature, and volume or density are related.

What you will learn

Explain what the compressibility factor measures and how it relates real-gas behaviour to the ideal-gas model.
Use a supplied compressibility factor to calculate a gas property while keeping units and the chosen gas constant consistent.
Identify when a value of the compressibility factor must come from supplied data or an identified chart or table.

1. The model and its meaning

For a fixed mass of gas, pressure, volume, and absolute temperature are linked by an equation of state. The ideal-gas equation assumes a particular relation among these properties. For a real gas, the compressibility factor

Z

adjusts that relation. It is dimensionless: it has no units.

When

Z=1

, the real-gas equation gives the ideal-gas relation. A value different from one signals a departure from that relation at the stated conditions. Do not assume that

Z

is always greater than one or that it is constant as a gas changes state. Use a value that applies to the particular gas and state.

pv=ZRT

$Z$ is a dimensionless correction to the ideal-gas equation.
Pressure and temperature in the equation must be absolute.
A supplied value of $Z$ applies only to the stated gas and state conditions.

2. Choosing the property basis and obtaining Z

In the mass-specific form,

v

is specific volume in cubic metres per kilogram and

R

is the gas-specific constant in joules per kilogram-kelvin. In the molar form, use molar volume and the universal gas constant. Do not mix these bases. For example, pairing a specific gas constant with molar volume gives an inconsistent result.

Rearranging the equation is often all that is needed. It can give specific volume from pressure and temperature, density from specific volume, or pressure when the other properties are known. Density is mass per volume, so it is the reciprocal of specific volume.

A problem may provide

Z

directly. Otherwise, obtain it from an identified compressibility chart or table if the course materials or problem provide one. Some such resources use reduced pressure and reduced temperature, formed using the gas’s critical properties. Use the definitions and data from that named source; do not guess chart readings or critical properties. If the required source data are absent, state that a numerical value cannot be determined from the information given.

\rho=\frac{1}{v}=\frac{p}{ZRT}

Choose either a mass basis or a molar basis and use it consistently.
Use kelvin and absolute pressure.
Read or use only a compressibility factor supported by supplied data or a named chart or table.

3. Applying the correction and checking the result

For a fixed mass in a rigid, closed tank, mass and tank volume are constant. The equation of state can then relate the tank’s pressure and temperature at different states, provided the appropriate value of

Z

is known at each state. Do not carry the initial value of

Z

to the final state unless the problem says that this is an acceptable assumption.

For a steady-flow control volume, the compressibility factor helps evaluate properties such as inlet and outlet specific volumes or densities. It does not replace the mass or energy balance. Write those balances for the device and use the real-gas equation only where a state property is needed. This lesson does not assume a particular device or energy transfer.

After calculating, check units, signs, and scale. The result must be consistent with the stated pressure, temperature, gas constant, and

Z

. Compare with the ideal-gas result by setting

Z=1

only as a reference. A difference in the expected direction for the specified

Z

is a useful arithmetic check, not a substitute for the supplied real-gas data.

p_1v_1=Z_1RT_1,\qquad p_2v_2=Z_2RT_2

Use state-specific $Z$ values when conditions change.
The compressibility factor modifies a property relation; it does not replace conservation balances.
Compare with the ideal-gas result only as a clearly labelled reference.

Worked example

Specific volume from supplied real-gas data

A gas at state 1 has pressure

8.00\,\mathrm{MPa}

and temperature

400\,\mathrm{K}

. The problem supplies

Z_1=0.920

and

R=0.2968\,\mathrm{kJ/(kg\cdot K)}

. Find the specific volume. Compare it with the ideal-gas reference at the same pressure and temperature.

Identify the state and basis
This is one gas state, not a cycle or a flow-device calculation. Use the mass-specific equation because the given gas constant is mass-specific. The stated pressure is absolute, and temperature is in kelvin.
Apply the real-gas equation
Convert pressure to kilopascals so that kilopascals times cubic metres per kilogram is kilojoules per kilogram. Then solve for specific volume using the supplied factor.
$v_1=\frac{Z_1RT_1}{p_1}=\frac{(0.920)(0.2968\,\mathrm{kJ/(kg\cdot K)})(400\,\mathrm{K})}{8000\,\mathrm{kPa}}=0.01365\,\mathrm{m^3/kg}$
Compare with the reference
The ideal-gas reference uses the same data but sets $Z$ to one. The real-gas value is smaller because the supplied factor is below one.
$v_{\mathrm{ideal}}=\frac{RT_1}{p_1}=0.01484\,\mathrm{m^3/kg}$

Answer:

v_1=0.01365\,\mathrm{m^3/kg}

; the ideal-gas reference is

0.01484\,\mathrm{m^3/kg}

.

Check: The factor is below one, so at fixed pressure and temperature the real-gas specific volume is below the ideal-gas reference. The units reduce to cubic metres per kilogram.

Worked example

Gas mass in a rigid vessel

A rigid vessel of volume

0.120\,\mathrm{m^3}

contains a gas at state 1:

p_1=500\,\mathrm{kPa}

and

T_1=300\,\mathrm{K}

. Use the supplied values

Z_1=0.980

and

R=0.287\,\mathrm{kJ/(kg\cdot K)}

to find the gas mass. Treat the vessel contents as a closed system at the stated equilibrium state.

Relate specific volume to vessel volume
The vessel’s volume is the total volume occupied by the gas. For a fixed mass, specific volume is total volume divided by mass. Rearranging gives mass as total volume divided by specific volume.
$m=\frac{V}{v_1}$
Find specific volume
Use the mass-specific real-gas equation. Since $1\,\mathrm{kPa\cdot m^3}=1\,\mathrm{kJ}$ , the given units are consistent.
$v_1=\frac{Z_1RT_1}{p_1}=\frac{(0.980)(0.287\,\mathrm{kJ/(kg\cdot K)})(300\,\mathrm{K})}{500\,\mathrm{kPa}}=0.1686\,\mathrm{m^3/kg}$
Calculate the mass
Divide the known vessel volume by the calculated specific volume. No energy balance is needed because the question asks only for the mass at one state.
$m=\frac{0.120\,\mathrm{m^3}}{0.1686\,\mathrm{m^3/kg}}=0.712\,\mathrm{kg}$

Answer:

m=0.712\,\mathrm{kg}

.

Check: The specific volume is about

0.169\,\mathrm{m^3/kg}

, so a vessel of

0.120\,\mathrm{m^3}

should contain less than one kilogram. The calculated mass is consistent with that scale.

Worked example

Pressure from a measured specific volume

At state 1, a gas has

T_1=350\,\mathrm{K}

and specific volume

v_1=0.0800\,\mathrm{m^3/kg}

. A supplied source gives

Z_1=0.950

for this state. With

R=0.2968\,\mathrm{kJ/(kg\cdot K)}

, determine the pressure and compare it with the ideal-gas reference.

Choose the pressure form
All supplied quantities are on a mass-specific basis. Rearrange the real-gas equation to solve for pressure.
$p_1=\frac{Z_1RT_1}{v_1}$
Substitute with consistent units
The product of the gas constant and temperature is in kilojoules per kilogram. Dividing by specific volume gives kilopascals because one kilojoule per cubic metre equals one kilopascal.
$p_1=\frac{(0.950)(0.2968\,\mathrm{kJ/(kg\cdot K)})(350\,\mathrm{K})}{0.0800\,\mathrm{m^3/kg}}=1234\,\mathrm{kPa}=1.234\,\mathrm{MPa}$
Check against the ideal-gas reference
At the same temperature and specific volume, setting the factor to one gives a slightly higher reference pressure because the supplied factor is below one.
$p_{\mathrm{ideal}}=\frac{RT_1}{v_1}=1299\,\mathrm{kPa}$

Answer:

p_1=1.234\,\mathrm{MPa}

; the ideal-gas reference is approximately

1.299\,\mathrm{MPa}

.

Check: The calculated pressure is positive and has pressure units. It is below the ideal-gas reference in proportion to the supplied factor

Z_1=0.950

.

Common mistakes and how to avoid them

Using Celsius directly in the real-gas equation.

Correction: Convert temperature to kelvin before substituting.

Using pressure in megapascals while keeping a gas constant in kilojoules per kilogram-kelvin, without converting units.

Correction: Make the pressure and energy units consistent; for these examples, kilopascals pair conveniently with kilojoules and cubic metres.

Assuming

Z

is always one or remains unchanged between states.

Correction: Use the value supplied for each stated condition, or obtain it from the identified source required by the problem.

Treating the compressibility factor as a replacement for a first-law or mass balance.

Correction: Use

Z

in the equation of state to evaluate properties; write conservation balances separately when a process or device requires them.

Lesson summary

The real-gas equation modifies the ideal-gas relation by the dimensionless factor $Z$ .
Keep the mass-specific or molar basis consistent, use absolute pressure and kelvin, and track SI units.
Use only a supplied or source-supported value of $Z$ ; do not invent chart readings or assume a value stays fixed between states.

Check your understanding

Question 1

At fixed pressure and temperature, a supplied factor is

Z=0.90

. Compared with the ideal-gas specific volume, what does the real-gas equation predict?

The real-gas specific volume is 90% of the ideal-gas value.
The real-gas specific volume is 110% of the ideal-gas value.
The real-gas specific volume is unchanged because $Z$ affects pressure only.
The real-gas specific volume cannot be compared with the ideal-gas value.

Show answer and explanation

The real-gas specific volume is 90% of the ideal-gas value.

At fixed pressure and temperature, specific volume is proportional to

Z

. Thus

Z=0.90

gives 90% of the ideal-gas reference.

Question 2

Which temperature belongs in

pv=ZRT

?

Temperature in kelvin.
Temperature in degrees Celsius.
Temperature in degrees Fahrenheit.
Any temperature scale, because $Z$ removes the unit dependence.

Show answer and explanation

Temperature in kelvin.

The equation requires absolute temperature. In SI calculations, use kelvin.

Question 3

A problem gives pressure, temperature, and gas constant, but no value of

Z

and no identified source for obtaining it. What is the sound course-level response?

State that a real-gas numerical result needs supplied or source-supported $Z$ data.
Assume $Z=0.80$ because real gases are always denser.
Choose a value of $Z$ that makes the calculation convenient.
Use the ideal-gas equation and report it as an exact real-gas result.

Show answer and explanation

State that a real-gas numerical result needs supplied or source-supported

Z

data.

A real-gas calculation needs a factor supported for the stated gas and conditions. Without it, a numerical real-gas result is not determined.

Key terms

Compressibility factor: A dimensionless factor that adjusts the ideal-gas equation to represent real-gas behaviour at a specified state.
Specific volume: Volume per unit mass of a substance, measured in cubic metres per kilogram in SI.
Gas-specific constant: The gas constant per unit mass of a particular gas, used with mass-specific properties.
Ideal-gas reference: The result obtained from the real-gas equation by setting the compressibility factor to one.

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6.4 · Use the compressibility factor for real gases

University of Alberta MEC E 340: Applied Thermodynamics

MEC E 340 Applied Thermodynamics — Study topic 6.4

What you will learn

1. The model and its meaning

2. Choosing the property basis and obtaining Z

3. Applying the correction and checking the result

Specific volume from supplied real-gas data

Gas mass in a rigid vessel

Pressure from a measured specific volume

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MEC E 340

About this lesson and its review