# 27 Ideal Gas Properties

Learning Objectives

By the end of this section, you should be able to:

Describe the relationship between variables of state for gases

Describe gas behaviour using ideal gas law and compressibility factor using equations of state

## Ideal Gas Law

Relates pressure ($P$), volume ($V$), temperature ($T$) and the number of moles ($n$) of an ideal gas species using the ideal gas constant ($R$):

 $PV=nRT$

Can also relate pressure, molar volume ($\hat{V}$) and temperature:

 $P\hat{V}=RT$

The ideal gas law is an approximation that works well under some conditions:

$\hat{V}\; or\; V_{m}=\frac{V}{n} \text{, with units of } \frac{volume}{mol}$

It is known experimentally that for gases at low density (such that their molecules occupy a negligible fraction of the total volume) and at temperatures well above the boiling point, these proportionalities hold to a good approximation. $^{[1]}$

There are different ways to estimate how well the ideal gas approximation applies to a system. We will not go over these estimates explicitly in this course. Generally for this course, unless otherwise noted, we will assume the ideal gas law applies. However, in future thermodynamic courses, you will see when the ideal gas approximation may not be appropriate and will see other methods of relating gas properties in non-ideal scenarios.

## Ideal Gas Mixtures

### Dalton’s Law

If two or more gases are mixed, they will come to a thermal equilibrium as a result of collisions between molecules. When the gases have the same temperature, their molecules have the same average kinetic energy. Thus, each gas obeys the ideal gas law separately and exerts the same pressure on the walls of a container that it would if it were alone.

Therefore, in a mixture of gases, the total pressure is the sum of partial pressures of the component gases, assuming ideal gas behavior and no chemical reactions between the components. $^{[2]}$

 $P=\sum_{i=1}^{n}p_{i}$

Using Dalton’s law, we can calculate the partial pressure of a gas component, which is defined by the pressure that individual component in the gas mixture exerts on the wall if it were alone.

 $p_{A}=y_{A}P$

where

$y_{A}=$ the mole fraction of the gas component $=\frac{\text{moles of the component}}{\text{total moles of gas in the container}}$

### Amgat’s Law

Amgat’s Law is analogous to Dalton’s Law, but is applied to the volume of the gas. The partial volume that each gas occupies will add up to the total system volume.

 $V=\sum_{i=1}^{n}v_{i}$ $v_{A}=y_{A}V$

These partial volumes are proportional to the molar fraction of each gas in a system.

## Standard Conditions

A word of caution that standard conditions may not always be the same in different industries or applications. This is shown below with two examples of standard conditions for gases from IUPAC (the International Union of Pure and Applied Chemistry) and NIST (the National Institute of Standards and Technology).

IUPAC NIST
Standard Temperature 273.15K or 0°C 20°C
Standard Pressure $1×10^5$Pa or 1 bar 1 atm or 101.325kPa

Using the IUPAC numbers we can calculate a standard volume from the ideal gas law. This is the amount of volume a mole of gas would occupy at these standard conditions:

$\hat{V}_{s}=\frac{RT_{s}}{P_{s}}=\frac{8.314\frac{m^3Pa}{Kmol}×273.15K}{1×10^5Pa}×\frac{1000mol}{1kmol}=22.7\frac{m^3(ST\!P)}{kmol}=22.7\frac{L(ST\!P)}{mol}$

In some applications, gas flows may be expressed in standard cubic metres (SCM) or feet (SCF). A flow of 18.2 SCMH means 18.2 $m^3/h$ at standard conditions (whatever T/P these are defined to be).

We can calculate the temperature, pressure, and volume by comparing fluids at standard and non-standard conditions using the fact that $R$ is constant. We usually use it to compare the same system under different conditions, which means the number of gas molecules (n) may be constant when we perform the calculation.

$R=\frac{V_{s}×P_{s}}{n×T_{s}}=\frac{V×P}{n×T}$

Exercise: Natural Gas Standard Flow

Natural gas (mostly $CH_4$ that we might use to heat our homes) is normally measured with a correction to a certain temperature and pressure (to ensure you are always getting the same number of gas molecules). If it is a very cold winter day in Quebec City, where natural gas is entering your home at -30°C at a pressure of 2 bar and a flow of 100 L per hour. What would the standard flow (in L/hour) be if our standard conditions were 0°C and 1 bar (the IUPAC conditions)?

### Solution

$\frac{V_{s}×P_{s}}{n×T_{s}}=\frac{V×P}{n×T}$

The number of moles of gas (n) is the same in both conditions. So we can cancel n from both sides and rearrange the equation to isolate $V_{s}$. Remember that T in the equation needs to be converted to an absolute scale.

$V_{s}=\frac{V×P×T_{S}}{T×P_{S}}=\frac{100\frac{L}{h}×2bar×273.15K}{243.15K×1bar}=224.676L/h$

### A Note on Notation

In thermodynamics, and generally in this course:

$p_{i}(T)$ means that the partial pressure $p_{i}$ is a function of temperature $T$

The brackets are not for multiplication, but are for showing what variables the variable of interest is a function of. This is confusing, but a common notation in thermodynamics. Think of f(x) you have seen in math, this means f is a function of x, rather than f is multiplied by x.

Some other examples are:

$\rho_{g}(T,P)$ means the density of gasses is a function of Temperature and Pressure

$\rho_{l}(T)$ means density of liquids is a function of Temperature

### Liquid and Solid Densities

In this course, we will consider solids and liquids incompressible, meaning their density does not change significantly with changes in pressure. We will consider liquid density to be a weak function of temperature.

$\rho_{𝑙𝑖𝑞𝑢𝑖𝑑} ≅ 𝑓(𝑇) ≠ 𝑓(𝑃)$

$\rho_{𝑠𝑜𝑙𝑖𝑑} ≠ 𝑓(𝑃, 𝑇)$

Density values can be found in:

• Perry’s Chemical Engineers’ Handbook – pp. 2-7 to 2-47 and 2-96 to 2-125, available in print as reference material in the library here:

Where to Find Thermodynamic Property Data:

• Introductory Chemical Engineering Thermodynamics – textbook for this course – available as an ebook through UBC library, link also available in Library course reserves on canvas:

• Perry’s Chemical Engineers’ Handbook – pp. 2-7 to 2-47, available through UBC Library in print:

### When are Gases Non-Ideal?

• At lower temperatures: as gas molecules collide, they may experience inelastic (sticky) collisions due to intermolecular interactions.
• At high pressure: the molecules are closely packed together, so there are more chance of gas molecules experience molecular interaction or inelastic collisions.
• Intermolecular forces vary with molecular structures and polarity. Longer chain or more polar molecules usually experience more intermolecular forces.

Review: Assumptions of the ideal gas law

• The volume of gas molecules is negligible.
• Collisions between gas molecules or between gas molecules and the walls of the container are elastic, which means no energy is lost due to collisions.
• There are no attractive or repulsive forces between gas molecules.
• The average kinetic energy is proportional to the temperature.

## Compressibility Factor

When rearranging the ideal gas law, we get that for ideal gases:

$\frac{P\hat{V}}{RT}=1$

For a real gas, we adjust the ideal gas law by the compressibility factor(Z):

$\frac{P\hat{V}}{RT}=Z$

We can generalize the value of Z based on the reduced temperature and reduced pressure, which can be obtained from charts or correlations:

 $\text{reduced temperature:}\;T_{r}=\frac{T}{T_{c}}$ $\text{reduced pressure:}\;P_{r}=\frac{P}{P_{c}}$

where $T_{c}$ and $P_{c}$ are the temperatrue at the critical point.

The charts to find compressibility factor Z can be found in the “module” tab from CHBE 220 course on canvas.

Exercise: Compressibility Factor

If our substance is propane at P = 20 bar and T = 100 °C, with critical parameters: $P_{c}$ = 4.23 MPa, $T_{c}$ = 97°C, find the compressibility factor $Z$.

### Solution

First, we calculate the reduced temperature and pressure:

Unit conversions used: $1 MPa\; (mega-Pascal) = 10^6 Pa = 10 bar$ , $T (K) = T (°C) + 273.15$:

$P_{r}=\frac{P}{P_{c}}=\frac{20bar×\frac{1MPa}{10bar}}{4.23MPa}=0.47$

$T_{r}=\frac{T}{T_{c}}=\frac{(100+273.15)K}{(97+273.15)K}=1.0$

Find the lines representing $P_{r}=0.47$ on the x-axis, and $T_{r}=1.0$ on the curves with solid lines. The compressibility factor $Z$ is the y-coordinate of the intersection.

$Z=0.83$

## References

[1] OpenStax University Physics Volume 2. 2016. 2.1 Molecular Model of an Ideal Gas. [online] <https://openstax.org/books/university-physics-volume-2/pages/2-1-molecular-model-of-an-ideal-gas> [Accessed 14 May 2020].

[2] OpenStax University Physics Volume 2. 2016. 2.2 Pressure, Temperature and RMS Speed. [online] <https://openstax.org/books/university-physics-volume-2/pages/2-2-pressure-temperature-and-rms-speed> [Accessed 14 May 2020].