Lecture 10 – Behaviors of Gases

Lecture Preview

• How do ideal and non-ideal gases behave, and how are they different?
• How do gases behave in a mixture?
• How do gases diffuse and effuse?
• What are clinical applications utilizing gas behavior?

Lecture Content (Part 1)

This lecture will be on the equation-heavy side. However, conceptually, it should be simpler compared to the previous lecture where we explored how IMFs can affect the phase of a substance.

Let’s get straight to the point. Here are some equations we will be learning this lecture:

• Definition of Pressure: $p = \frac{F}{A}$

• The Ideal Gas Law: $pV = nRT$
• The Not-so-ideal Gas Law: $(p + \frac{n^2a}{V^2})(V – nb) = nRT$

• Density of Gas: $\rho = \frac{Mp}{RT}$ (Ideal gas law derivative)
• Molar Mass of Gas: $M = \frac{mRT}{PV}$ (Ideal gas law derivative)

• Dalton’s Law of Partial Pressures: $p_{total} = p_A + p_B + p_C + … = \sum_i p_i$

• Diffusion: $\text{rate of diffusion} = \frac{\text{amount of gas passed through an area}}{t}$
• Effusion: $\frac{\text{rate of effusion for gas A}}{\text{rate of effusion for gas B}}= \frac{\sqrt{M_B}}{\sqrt{M_A}}$

We will basically work our way down this equation list, explaining each equation and its applications. Most of the questions asked will be pretty much chugging numbers into the equations, as long as we use the right ones.

Also, there is something we should remember: we use temperature in Kelvins, which can be converted from Celsius using the following formula:

$$\text{T in Kelvin} = \text{T in Celsius}\ +\ 273.15$$

Gas Pressure

Pressure and force are different concepts. In physics, a force is an action (usually a push or a pull) that can cause an object to change its velocity or its shape, or to resist other forces, or to cause changes of pressure in a fluid. Pressure, on the other hand, is force, but over an area.

Pressure is defined as the amount of force exerted over an area.

$$p = \frac{F}{A}$$

where:

• $p$ denotes pressure
• $F$ denotes force
• $A$ denotes area

The unit of pressure is Pascals (Pa). 1 Pascal of pressure denotes the pressure of 1 Newton of force evenly distributed over 1 square meter of area.

$$1\ Pa = \frac{1\ N}{1\ m^2}$$

Pressure is an important quantity. For example, think about hammering a nail through a wooden plank. If the nail is not sharp at all, it would be difficult to hammer the nail through. However, if the nail is very sharp, using the same force, we can easily hammer it through the wooden plank.

To visualize this quantity:

• The atmospheric pressure (at sea level) is $1\ atm = 1.01325\times 10^5\ Pa$ (that is a lot of pressure but we’re used to it).
• The air pressure on top of Mount Everest (8848 meters above sea level) is $\frac{1}{3}\ atm$. As a result, oxygen levels there are extremely low.
• The average pressure exerted by our lungs when we play the trumpet is around $7000\ Pa$.
• By diving into 1-meter deep water, we will experience around an additional $10000\ Pa$ worth of pressure.

There are other common units for pressure. Here is the conversion chart.

Pressure will be explained in detail in Physics. For General Chemistry, a conceptual understanding will suffice.

The Ideal Gas Law and Its Derivatives

As listed above, the ideal gas law manifests itself in the form of an equation. However, the discovery of this equation can be attributed to multiple observations from multiple scientists.

Deriving the Ideal Gas Law

There will be a lot of names involved in this section. Memorizing them is not necessary, but memorizing the relations between quantities is.

Scientists Guillaume Amontons (~1700) and Joseph Louis Gay-Lussac (~1800) discovered the relationship between pressure and temperature of a gas. They discovered that, given everything else constant, the pressure of the gas is proportional to its temperature.

The Amontons’s or Gay-Lussac’s Law states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale under isovolumetric (same volume) conditions.

$$p \propto T\ \text{or}\ p = k_1 T$$

where

• $p$ denotes pressure
• $T$ denotes temperature in Kelvins
• $k_1$ denotes a constant

Scientists also discovered the relationship between volume and temperature of a gas, which they named it after French scientist Jacques Alexandre César Charles.

The Charles’s Law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale under isobaric (same pressure) conditions.

$$V \propto T\ \text{or}\ V = k_2 T$$

where

• $V$ denotes volume
• $T$ denotes temperature in Kelvins
• $k_2$ denotes a constant

Scientist Robert Boyle discovered the relationship between pressure and volume of a gas. This time, they are not directly proportional but inversely proportional (or reciprocal).

The Boyle’s Law states that the volume of a given amount of gas is inversely proportional to the pressure under which it is measured under isothermal (same temperature) conditions.

$$p \propto \frac{1}{V}\ \text{or}\ pV = k_3$$

where

• $p$ denotes pressure
• $V$ denotes volume
• $k_3$ denotes a constant

The same person who is behind the number $N_A = 6.022\times 10^{23}$ – Italian scientist Amedeo Avogadro – discovered the relationship between the number of moles and volume of a gas.

The Avogadro’s Law states that for a confined gas, the volume and number of moles are directly proportional under isobaric and isothermal conditions.

$$V \propto n\ \text{or}\ V = k_4 n$$

where

• $V$ denotes volume
• $n$ denotes moles
• $k_4$ denotes a constant

To summarize, we have 4 relationships among pressure, volume, temperature, and number of moles of a gas.

Using some algebra, we can also derive the following equations. These equations are easy to derive and can save some time in calculations.

So, by combining theses 4 relationships, we basically end up with the ideal gas law.

The ideal gas law summarizes the relationships between the pressure, volume, temperature, and number of moles of a gas into a nice equation.

$$pV = nRT$$

where

• $p$ denotes pressure
• $V$ denotes volume
• $n$ denotes number of moles
• $T$ denotes temperature
• $R$ denotes the universal gas constant (the combination of $k_1, k_2, k_3, k_4$ above), where $$R = 8.314\ \text{Pa}\cdot\text{m}^3\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$$

The universal gas constant $R$ will take on different values depending on the units used for $p, V, n, T$. The table shows some values of $R$ with different units.

Other values of R can be found here (Wikipedia). Let’s look at an example.

To be frank, questions involving the ideal gas law are calculation-heavy, because the numbers are all over the place, and sometimes we need to convert. For example, if we are given the number of molecules, we need to convert it to number of moles first before applying the ideal gas law. Remember to check if we’re using the right value of R.

Derivatives of the Ideal Gas Law

Here, we reintroduce 2 equations from Chapter 1 (check our equation sheet) – the density and molar mass equations…

…where $rho$ denotes density, $m$ denotes mass, $V$ denotes volume, $M$ denotes molar mass, and $n$ denotes number of moles.

We can substitute some quantities in the ideal gas law.

• Substituting $V$ with $\frac{m}{\rho}$ and $\frac{m}{n}$ with $M$ yields $$\rho = \frac{pm}{nRT} = \frac{Mp}{RT}$$
• Substituting $n$ with $\frac{m}{M}$ yields $$M = \frac{mRT}{pV}$$

The ideal gas law can also express properties such as density and molar mass of a gas.

$$\rho = \frac{Mp}{RT}$$

$$M = \frac{mRT}{pV}$$

where

• $\rho$ denotes density
• $M$ denotes molar mass
• $m$ denotes mass
• $p, V, T$ denotes pressure, volume, and temperature, respectively
• $R$ denotes the universal gas constant, where $$R = 8.314\ \text{Pa}\cdot\text{m}^3\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$$

Ideal vs. Not-so-ideal Gases

The word “ideal” in the ideal gas law suggests that there are some assumptions in place to use the ideal gas law.

• Assumption 1: We assume that ideal gas molecules have negligible volume, which is false in real life considering that under increasing pressure, gas molecules repel each other more vehemently due to their volumes, effectively invalidating Boyle’s Law ($p \propto \frac{1}{V}$).
• Assumption 2: We assume that ideal gas molecules do not attract each other, which is false in real life considering that under isovolumetric conditions, non-ideal gases would attract while ideal gases won’t, making pressure readings from non-ideal gases smaller.

To account for molecule volume and molecular attraction, we introduce extra terms in the ideal gas law, creating a new equation, called the van der Waals equation.

$$(p + \frac{an^2}{V^2})(V – nb) = nRT$$

where

• $\frac{an^2}{V^2}$ accounts for molecular attraction, where
  • $a$ denotes the van der Waals constant, dependent on the molecules
  • $n$ denotes number of moles
  • $V$ denotes volume
• $nb$ accounts for molecule volume, where
  • $n$ denotes number of moles
  • $b$ denotes the van der Waals constant, dependent on the molecules
• $p, V, n, T$ denote pressure, volume, number of moles, and temperature, respectively
• $R$ denotes the universal gas constant

We will not be using this equation. We will assume ideal gas in this course unless otherwise specified. However, we should be familiar with how molecular attraction and inherent volume can affect pressure or volume readings compared to ideal gas counterparts.

Partial Pressures

Using the ideal gas law, we can calculate properties of a gas. What about a mixture of gases? Turns out, unless those gases can react chemically, they act independently. This basically means that:

• Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container. We call that pressure “partial pressure”.

The total pressure in that container will simply be the sum of all partial pressures. The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction, a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components.

$$p_A = X_A \times p_{\text{total}} \quad \text{where} \quad X_A = \frac{n_A}{n_{\text{total}}}$$

where

• $p_A, n_A$ denote the partial pressure and number of moles of gas A
• $p_{\text{total}}, n_{\text{total}}$ denote the total pressure and total number of moles
• $X_A$ denotes the mole fraction of gas A

Diffusion and Effusion

Diffusion and effusion both describe gas movements.

• Diffusion is a process by which molecules disperse in space in response to differences in concentration. The rate of which is defined as… $$\text{rate of diffusion} = \frac{\text{amount of gas passing through an area}}{t}$$
• Effusion is the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum. The rate of which is inversely proportional to the square root of the gas’s molar mass. $$\text{rate of effusion} \propto \frac{1}{\sqrt{M}}$$

That sums up the new content for this lecture. For the rest of this lecture, we will explore clinical applications using the equations we learned today.

Lecture Content (Part 2)

Here, we will discuss clinical applications using the equations we learned today. Since the nature of practical applications are more complicated than that of simple plug-and-chug questions, the solutions won’t be very straightforward.

Applications of Gas Behavior

Application 1: BTPS Correction Calculations

Since the number of moles of a gas stays constant whether it’s inside our lungs or passed through a spirometer, we have the following equation using the idea gas law $pV = nRT$.

$$\frac{p_{\text{BTPS}}\ V_{\text{BTPS}}}{R\ T_{\text{BTPS}}} = n = \frac{p_{\text{ATPH}}\ V_{\text{ATPH}}}{R\ T_{\text{ATPH}}}$$ $$\Leftrightarrow V_{\text{BTPS}} = V_{\text{ATPH}}\ \times\ \frac{T_{\text{BTPS}}}{T_{\text{ATPH}}}\ \times\ \frac{p_{\text{ATPH}}}{p_{\text{BTPS}}}$$

We can further express $p_{\text{BTPS}}$ and $p_{\text{ATPH}}$ by subtracting the water vapor pressure from ambient pressure using Dalton’s law of partial pressure.

$$p_{\text{BTPS}} = p_{\text{ambient}}\ -\ p_{H2O \text{, body}}$$ $$p_{\text{ATPH}} = p_{\text{ambient}}\ -\ p_{H2O \text{, ambient}}$$

In the end, we arrive at the following equation.

BTPS corrections for expiratory and inspiratory air flow and volume measurements can be expressed using the following equation.

$$V_{\text{BTPS}} = V_{\text{ATPH}}\ \times\ \frac{T_{\text{BTPS}}}{T_{\text{ATPH}}}\ \times\ \frac{p_{\text{ambient}}\ -\ p_{H2O \text{, ambient}}}{p_{\text{ambient}}\ -\ p_{H2O \text{, body}}}$$

where the BTPS correction factor is defined as $$\text{BTPS correction factor} = \frac{T_{\text{BTPS}}}{T_{\text{ATPH}}}\ \times\ \frac{p_{\text{ambient}}\ -\ p_{H2O \text{, ambient}}}{p_{\text{ambient}}\ -\ p_{H2O \text{, body}}}$$

• BTPS stands for “body temperature and pressure, saturated with water vapor”
• ATPH stands for “ambient temperature, pressure, and humidity”, and it is assumed that ambient humidity is equivalent to 100% saturation for convenience

$p_{H2O}$ at various temperatures can be found here (Wired Chemist). Most notably, $p_{H2O, 37^{\circ} C} = 47.1 \text{torr}$.

This equation is not required in this course and will be provided if asked in a quiz or exam separate from the equation sheet.

Example 2: BTPS correction factor under various conditions

Assume body temperature is at $37^{\circ} C$. Thus, $p_{H2O \text{, body}} = 47.1 \text{torr}$.

The BTPS correction factor at sea level with ambient temperature at $20^{\circ}C$ will be
$$\begin{aligned}
\text{BTPS correction factor}
&= \frac{T_{\text{BTPS}}}{T_{\text{ATPH}}}
\times \frac{p_{\text{ambient}} – p_{H_2O,\,\text{ambient}}}{p_{\text{ambient}} – p_{H_2O,\,\text{body}}} \\
&= \frac{310.15~\text{K}}{293.15~\text{K}}
\times \frac{760~\text{torr} – 17.5~\text{torr}}{760~\text{torr} – 47.1~\text{torr}} \\
&= 1.101
\end{aligned}$$

In other words, the volume of gas inside our lungs will be 10.1% more than that outside our lungs at 20 degrees Celsius at sea level .

As ambient temperature increases, ambient water vapor pressure increases. Both changes cause the BTPS correction factor to decrease. At $T = 37^{\circ}$, the correction factor becomes 1. And as temperature continues to increase, the correction factor will drop below 1.

As altitude increases, ambient pressure decreases. At first glance, it’s a bit hard to figure out how BTPS changes. However, we can calculate it and see that it increases as altitude increases. In this case, we approximate ambient pressure in terms of height using the following formula: $$p = p_{\text{atm}}(1 – 2.25577\times 10^{-5} h)^{5.25588}$$ where $p_{\text{atm}}$ denotes atmospheric pressure (101325 Pa, 1 atm, or 760 torr), and $h$ denotes altitude.

Application 2: Hypoxia at High Altitudes

Using the formula and some other data, we can actually calculate how scarce oxygen is at high altitudes. Let’s go for the extreme and look at oxygen levels at the top of Mount Everest (8848m above sea level).

Example 3: Oxygen Levels at the Top of Mount Everest

We will assume that at both atmospheric and Mount Everest conditions, air consists of 21% oxygen. Since we want to look at solely oxygen levels, nitrogen levels and levels of other gases don’t matter for this example.

At the top of Mount Everest, the air pressure will be $$p = 760\ \text{torr} \times (1 – 2.25577 \times 10^{-5} \times 8848)^{5.25588} = 236\ \text{torr}$$

Since $X_{O_2} = 0.21$, we can calculate the partial pressure of oxygen. $$p_{O_2} = X_{O_2} \times p = 0.21 \times 236\ \text{torr} = 49.6\ \text{torr}$$

Using that value, we can calculate the mole fraction of oxygen under atmospheric pressure, which essentially simulates how low oxygen levels are at that altitude. $$X’_{O_2} = \frac{p_{O_2}}{p_{\text{atm}}} = \frac{49.6\ \text{torr}}{760\ \text{torr}} = 6.53\%$$

Our calculated oxygen levels at the top of Mount Everest is a third of normal oxygen levels. $$\frac{6.53\%}{21\%} = 0.31$$

Actual data of oxygen levels at the top of Mount Everest yields 6.9%, which is not far from our approximation. The reason we have different values is because our formula $p = p_{\text{atm}}(1 – 2.25577\times 10^{-5} h)^{5.25588}$ is an approximation.

Using the logic above, we can calculate effective oxygen levels at any altitude, despite it being more inaccurate as altitude becomes larger. There is a clear decreasing trend as altitude increases.

Application 3: Air Narcosis at Deep Waters

Using this, let’s figure out how much nitrogen gas will increase in pressure when a diver proceeds to a depth of 20 meters – the maximum depth most recreational training agencies will certify for entry level divers.

To be honest, today’s lecture materials are mostly secondary tools for other concepts we will encounter. Not many clinical applications require specifically ideal gas law calculations, but, at the same time, part of the ideal gas law is conceptually needed to visualize a problem. In the end, get used to using some parts of the ideal gas law and other concepts from the back of our minds – sometimes they are randomly needed.