Processing math: 100%

$$\require{mhchem}$$

HOME

General Chemistry Study Guide

Chapter 5. Gases

---

Yu Wang

OpenStax 9 Gases. Brown 10 Gases.

1. General Concepts

All gases have the following physical characteristics:

• Gases assume the volume and shape of their containers.
• Gases are the most compressible of the states of matter.
• Gases will mix evenly and completely when confined to the same container.
• Gases have much lower densities than liquids and solids.

1.1. Substances That Exist as Gases

Elements that exist as gases at $25\,^\circ\text{C}$ and $1\,\text{atmosphere}$ are: noble gases ($\ce{He}$, $\ce{Ne}$, $\ce{Ar}$, $\ce{Kr}$, $\ce{Xe}$, $\ce{Rn}$), two halogens ($\ce{F2}$, $\ce{Cl2}$), oxygen ($\ce{O2}$), nitrogen ($\ce{N2}$) and hydrogen ($\ce{H2}$)

1.2. Pressure

$$\text{Pressure} = \frac{\text{Force}}{\text{Area}}$$ $$\text{Force}=\text{Mass}\times\text{Accelaration}\qquad\text{Newton's Second Law}$$

The unit for force is $\text{N}$ which is defined as $\text{kg}\cdot\text{m/s}^2$.

The unit for pressure is $\text{Pa}$ which is defined as $\text{N/m}^2$.

Other commonly used units for pressure are

1. $\text{atm}$ which means atomospheric pressure and $1\,\text{atm}$ is equal to $101325\,\text{Pa}$.
2. $\text{mmHg}$ or $\text{cmHg}$ which is the pressure of mercury with the hight of $1\,\text{mm}$ or $1\,\text{cm}$, respectively. $1\,\text{atm} = 760\,\text{mmHg}$.
3. $\text{torr}$. $1\,\text{torr} = 1\,\text{mmHg}$.

Requirements

1. Understand and remember the general properties of gases.
2. Remember elements exist as gases at $25\,^\circ\text{C}$ and $1\,\text{atmosphere}$.
3. Be familiar with the substances exist as gases at $25\,^\circ\text{C}$ and $1\,\text{atmosphere}$.
4. Understand the defination of pressure and learn how to convert the units.

2. The Gas Laws

Boyle's Law: $P \propto \frac{1}{V}$ or $V \propto \frac{1}{P}$ (at constant $n$ and $T$)

Charles's Law: $V \propto T$ (at constant $n$ and $P$)

Avogadro's Law: $V \propto n$ (at constant $P$ and $T$)

Combine all three expressions

$$V \propto \frac{nT}{P}$$

Ideal Gas Equation:

An ideal gas is a theoretical gas composed of many randomly moving point particles whose only interactions are perfectly elastic collisions. In another word, the molecules of an ideal gas do not attract or repel one another, and their volume is negligible.

$$V = R\frac{nT}{P}$$

or

$$PV = nRT$$

where $R$, the proportionality constant, is called the gas constant.

$$R = 8.314\,\text{J/K mol} = 8.314\,\text{N m/K mol} = 8.314\times10^{-3}\,\text{kJ/K mol} = 0.0821\,\text{L atm/K mol}$$

Note

$$1\,\text{J} = 1\,\text{N m}$$ $$1\,\text{Pa} = 1\,\text{N/m}^2$$

For an ideal gas, at $0\,^\circ\text{C}$ ($273.15\,\text{K}$) and $1\,\text{atm}$, it occupies $22.14\,\text{L}$ volume. The conditions $0\,^\circ\text{C}$ and $1\,\text{atm}$ are called standard temperature and pressure,

Example: Calculate the pressure of 1.92 mol of ideal gas in a steel vessel of volume 4.50 L at $25.6\,^\circ\text{C}$
Answer:
$$\begin{align*} P & = \frac{nRT}{V} \\ & = \frac{1.92\,\text{mol}\times 0.0821\,\text{L atm/K mol}\times (25.6+273.2)\,\text{K}}{4.50\,\text{L}} \\ & = 10.5\,\text{atm} \end{align*}$$ or $$\begin{align*} P & = \frac{nRT}{V} \\ & = \frac{1.92\,\text{mol}\times 8.314\,\text{N m/K mol}\times (25.6+273.2)\,\text{K}}{4.50\times10^{-3}\,\text{m}^3} \\ & = 1.06\times10^6\,\text{N/m}^2= 1.06\times10^6\,\text{Pa} \end{align*}$$

Example: An air bubble with an initial volume 2.54 mL rises from the bottom of a lake. The temperature and the pressure at the bottom of the lake are $6.42\,^\circ\text{C}$ and 4.56 atm. The temperature and pressure of the lake surface are $24.8\,^\circ\text{C}$ and 1.00 atm. Calculat the final volume of the air bubble.
Answer:
Key idea: the moles of the gas does not change during this process. Thus, at the beginning $P_iV_i=nRT_i$; and when the air bubble reaches the lake surface $P_fV_f=nRT_f$. $$\begin{align*} n & =\frac{P_iV_i}{RT_i}=\frac{P_fV_f}{RT_f} \\ V_f & =\frac{P_iV_i}{T_i}\times\frac{T_f}{P_f} \\ & = \frac{4.56\,\text{atm}\times 2.54\,\text{mL}}{(6.42+273.15)\,\text{K}}\times\frac{(24.8+273.15)\,\text{K}}{1.00\,\text{atm}} \\ & = 12.3\,\text{mL} \end{align*}$$

Example: In the reaction $$\ce{2NaN3 (s) -> 2Na(s) + 3 N2(g)}$$ What is the volume of $\ce{N2}$ generated by 38.2 g $\ce{NaN3}$ at $20.5\,^\circ\text{C}$ and 0.982 atm?
Answer:
$$\begin{align*} \text{moles of }\ce{N2} & =38.2\,\text{g }\ce{NaN3}\times\frac{1\,\text{mol }\ce{NaN3}}{65.02\,\text{g }\ce{NaN3}}\times\frac{3\,\text{mol }\ce{N2}}{2\,\text{mol }\ce{NaN3}} \\ & = 0.881\,\text{mol }\ce{N2} \end{align*}$$ $$\begin{align*} \text{volume of }\ce{N2} & =\frac{nRT}{P} \\ & = \frac{0.881\,\text{mol}\times 0.0821\,\text{L atm/K mol}\times(20.5+273.2)\,\text{K}}{0.982\,\text{atm}} \\ & = 21.6\,\text{L} \end{align*}$$

Density and Molar Mass of Gaseous Substance

$$n = \frac{m}{\mathcal{M}}$$

where $m$ is the mass, $\mathcal{M}$ is the molar mass.

Thus,

$$PV = \frac{m}{\mathcal{M}}RT$$

or

$$\frac{m}{\mathcal{M}V} = \frac{P}{RT}$$

Since

$$\text{d} = \frac{m}{V}$$

we get

$$\frac{d}{\mathcal{M}}= \frac{P}{RT}$$

or

$$\mathcal{M}=\frac{dRT}{P}$$

where $d$ is the density.

Example: A gas has a density of 7.71 g/L at $36.0\,^\circ\text{C}$ and 2.88 atm. What is the molar mass of this gas?
Answer:
$$\begin{align*} \mathcal{M} & =\frac{dRT}{P} \\ & = \frac{7.71\,\text{g/L}\times 0.0821\,\text{L atm/K mol}\times(36.0+273.2)\,\text{K}}{2.88\,\text{atm}} \\ & = 68.0\,\text{g/mol} \end{align*}$$

from IPython.display import YouTubeVideo
YouTubeVideo("88uEsGgXJrU", width=700, height=550)

# Watch the following video for a better understanding 
# on gas laws.

Requirements

1. Understand and remember the ideal gas equation. Konw the meaning of gas constant. Use ideal gas equation to do calculations.
2. Learn how to calculate the molar mass or density of a gaseous substance.

3. Partial Pressures

In a mixture of several gaseous substances, the total moles of gasses is expressed as $n$, and the moles of each gas is expressed as $n_1$, $n_2$, etc. According to ideal gas equation

$$P = \frac{nRT}{V}$$

where $P$ is the total pressure.

Define

$$P_1 = \frac{n_1RT}{V}$$ $$P_2 = \frac{n_2RT}{V}$$

and so on ...

The values of $P_1$, $P_2$, ... are regarded as the partial pressures of each gaseous substance.

The sum of all partial pressures must equal to the total pressure.

$$P = P_1 + P_2 + \dots$$

Define

$$X_i = \frac{n_i}{n}$$

as the molar fraction of one gas substance in the mixture. The sum of the fractions of every gaseous substances must equal to $1$.

We have

$$P_i = X_iP$$

Example: A mixture of three gases A, B and C. The total pressure is 3.46 atm. The moles of each components are 1.53 mol A, 1.64 mol B, and 2.64 mol C. What are the partial pressures of each gases?
Answer:
$$\begin{align*} P_\text{A} & = P_\text{total}\times\frac{n_\text{A}}{n_\text{A}+n_\text{B}+n_\text{C}} \\ & = 3.46\,\text{atm}\times\frac{1.53\,\text{mol}}{1.53\,\text{mol}+1.64\,\text{mol}+2.64\,\text{mol}} \\ & = 0.911\,\text{atm} \end{align*}$$ The partial pressures of B and C can be calculated the same way. The sum of the partial pressures must equal the total pressure.

Example: In one reaction 153 mL $\ce{H2}$ gas is collected. The temperature is $23.7\,^\circ\text{C}$, the atomspheric pressure is 758 mmHg. Knowing the water vapor pressure is 22.4 mmHg, calculate the moles of $\ce{H2}$ obtained.
Answer:
The partial pressure of $\ce{H2}$ is $758 - 22.4 = 736$ mmHg = 0.968 atm. $$\begin{align*} n & = \frac{PV}{RT} \\ & = \frac{0.968\,\text{atm}\times 0.153\,\text{L}}{0.0821\,\text{L atm/K mol}\times (23.7+273.2)\,\text{K}} \\ & = 6.08\times 10^{-3}\,\text{mol} \end{align*}$$

from IPython.display import YouTubeVideo
YouTubeVideo("A2xi-jEqpmg", width=700, height=550)

# Watch the following video for a better understanding 
# on partial pressures.

Requirements

1. Understand what is partial pressure and learn how to do calculation with it.

4. The Kinetic Molecular Theory of Gases

Energy is defined as the capacity to do work or to produce change. In mechanics, work is defined as force times distance. Because energy can be measured as work, we can write

$$\text{energy}=\text{word done}=\text{force}\times\text{distance}$$

The SI unit of energy is joule (J)

$$1\,\text{J} = 1\,\text{kg m}^2\text{/s}^2 = 1\,\text{N m}$$

Kinetic energy (KE) is the type of energy expended by a moving object, or energy of motion.

The assumptions of the kinetic molecular theory of gases

1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
2. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic.
3. Gas molecules exert neither attractive nor repulsive forces on one another.
4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy

$$\overline{KE}=\frac{1}{2}m\overline{u^2}\propto T$$

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter.

Explanation of the gas laws

• Compressibility of Gases. Because molecules in the gas phase are separated by large distances (assumption 1), gases can be compressed easily to occupy less volume.
• Boyle's Law. The pressure exerted by a gas results from the impact of its molecules on the walls of the container. The collision rate, or the number of molecular collisions with the walls per second, is proportional to the number density (that is, number of molecules per unit volume) of the gas. Decreasing the volume of a given amount of gas increases its number density and hence its collision rate. For this reason, the pressure of a gas is inversely proportional to the volume it occupies; as volume decreases, pressure increases and vice versa.
• Charles's Law. Because the average kinetic energy of gas molecules is proportional to the sample's absolute temperature (assumption 4), raising the temperature increases the average kinetic energy. Consequently, molecules will collide with the walls of the container more frequently and with greater impact if the gas is heated, and thus the pressure increases. The volume of gas will expand until the gas pressure is balanced by the constant external pressure
• Avogadro's Law. For two gases under the same conditions of pressure, volume, and temperature (that is, when $P_1 = P_2$, $T_1 = T_2$, and $V_1 = V_2$), it follows that $n_1 = n_2$, which is a mathematical expression of Avogadro's law.
• Dalton's Law of Partial Pressures. If molecules do not attract or repel one another (assumption 3), then the pressure exerted by one type of molecule is unaffected by the presence of another gas. Consequently, the total pressure is given by the sum of individual gas pressures.

Root-Mean-Square Speed

$$u_{rms}=\sqrt{\frac{3RT}{\mathcal{M}}}$$

where $u_{rms}$is the root-mean-square speed which is an average molecular speed.

Gas diffusion

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

Gas effusion

is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening.

In both diffusion and effusion the correlations among rate, time and molar mass are:

$$\frac{r_1}{r_2}=\frac{t_2}{t_1}=\sqrt{\frac{\mathcal{M}_2}{\mathcal{M}_1}}$$

Requirements

1. Understand the concepts and explanations of the kinetic molecular theory.
2. Understand the square root rule. If the gas molecule is 100 times heavier, the average speed would reduce to 1/10.

5. Deviations from Ideal Behavior

Real gas molecules do have certain sizes and interactions! At higher pressure and/or lower temperature the deviations from ideal behavior become more significant.

• Pressure of a real gas is lower than an ideal gas counterpart. Consider the approach of a particular molecule toward the wall of a container. The intermolecular attractions exerted by its neighbors tend to soften the impact made by this molecule against the wall.
• Free volume of a real gas is less than the volume of the container since the gas molecules occupy certain volume.

One can use van der Waals equation to deal with real gases.

$$\left( P + \frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$$

The parameters $a$ and $b$ can be found in Table 5.3.

Requirements

1. Understand why real gases do not behave exactly like ideal gases.

---

Please ignore the cell below. It just loads our style for the notebook.

from IPython.core.display import HTML
def css_styling():
    styles = open("custom.css", "r").read()
    return HTML(styles)
css_styling()