Introductory Chemistry Online/Gaseous State

  Chapter 9. The Gaseous State

Some of the first real breakthroughs in the study of chemistry happened in the study of the gaseous state. In gases, the volume of the actual gas particles is but a tiny fraction of the total volume that the gas occupies. This allowed early chemists to relate parameters such as volume and the number of gas particles, leading to the development of the mole concept. As we have seen in previous chapters, the notion of a chemical mole allows us to do quantitative chemistry and lead us to the point where we can routinely address reaction stoichiometry, etc. In this chapter, we will visit some of the early observations that lead to our current understand of gasses and how they behave. We will see how the relationships between pressure and volume; volume and temperature and volume and moles lead to the ideal gas laws and how these simple rules can allow us to do quantitative calculations in the gas phase.

  9.1 Gasses and Atmospheric Pressure

In Chapter 2, we learned about the three principle states of matter; solids, liquids and gasses. We explained the properties of the states of matter using the kinetic molecular theory (KMT). Substances in the gaseous state, according to the KMT, have enough kinetic energy to break all of the attractive forces between the individual gas particles and are therefore free to separate and rapidly move throughout the entire volume of their container. Because there is so much space between the particles in a gas, a gas is highly compressible. High compressibility and the ability of gases to take on the shape and volume of its container are two of the important physical properties of gasses.

The gas that we are all most familiar with is the mixture of elements and compounds that we call the “atmosphere”. The air that we breath is mostly nitrogen and oxygen, with much smaller amounts of water vapor, carbon dioxide, noble gasses and the organic compound, methane (Table 9.1).

Table 9.1. Approximate Composition of the Atmosphere

Gas	Concentration, Parts per Billion	Percentage
N2	7.8 × 108	78%
O2	2.0 × 108	20%
H2O	About 106 – 107	< 1%
Ar	9.3 × 106	< 1%
CO2	3.5 × 105	< 0.05%
Ne	1.8 × 104	trace
He	5.2 × 103	trace
CH4	1.6 × 103	trace

A gas that is enclosed in a container exerts a pressure on the inner walls of that container. This pressure is the result of the countless collisions of the gas particles with the container wall (Figure 9.1)(. As each collision occurs, a small amount of energy is transferred, generating a net pressure. Although we are generally unaware of it, the gasses in the atmosphere generate a tremendous pressure on all of us. At sea level, atmospheric pressure is equal to 14.7 pounds per square inch. Putting this in perspective, for a person of average height and build, the total pressure from the atmosphere pressing on their body is about 45,000 pounds! Why aren’t we squashed? Remember, we also have air inside our bodies and the pressure from the inside balances the pressure outside, keeping us nice and firm, not squishy!

The proper SI unit for pressure is the Pascal (Pa), where 1 Pa = 1 kg m^-1 s^-2. In chemistry, however, it is more common to measure pressure in terms of atmospheres (atm) where 1 atm is atmospheric pressure at sea level, or 1 atm = 14.7 pounds per square inch (1 atm = 101,325 Pa). Atmospheric pressure is typically measured using a device called a barometer. A simple mercury barometer (also called a Torricelli barometer, after its inventor) consists of a glass column, about 30 inches high, closed at one end and filled with mercury. The column is inverted and placed in an open, mercury-filled reservoir. The weight of the mercury in the tube causes the column to drop to the point that the mass of the mercury column matches the atmospheric pressure exerted on the mercury in the reservoir. The atmospheric pressure is then read as the height of the mercury column. Again, working at sea level, 1 atmosphere is exactly equal to a column height of 760 mm of mercury. The units for the conversion are 1 atm = 760 mm Hg, and this is an exact relationship with regard to significant figures. The unit torr (after Torricelli) is sometimes used in place of mm Hg. A schematic of a mercury barometer is shown in Figure 9.2.

  9.2 The Pressure-Volume Relationship: Boyle’s Law

The kinetic molecular theory is useful when we are trying to understand the properties and behaviors of gases. The KMT (and related theories) tell us that:

• There is a tremendous amount of distance between individual particles in the gas phase.
• Gas particles move randomly at various speeds and in every possible direction.
• Attractive forces between individual gas particles are negligible.
• Collisions between gas particles are fully elastic.
• The average kinetic energy of particles in the gas phase is proportional to the temperature of the gas.

In reality, these predictions only apply to “ideal gases”. An ideal gas has perfectly elastic collisions and has no interactions with its neighbors or with the container. Real gases deviate from these predictions, but at common temperatures and pressures, the deviations are generally small and in this text we will treat all gases as if they were “ideal”.

Because there is so much empty space between gas molecules, it is easy to see why a gas is so compressible. If you have a container filled with a gas, you can squeeze it down to a smaller volume by applying pressure. The harder you squeeze (the more pressure you apply) the smaller the resulting volume will be. Imagine a bicycle pump compressing air into a tire (Figure 9.3). As pressure is applied to the pump, the same number of gas molecules are squeezed into a smaller volume.

The dependence of volume on pressure is not linear. In 1661, Robert Boyle systematically studied the compressibility of gasses in response to increasing pressure. Boyle found that the dependence of volume on pressure was non-linear (Figure 9.4) but that a linear plot could be obtained if the volume was plotted against the reciprocal of the pressure, 1/P. This is stated as Boyle’s law:

The volume (V) of an ideal gas varies inversely with the applied pressure (P) when the temperature (T) and the number of moles (n) of the gas are constant.

Mathematically, Boyle’s law can be stated as:

$V\propto {\frac {1}{P}}{\text{ }}\left({\text{at constant }}T{\text{ and }}n\right)$

, and

$V={\text{ }}constant{\text{ }}\left({\frac {1}{P}}\right){\text{, or, }}PV{\text{=}}constant$

We can use Boyle’s law to predict what will happen to the volume of a sample of gas as we change the pressure. Because PV is a constant for any given sample of gas (at constant T), we can imagine two states; an initial state with a certain pressure and volume (P₁V₁), and a final state with different values for pressure and volume (P₂V₂). Because PV is always a constant, we can equate the two states and write:

$P_{1}V_{1}=P_{2}V_{2}$

Now imagine that we have a container with a piston that we can use to compress the gas inside (for example, Figure 9.5). You are told that, initially, the pressure in the container is 765 mm Hg and the volume is 1.00 L. The piston is then adjusted so that the volume is now 0.500 L; what is the final pressure?

We substitute into our Boyle’s law equation:

$P_{1}V_{1}=P_{2}V_{2}$ , or, $\left(765{\text{ mm Hg}}\right)\left({\text{1}}{\text{.00 L}}\right)=P_{2}\left(0.500{\text{ L}}\right)$

$P_{2}=\left({\frac {\left(765{\text{ mm Hg}}\right)\left({\text{1}}{\text{.00 L}}\right)}{0.500{\text{ L}}}}\right)=1530{\text{ mm Hg}}$

Example 9.1 Pressure-Volume Relationships

A container with a piston contains a sample of gas. Initially, the pressure in 
the container is exactly 1 atm, but the volume is unknown. The piston is 
adjusted so that the volume is 0.155 L and the pressure is 956 mm Hg; what was 
the initial volume?

Exercise 9.1 Pressure-Volume Relationships

The pressure of 12.5 L of a gas is 0.82 atm. If the pressure changes to 1.32 atm, 
what will the final volume be?
A sample of helium gas has a pressure of 860.0 mm Hg. This gas is transferred to a 
different container having a volume of 25.0 L; in this new container, the 
pressure is determined to be 770.0 mm Hg. What was the initial volume of the gas?

  9.3 The Temperature-Volume Relationship: Charles’s Law

Figure 9.6 shows a fun laboratory demonstration in which a fully inflated balloon is placed in liquid nitrogen (at a temperature of –196 ˚C) and it shrinks to about 1/1000^th of its former size. If the balloon is carefully removed and allowed to warm to room temperature, it will again be fully inflated.

This is a simple demonstration of the effect of temperature on the volume of a gas. In 1787, Jacques Charles performed a systematic study of the effect of temperature on gases. Charles took samples of gases at various temperatures, but at the same pressure, and measured their volumes. A plot showing representative results is shown in Figure 9.7.

The first thing to note is that the plot is linear. When the pressure is constant, volume is a direct linear function of temperature. This is stated as Charles’s law:

The volume (V) of an ideal gas varies directly with the temperature of the gas (T) when the pressure (P) and the number of moles (n) of the gas are constant.

We can express this mathematically as:

$V\propto T{\text{ }}\left({\text{at constant }}P{\text{ and }}n\right)$

, and

$V={\text{ }}constant{\text{ }}\left(T\right){\text{, or, }}{\frac {V}{T}}{\text{=}}constant$

Figure 9.7 shows data for three different samples of the same gas; 0.25 moles, 0.50 moles and 0.75 moles. All of these samples behave as predicted by Charles’s law (the plots are all linear), but , if you extrapolate each of the lines back to the y-axis (the temperature axis), all three lines intersect at the same point! This point, with a temperature of –273.15 ˚C, is the theoretical point where the samples would have “zero volume”. This temperature, -273.15 ˚C, is called absolute zero. An even more intriguing thing is that the value of absolute zero is independent of the nature of the gas that is used. Hydrogen, oxygen, helium, argon, (or whatever), all gases show the same behavior and all intersect at the same point.

The temperature of this intersection point is taken as “zero” on the Kelvin temperature scale. The abbreviation used in the Kelvin scale is K (no degree sign) and there are never negative values in degrees Kelvin. The size of the degree increment in Kelvin is identical to that in Centigrade and Kelvin and centigrade scales are related by the simple conversion:

${\text{Kelvin=centigrade+273}}{\text{.15}}$

Please note that whenever you work gas law problems where temperature is one variable, you MUST use the Kelvin scale.

Just like we did for pressure-volume problems, we can use Charles’s law to predict what will happen to the volume of a sample of gas as we change the temperature. Because ${}^{V}\!\!\diagup \!\!{}_{T}\;$ is a constant for any given sample of gas (at constant P), we can again imagine two states; an initial state with a certain temperature and volume ( ${}^{V_{1}}\!\!\diagup \!\!{}_{T_{1}}\;$ ), and a final state with different values for pressure and volume ([ ${}^{V_{2}}\!\!\diagup \!\!{}_{T_{2}}\;$). Because ${}^{V}\!\!\diagup \!\!{}_{T}\;$ is always a constant, we can equate the two states and write:

${\frac {V_{1}}{T_{1}}}={\frac {V_{2}}{T_{2}}}$

Again, we have a container with a piston that we can use to adjust the pressure on the gas inside. You are told that, initially, the temperature of the gas in the container is 175 K and the volume is 1.50 L. The temperature is changed to 76 K and the piston is then adjusted so that the pressure is identical to the pressure in the initial state; what is the final volume?

We substitute into our Charles’s law equation:

${\frac {V_{1}}{T_{1}}}={\frac {V_{2}}{T_{2}}}$, or, ${\frac {\left({\text{1}}{\text{.50 L}}\right)}{175{\text{ K}}}}={\frac {V_{2}}{76{\text{ K}}}}$

$V_{2}=\left({\frac {\left(76{\text{ K}}\right)\left({\text{1}}{\text{.50 L}}\right)}{\text{175 K}}}\right)=0.65{\text{ L}}$

Example 9.2 Temperature-Volume Relationships

A container with a piston contains a sample of gas. Initially, the pressure in the 
container is exactly 1 atm, the temperature is 14.0 ˚C and the volume is 997 mL. 
The temperature is raised to 100.0 ˚C and the piston is adjusted so that 
the pressure is again exactly 1 atm What is the final volume?

Exercise 9.2 Temperature-Volume Relationships

A 50.0 mL sample of gas at 26.4˚ C, is heated at constant pressure until its 
volume is 62.4 mL . What is the final temperature of the gas?
A sample container of carbon monoxide occupies a volume of 435 mL at a temperature 
of 298 K. What would its temperature be if the pressure remained constant and 
the volume was changed to 265 mL? (182 K)

  9.4 The Mole-Volume Relationship: Avogadro’s Law

In Figure 9.7, we made a plot of the effect of temperature on the volume of a gas at constant pressure. The plot shows three lines, one for 0.025 moles of gas, one for 0.50 moles and one for 0.75 moles. If you take these data and combine them with data for other molar quantities, and plot them as a function of the volume at constant temperature and constant pressure, you can obtain a plot similar to that shown in Figure 9.8. The data in this plot form a linear series, passing through the origin. What this simply says is that the volume of a gas is directly proportional to the number of moles of that gas. This is stated as Avogadro’s law:

The volume (V) of an ideal gas varies directly with the number of moles of the gas (n) when the pressure (P) and the number of temperature (T) are constant.

We can express this mathematically as:

$V\propto n{\text{ }}\left({\text{at constant }}P{\text{ and }}T\right)$

, and

$V=constant\times \left(n\right),or,{\frac {V}{n}}=constant$

As before, we can use Avogadro’s law to predict what will happen to the volume of a sample of gas as we change the number of moles. Because ${}^{V}\!\!\diagup \!\!{}_{n}\;$ is a constant for any given sample of gas (at constant P and T), we can again imagine two states; an initial state with a certain number of moles and volume ( ${}^{V_{1}}\!\!\diagup \!\!{}_{n_{1}}\;$), and a final state with values for a different number of moles and volume (${}^{V_{2}}\!\!\diagup \!\!{}_{n_{2}}\;$ ). Because ${}^{V}\!\!\diagup \!\!{}_{n}\;$ is always a constant, we can equate the two states and write:

${\frac {V_{1}}{n_{1}}}={\frac {V_{2}}{n_{2}}}$

As an example of this type of problem, we have a container with a piston that we can use to adjust the pressure on the gas inside, and we can control the temperature. You are told that, initially, the container contains 0.20 moles of hydrogen gas and 0.10 mole of oxygen in a volume is 2.40 L. The two gases are allowed to react (a spark ignites the mixture) and the piston is then adjusted so that the pressure is identical to the pressure in the initial state and the container is cooled to the initial temperature; what is the final volume of the product of the reaction?

First, we need to look at the reaction involved. Hydrogen and oxygen react to form water. Two moles of hydrogen react with one mole of oxygen to give two moles of water, as shown below:

2H₂ (g) + O₂ (g) → 2 H₂O (g)

Initially we have three moles of gas and, after reaction, we have two moles. We can now substitute into Avogadro’s law:

${\frac {V_{1}}{n_{1}}}={\frac {V_{2}}{n_{2}}}$, or, ${\frac {{\text{2}}{\text{.40 L}}}{\text{3 moles}}}={\frac {V_{2}}{2{\text{ moles}}}}$

$V_{2}=\left({\frac {\left(2.{\text{40 L}}\right)\left({\text{2 moles}}\right)}{\text{3 moles}}}\right)=1.60{\text{ L}}$

Thus we have described the dependence of the volume of a gas on the pressure (Boyle’s law), the temperature (Charles’s law) and the number of moles of the gas (Avogadro’s law). In the following section, we will combine these to generate the Ideal Gas Law, in which all three variables (pressure, temperature and number of moles) can vary independently.

  9.5 The Ideal Gas Law

The three gas laws that we covered in Section 9.2 – 9.5 describe the effect of pressure, temperature and the number of moles of a gas on volume. The three independent gas laws are:

• Boyle’s law: $V\propto {}^{1}\!\!\diagup \!\!{}_{P}\;$
• Charles’s law: $V\propto T{\text{ }}$
• Avogadro’s law: $V\propto n{\text{ }}$

If volume (V) is proportion to each of these variables, it must also be proportional to their product:

$V\propto \left({}^{Tn}\!\!\diagup \!\!{}_{P}\;\right)$

If we replace the proportionality symbol with a constant (let’s just choose R to represent our constant), we can re-write the equation as:

${\mathsf {V=R}}\left({}^{\mathsf {nT}}\!\!\diagup \!\!{}_{\mathsf {P}}\;\right)$, or rearranging, $PV=nRT$.

The value of the proportionality constant R, can be calculated from the fact that exactly one mole of a gas at exactly 1 atm and at 0 ˚C (273 K) has a volume of 22.414 L (Figure 9.9). Substituting in the equation:

$PV=nRT$, or, $R={\frac {PV}{nT}}$ and $R={\frac {\left({\text{1 atm}}\right)\left(22.414{\text{ L}}\right)}{\left({\text{1 mole}}\right)\left(273{\text{ K}}\right)}}=0.082057{\text{ L atm mol}}^{\text{-1}}{\text{ K}}^{\text{-1}}$

The uncomfortable and somewhat obnoxious constant is called the universal gas constant, and you will need to know it (or look it up) whenever you solve problems using the combined ideal gas law.

Example 9.3 Ideal Gas Law Calculations

What volume will 17.5 grams of N2 occupy at a pressure of 876 mm Hg and 
at 123 ˚C?

Many of the problems that you will encounter when dealing with the gas laws can be solved by simply using the “two-state” approach. Because R is a constant, we can equate an initial and a final state as:

$R={\frac {P_{1}V_{1}}{n_{1}T_{1}}}$ for the initial state, and $R={\frac {P_{2}V_{2}}{n_{2}T_{2}}}$

for the final state, or, 

${\frac {P_{1}V_{1}}{n_{1}T_{1}}}={\frac {P_{2}V_{2}}{n_{2}T_{2}}}$.

Using this equation, you can solve for multiple variables within a single problem.

Example 9.4 Ideal Gas Law Calculation; Determining Volume

A sample of oxygen occupies 17.5 L at 0.75 atm and 298 K. The temperature is raised to 303 K and the pressure 
is increased to 0.987 atm. What is the final volume of the sample?

If you noticed, we calculated the value of the proportionality constant R based on the fact that exactly one mole of a gas at exactly 1 atm and at 0 ˚C (273 K) has a volume of 22.414 L. This is one of the “magic numbers in chemistry; exactly one mole of any gas under these conditions will occupy a volume of 22.414 L. The conditions, 1 atm and 0 ˚C, are called standard temperature and pressure, or STP. The fact that all gases occupy this same molar volume can be rationalized by realizing that 99.999% of a gas is empty space, so it really doesn’t matter what’s in there, it all occupies the same volume. This realization is attributed to Amedeo Avogadro and Avogadro’s hypothesis, published in 1811, suggested that equal volumes of all gases at the same temperature and pressure contained the same number of molecules. This is the observation that led to the measurement of Avogadro’s number (6.0221415 × 10²³), the number of things in a mole. The importance of the “magic number” of 22.414 L per mole (at STP) is that, when combined with the ideal gas laws, any volume of a gas can be easily converted into the number of moles of that gas. This is shown in the following example:

Example 9.5 Ideal Gas Law Calculation; Determining Moles

A sample of methane has a volume of 17.5 L at 100.0 ˚C and 1.72 atm. How many 
moles of methane are in the sample?

Exercise 9.3 Ideal Gas Law Calculations

A 0.0500 L sample of a gas has a pressure of 745 mm Hg at 26.4˚ C. The temperature 
is now raised to 404.4 K and the volume is allowed to expand until a final 
pressure of 1.06 atm is reached. What is the final volume of the gas?

When 128.9 grams of cyclopropane (C3H6) are placed into an 
8.00 L cylinder at 298 K, the pressure is observed to be 1.24 atm. A piston in 
the cylinder is now adjusted so that the volume is now 12.00 L and the pressure 
is 0.88 atm. What is the final temperature of the gas?

  9.6 Combining Stoichiometry and the Ideal Gas Laws

With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:

2 HCl (aq) + Zn (s) → ZnCl₂ (aq) + H₂ (g)

A sample of pure zinc with a mass of 5.98 g is reacted with excess hydrochloric acid and the (dry) hydrogen gas is collected at 25.0 ˚C and 742 mm Hg. What volume of hydrogen gas would be produced?

This is a “single state” problem, so we can solve it using the ideal gas law, PV = nRT. In order to find the volume of hydrogen gas (V), we need to know the number of moles of hydrogen that will be produced by the reaction. Our stoichiometry is simply one mole of hydrogen per mole of zinc, so we need to know the number of moles of zinc that are present in 5.98 grams of zinc metal. The temperature is given in centigrade, so we need to convert into Kelvin, and we also need to convert mm Hg into atm.

Conversions:

${\text{25}}{\text{.0 C+273=298 K}}$; $\left(742{\text{ mm Hg}}\right)\times \left({\frac {\text{1 atm}}{\text{760 mm Hg}}}\right){\text{=0}}{\text{.976 atm}}$

$\left(5.{\text{98 g Zn}}\right)\times \left({\frac {{\text{1}}{\text{.00 mol}}}{{\text{65}}{\text{.39 g Zn}}}}\right){\text{=0}}{\text{.0915 mol}}$

Substituting:

$PV=nRT$, or, $\left({\text{0}}{\text{.976 atm}}\right)\times V=\left(0.0915{\text{ mol}}\right)\left(0.0821{\text{ L atm mol}}^{\text{-1}}{\text{ K}}^{\text{-1}}\right)\left(2{\text{98 K}}\right)$

$V={\frac {\left(0.0915{\text{ mol}}\right)\left(0.0821{\text{ L atm mol}}^{\text{-1}}{\text{ K}}^{\text{-1}}\right)\left(2{\text{98 K}}\right)}{\left({\text{0}}{\text{.976 atm}}\right)}}=2.29{\text{ L}}$

We can also use the fact that one mole of a gas occupies 22.414 L at STP in order to calculate the number of moles of a gas that is produced in a reaction. For example, the organic molecule ethane (CH₃CH₃) reacts with oxygen to give carbon dioxide and water according to the equation shown below:

2 CH₃CH₃ (g) + 7 O₂ (g) → 4 CO₂ (g) + 6 H₂O (g)

A sample problem might read like this; an unknown mass of ethane is allowed to react with excess oxygen and the carbon dioxide produced is separated and collected. The carbon dioxide collected is found to occupy 11.23 L at STP; what mass of ethane was in the original sample?

Because the volume of carbon dioxide is measured at STP, the observed value can be converted directly into moles of carbon dioxide by dividing by 22.414 L mol^–1. Once moles of carbon dioxide are known, the stoichiometry of the problem can be used to directly give moles of ethane (molar mass 30.07 g mol^-1), which leads directly to the mass of ethane in the sample.

$\left(1{\text{1}}{\text{.23 L CO}}_{\text{2}}\right)\times \left({\frac {\text{1 mol}}{{\text{22}}{\text{.414 L}}}}\right){\text{=0}}{\text{.501 mol CO}}_{\text{2}}$

Reaction stoichiometry:

$\left({\text{0}}{\text{.501 mol CO}}_{\text{2}}\right)\times \left({\frac {{\text{2 mol CH}}_{\text{3}}{\text{CH}}_{\text{3}}}{{\text{4 mol CO}}_{\text{2}}}}\right){\text{=0}}{\text{.250 mol CH}}_{\text{3}}{\text{CH}}_{\text{3}}$

The ideal gas laws allow a quantitative analysis of whole spectrum of chemical reactions. When you are approaching these problems, remember to first decide on the class of the problem:

• If it is a “single state” problem (a gas is produced at a single, given, set of conditions), then you want to use PV = nRT.
• If it is a “two state” problem (a gas is changed from one set of conditions to another) you want to use

${\frac {P_{1}V_{1}}{n_{1}T_{1}}}={\frac {P_{2}V_{2}}{n_{2}T_{2}}}$ .

• If the volume of gas is quoted at STP, you can quickly convert this volume into moles with by dividing by 22.414 L mol^-1.

Once you have isolated your approach ideal gas law problems are no more complex that the stoichiometry problems we have addressed in earlier chapters.

Example 9.6 Ideal Gas Law Calculation: Reaction Stoichiometry

An automobile air bag requires about 62 L of nitrogen gas in order to inflate. The nitrogen gas is produced by 
the decomposition of sodium azide, according to the equation shown below (Figure 9.10):

2 NaN₃ (s) → 2 Na (s) + 3 N₂ (g)

What mass of sodium azide is necessary to produce the required volume of nitrogen 
at 25 ˚C and 1 atm?

Exercise 9.4 Ideal Gas Law Calculations: Reaction Stoichiometry

When Fe2O3 is heated in the presence of carbon, 
CO2 gas is produced, according to the equation shown below. A sample 
of 96.9 grams of Fe2O3 is heated in the presence of 
excess carbon and the CO2 produced is collected and measured at 1 atm 
and 453 K. What volume of CO2 will be observed?

2 Fe₂O₃(s) + 3 C (s) → 4 Fe (s) + 3 CO₂ (g)

The reaction of zinc and hydrochloric acid generates hydrogen gas, according to 
the equation shown below. An unknown quantity of zinc in a sample is observed 
to produce 7.50 L of hydrogen gas at a temperature of 404 K and a pressure of 
1.75 atm. How many moles of zinc were in the sample?

Zn (s) + 2 HCl (aq) → ZnCl₂ (aq) + H₂ (g)

  Study Points

• Gasses are compressible because there is so much space between individual gas particles. Energy transferred from the collision of gas particles with their container exerts a gas pressure. In chemistry, we typically measure gas pressure using units of atmospheres (atm) or in mm Hg (also referred to as torr). One atmosphere of pressure equals exactly 760 mm Hg.
• The volume of a gas varies inversely with the applied pressure; the greater the pressure, the smaller the volume. This is relationship is referred to as Boyles’s Law. For a two-state system where the number of moles of gas and the temperature remain constant, Boyle’s Law can be expressed as $P_{1}V_{1}=P_{2}V_{2}$

.

• The volume of a gas varies directly with the absolute temperature; the higher the temperature, the larger the volume. This is relationship is referred to as Charles’s Law. For a two-state system where the number of moles of gas and the pressure remain constant, Charles’s Law can be expressed as:

${\frac {V_{1}}{T_{1}}}={\frac {V_{2}}{T_{2}}}$

• In this equation, the absolute temperature in Kelvin (K) must be used. Kelvin is defined as (degrees centigrade + 273.15). Zero degrees Kelvin is referred to as “absolute zero” and it is the temperature at which (theoretically) all molecular motion would cease.
• The volume of a gas varies directly with the number of moles of the gas that are present; the greater the number of moles, the larger the volume. This is relationship is referred to as Avogadro’s Law. For a two-state system where the temperature and the pressure of a gas remain constant, Avogadro’s Law can be expressed as:

${\frac {V_{1}}{n_{1}}}={\frac {V_{2}}{n_{2}}}$

• Because the volume of a gas varies directly with the number of moles of the gas that are present and with the absolute temperature (in Kelvin), and inversely with the pressure, the gas laws can be combined into a single proportionality;

$V\propto \left({}^{Tn}\!\!\diagup \!\!{}_{P}\;\right)$

.

This proportionality can be converted to an equality by inserting the proportionality constant R (the universal gas constant), where R = 0.082057 L atm mol^-1 K^-1, and can be re-written as:

${\mathsf {V=R}}\left({}^{\mathsf {nT}}\!\!\diagup \!\!{}_{\mathsf {P}}\;\right)$ , or rearranging, $PV=nRT$

.

This is referred to as the Ideal Gas Law and is valid for most gasses at low concentrations. For a two-state system where the identity of the gas does not change, the Ideal Gas Law can be expressed as:

${\frac {P_{1}V_{1}}{n_{1}T_{1}}}={\frac {P_{2}V_{2}}{n_{2}T_{2}}}$

.

The gas constant R, is calculated based on the experimental observation that exactly one mole of any gas at exactly 1 atm and at 0 ˚C (273 K) has a volume of 22.414 L. The conditions, 1 atm and 0 ˚C, are called standard temperature and pressure, or STP.

• The ideal gas laws allow a quantitative analysis of whole spectrum of chemical reactions involving gasses. When you are approaching these problems, remember to first decide on the class of the problem:

• 1. If it is a “single state” problem (a gas is produced at a single, given, set of conditions), then you want to use PV = nRT.
  2. If it is a “two state” problem (a gas is changed from one set of conditions to another) you want to use

${\frac {P_{1}V_{1}}{n_{1}T_{1}}}={\frac {P_{2}V_{2}}{n_{2}T_{2}}}$ .

• 1. If the volume of gas is quoted at STP, you can quickly convert this volume into moles with by dividing by 22.414 L mol^-1.