University High School Chemistry: Review of Gas Laws

The Three States of Matter

Lesson Objectives

The student will:

describe the differences in molecular arrangement among solids, liquids, and gases.
describe the basic characteristic differences among solids, liquids, and gases.

Vocabulary

$&\mathbf{ideal \ gas} & &\mathbf{phase} &\mathbf{real \ gas}$

Introduction

The kinetic molecular theory allows us to explain the existence of the three phases of matter. In addition, it helps explain the physical characteristics of each phase and why phases change from one to another. The kinetic molecular theory is essential for the explanations of gas pressure, compressibility, diffusion, and mixing. Our explanations for reaction rates and equilibrium in later chapters also rest on the concepts of the kinetic molecular theory.

The Assumptions of the Kinetic Molecular Theory

According to the kinetic molecular theory, all matter is composed of tiny particles that are in constant, random, straight-line motion. This motion is constantly interrupted by collisions between between the particles and surfaces, as well as collisions between the particles themselves. The rate of motion for the particles is related to the temperature. The velocity of the particles is greater at higher temperatures and lower at lower temperatures.

In our discussions of gases, we will be referring to what are called ideal gases. In an ideal gas, we assume that the molecules are points that do not take up any space. We also assume that there are no attractions between molecules. In real gases, however, the gas molecules do take up a small amount of space, and there can be slight attractions among the molecules. Later in this chapter, real and ideal gases will be discussed in more detail.

Phases of Matter

Earlier in the chapter “Matter and Energy,” we had discussed how matter can be categorized as mixtures or pure substances. We did not, however, discuss how substances can be found in different phases. Depending on the temperature and pressure, matter can exist in different forms or states known as phases. You are most likely familiar with the phases of hydrogen dioxide $(\text{H}_2\text{O})$ , known commonly as water. Besides existing as a liquid that we can drink, water can also exist as ice, which is a solid. When we boil water, we are producing water vapor, which is a form of gas.

In a solid, the molecules are held in a tightly packed pattern, as seen in the figure above. As a result, the molecules hold a set position in spite of random motion. Molecular motion is reduced to vibrating in place. In comparison, the molecules in a liquid touch each other but are not held in a pattern. The liquid structure has holes in it, which allow molecules to pass each other and change positions in the structure. In a gaseous substance, the molecules are completely separated from each other and move around independently. Most of the volume of a gas is empty space. Scientists also recognize one more phase of matter called plasma. A plasma is a type of ionized gas with unique properties that distinguishes it as a fourth phase of matter. In this text, however, we will be primarily concerned with the solid, liquid, and gas phases.

Characteristics of Solids

The molecular arrangements in the three phases account for the various characteristics that differentiate the phases. For example, the mixing of particles is almost non-existent in solids. This is because the molecules cannot pass by one another in the tightly packed pattern. Solids are essentially incompressible because when a substance is compressed, it is the spaces between molecules that are compressed, not the molecules themselves. Since solids have almost no empty space in their structure, they do not compress. Solids maintain their shape and volume, as seen in the figure below. A $25 \ \mathrm{mL}$ rectangular piece of copper has the same shape and volume when it is resting on the table top as it does inside a beaker.

Characteristics of Liquids

In liquids, mixing occurs more readily because there are spaces between the molecules that allow the molecules to pass each other. The spaces between the molecules in liquids are small, so liquids have very little compressibility. Liquids maintain their own volume, but they take the their containers, as seen in the illustration below.

A $25 \ \mathrm{mL}$ sample of liquid in a graduated cylinder has a volume of $25 \ \mathrm{mL}$ and the shape of a cylinder. If the $25 \ \mathrm{mL}$ sample is placed in a beaker, the liquid still has a volume of $25 \ \mathrm{mL}$ , but now it has the shape of the beaker. The structure of the liquid keeps the particles in touch with each other so that the volume does not change, but because the particles can slide by each other, the particles can flow to fill the shape of the container.

Characteristics of Gases

Mixing in gases is almost instantaneous because there are no inhibitions for particles to pass one another. The volume of a gas is nearly all empty space, so particles are able to move freely. Gases are highly compressible because of the great amount of empty space, which allows the particles to be pushed closer together. Gases do not have either their own volume or their own shape. They take both volume and shape from their container.

Lesson Summary

In the solid phase, the molecules are held in a highly organized, tightly packed pattern.
Due to the tightly packed pattern of molecules in a solid, solids maintain their own shape and volume and do not mix readily.
In the liquid phase, molecules are in touch with each other but they are loosely packed and may move past each other easily.
Due to the loosely packed structure of a liquid, liquids maintain their own volume but take the shape of their container. They are also able to mix readily.
In the gaseous phase, molecules are completely separate from each other.
The volume of a gas is mostly empty space.
Due to the structure of gases, they take both the volume and the shape of their container, and they mix almost instantaneously.

Review Questions

Automobile brakes have a hose full of liquid connecting your brake pedal on one end to the brake pads on the other end. When you press on the brake pedal, the force is transferred through the liquid and presses the brake pads against the wheels to slow or stop them. Brakes that use liquid in this fashion are called hydraulic brakes and the liquid is called hydraulic brake fluid. What don't they use air in the brake lines instead of liquid?
Why would it cause a problem if some air got into your liquid-filled brake lines?
If you had a $250 \ \mathrm{mL}$ container full of helium gas, and you transferred all of the gas into an empty $1.00 \ \mathrm{liter}$ container, would the $1.00 \ \mathrm{liter}$ container be full or only $1/4$ full?

Gas Laws

Lesson Objectives

The student will:

state Boyle's law, Charles's law, and Gay-Lussac's law.
solve problems using Boyle's law, Charles's law, and Gay-Lussac's law.
state the combined gas law.
solve problems using the combined gas law.

Vocabulary

$&\mathbf{Boyle's \ law} & &\mathbf{Charles's \ law} & &\mathbf{combined \ gas \ law} \\ &\mathbf{Gay}\text{-}\mathbf{Lussac's \ law}$

Introduction

The gas laws are mathematical expressions that relate the volume, pressure, temperature, and quantity of gas present. They were determined from the results of over 100 years of experimentation. They can also be derived logically by examining the present day definitions of pressure, volume, and temperature.

Boyle’s Law

Gases are often characterized by their volume, temperature, and pressure. These characteristics, however, are not independent of each other. Gas pressure is dependent on the force exerted by the molecular collisions and the area over which the force is exerted. In turn, the force exerted by these molecular collisions is dependent on the absolute temperature. The relationships between these characteristics can be determined both experimentally and logically from their mathematical definitions.

The relationship between the pressure and volume of a gas was first determined experimentally by an Irish chemist named Robert Boyle (1627-1691). The relationship between the pressure and volume of a gas is commonly referred to as Boyle’s law.

When we wish to observe the relationship between two variables, it is absolutely necessary to keep all other variables constant so that the change in one variable can be directly related to the change in the other. Therefore, when the relationship between the volume and pressure of a gas is investigated, the quantity and temperature of the gas must be held constant so that these factors do not contribute to any observed changes.

You may have noticed that when you try to squeeze a balloon, the resistance to squeezing becomes greater as the balloon becomes smaller. That is, the pressure inside the balloon becomes greater when the volume is reduced. This phenomenon can be studied more carefully with an apparatus like the one shown below. This device is a cylinder with a tightly fitted piston that can be raised or lowered. There is also a pressure gauge fitted to the cylinder so that the gas pressure inside the cylinder can be measured. The amount of gas of gas inside the cylinder cannot change, and the temperature of the gas is not allowed to change.

In the picture on the left, a 4.0-liter volume of gas is exerts a pressure of $2.0 \ \mathrm{atm}$ . If the piston is pushed down to decrease the volume of the gas to $2.0 \ \mathrm{liters}$ , the pressure of the gas is found to be $4.0 \ \mathrm{atm}$ . The piston can be moved up and down to positions for several different volumes, and the pressure of the gas can be for each of the volumes. Several trials would generate a data set like that shown in Table below.

PV Data
Trial	Volume	Pressure
1	$8.0\;\mathrm{liters}$	$1.0\;\mathrm{atm}$
2	$4.0\;\mathrm{liters}$	$2.0\;\mathrm{atm}$
3	$2.0\;\mathrm{liters}$	$4.0\;\mathrm{atm}$
4	$1.0\;\mathrm{liters}$	$8.0\;\mathrm{atm}$

We might note from casual observation of the data that doubling volume is associated with the pressure being reduced by half. Likewise, if we move the piston to cause the pressure to double, the volume is halved. We can analyze this data mathematically by adding a fourth column to our table – namely, a column showing the product of multiplying pressure times volume for each trial (see Table below).

PV Data
Trial	Volume	Pressure	Pressure $\times$ Volume
1	$8.0\;\mathrm{liters}$	$1.0\;\mathrm{atm}$	$8.0\;\mathrm{liters} \cdot \mathrm{atm}$
2	$4.0\;\mathrm{liters}$	$2.0\;\mathrm{atm}$	$8.0\;\mathrm{liters} \cdot \mathrm{atm}$
3	$2.0\;\mathrm{liters}$	$4.0\;\mathrm{atm}$	$8.0\;\mathrm{liters} \cdot \mathrm{atm}$
4	$1.0\;\mathrm{liters}$	$8.0\;\mathrm{atm}$	$8.0\;\mathrm{liters} \cdot \mathrm{atm}$

The data in the last column shows that with constant temperature and quantity of gas, the pressure times the volume for this sample of gas yields a constant. A mathematical constant (often represented by $k$ ) is a number that does not change even when other quantities in the formula do change. The value of $k_1$ will change if a different quantity of gas is used or if the trials are carried out at a different temperature, but for a particular mass of a particular gas at a particular temperature, the value of $k_1$ will always be the same. A subscript 1 is used to distinguish this constant from the constants of other gas laws. This relationship can be shown in a mathematical equation.

: $PV = k_1$

This equation is a mathematical statement of Boyle’s law. This particular equation demonstrates what is called an inverse proportionality. When one of the variables is increased, the other variable will decrease by exactly the same factor. This relationship can be easily seen in a graph like the one shown below.

This result matches our logic intuition. If the pressure a gas exerts is equal to the force divided by the area over which it is exerted, we would expect the pressure to increase when we decrease the area but keep the force constant. Similarly, if we maintain the same number of molecules of gas and we keep the same temperature, we expect the total force exerted by the molecules to be the same. As a result, if we expand the volume of the gas, which increases the area over which the force is exerted, we would expect the pressure to decrease.

This video is a laboratory demonstration of Boyle's Law (4c): http://www.youtube.com/watch?v=J_I8Y-i4Axc (1:38).

Charles’s Law

The relationship between the volume and temperature of a gas was investigated by a French physicist, Jacques Charles (1746-1823). The relationship between the volume and temperature of a gas is often referred to as Charles’s law.

An apparatus that can be used to study the relationship between the temperature and volume of a gas is shown below. Once again, the sample of gas trapped inside a cylinder so that no gas can get in or out. Thus, we would have a constant mass of gas inside the cylinder. In this setup, we would also place a mass on top of a movable piston to keep a constant force pushing against the gas. This guarantees that the gas pressure in the cylinder will be constant. If the pressure inside increases, the piston will be pushed up until the inside pressure becomes equal to the outside pressure. Similarly, if the inside pressure decreases, the outside pressure will push the cylinder down, decreasing the volume until the two pressures again become the same. This system guarantees constant gas pressure inside the cylinder.

With this set up, we can adjust the temperature and measure the volume at each temperature to produce a data table similar to the one we created for comparing pressure and volume. The picture on the left in the diagram above shows the volume of a sample of gas at $250\;\mathrm{K}$ , while the picture on the right shows the volume when the temperature has been raised to $500\;\mathrm{K}$ . After two more trials, the collected data is shown in Table below.

Charles's Law Data
Trial	Volume	Temperature	Volume/Temp
1	$1000.\;\mathrm{mL}$	$250.\;\mathrm{K}$	$4.00\;\mathrm{mL/K}$
2	$1200.\;\mathrm{mL}$	$300.\;\mathrm{K}$	$4.00\;\mathrm{mL/K}$
3	$2000.\;\mathrm{mL}$	$500.\;\mathrm{K}$	$4.00\;\mathrm{mL/K}$
4	$2400.\;\mathrm{mL}$	$600.\;\mathrm{K}$	$4.00\;\mathrm{mL/K}$

In order to find a constant from this data, it was necessary to divide each volume with the corresponding Kelvin temperature. The mathematical expression for Charles’s Law is:

$\frac {V} {T} = k_2$

This relationship is to be expected if we recognize that we are increasing molecular collisions with the walls by raising the temperature. The only way to keep the pressure from increasing is to increase the area over which that force is exerted. This mathematical relationship is known as a direct proportionality. When one variable is increased, the other variable also increases by exactly the same factor.

Historical note: In addition to exploring the relationship between volume and temperature for gases, Jacques Charles was also the first person to fill a large balloon with hydrogen gas and take a solo balloon flight.

This video is a laboratory demonstration of Charle's Law (4c): http://www.youtube.com/watch?v=IkRIKGN3i0k (4:02).

Gay-Lussac’s Law

The relationship between temperature and pressure was investigated by the French chemist, Joseph Gay-Lussac (1778-1850). An apparatus that could be used for this investigation is shown below. In this case, the cylinder does not have a movable piston because it is necessary to hold the volume, as well as the quantify of gas, constant. This apparatus allows us to alter the temperature of a gas and measure the pressure exerted by the gas at each temperature.

After a series of temperatures and pressures have been measured, Table below can be produced. Like Charles’s Law, in order to produce a mathematical constant when operating on the data in the table, we must divide pressure by temperature. The relationship, again like Charles’s Law, is a direct proportionality.

Pressure vs. Temperature Data
Trial	Temperature	Pressure	Pressure/Temp
1	$200.\;\mathrm{K}$	$600.\;\mathrm{torr}$	$3.00\;\mathrm{torr/K}$
2	$300.\;\mathrm{K}$	$900.\;\mathrm{torr}$	$3.00\;\mathrm{torr/K}$
3	$400.\;\mathrm{K}$	$1200.\;\mathrm{torr}$	$3.00\;\mathrm{torr/K}$
4	$500.\;\mathrm{K}$	$1500.\;\mathrm{torr}$	$3.00\;\mathrm{torr/K}$

The mathematical form of Gay-Lussac’s Law is:

$\frac {P} {T} =k_3K$

This relationship demonstrates that when the temperature is increased, the pressure must also increase to maintain the value of the constant, $k_3$ . If the area the molecules are occupying is kept the same, the collisions of the molecules with the surroundings will increases as the temperature increases. This results in a higher pressure.

Standard Conditions for Temperature and Pressure (STP)

It should be apparent by now that expressing a quantity of gas simply by stating its volume is inadequate. Ten liters of oxygen gas could contain any mass of oxygen from $4,000\;\mathrm{grams}$ to $0.50\;\mathrm{grams}$ depending on the temperature and pressure of the gas. Chemists have found it useful to choose a standard temperature and pressure with which to express gas volume. The standard conditions for temperature and pressure (STP) were chosen to be $0^\circ \text{C} \ (273\;\mathrm{K})$ and $1.00\;\mathrm{atm}$ ( $760\;\mathrm{mm \ of \ Hg}$ ). You will commonly see gas volumes expressed as $1.5\;\mathrm{liters}$ of gas under standard conditions or $1.5\;\mathrm{liters}$ of gas at STP. Once you know the temperature and pressure conditions of a volume of gas, you can calculate the volume under other conditions.

This video is a black board presentation of some ideal gas law calculations and it includes the definition of standard temperature and pressure (4d):http://www.youtube.com/watch?v=GwoX_BemwHs (13:01).

The Combined Gas Law

Boyle’s law shows how the volume of a gas changes when its pressure is changed with the temperature held constant, while Charles’s law shows how the volume of a gas changes when the temperature is changed with the pressure held constant. Is there a formula we can use to calculate the change in volume of a gas if both pressure and temperature change? The answer is yes, as we can use a formula that combines both Boyle’s law and Charles’s law.

Boyle's law states that for a sample of gas at constant temperature, every volume times pressure trial will yield the same constant. We use the subscript 1 to represent one set of conditions, and the subscript 2 to represent a second set of conditions,

$P_1 V_1= P_2 V_2 \ \ \ \ \ \text{so} \ \ \ \ \ V_2 = V_1 \cdot \frac {P_1} {P_2}$

We can find a similar expression for Charles's law:

$\frac{V_1} {T_1} = \frac{V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ V_2 = V_1 \cdot \frac{T_2} {T_1}$

Combining the two equations yields:

$V_2 = V_1 \cdot \frac {P_1} {P_2} \cdot \frac {T_2} {T_1}$

The terms in this equation are rearranged and are commonly written in the form shown below. This equation is also known as the combined gas law.

$\frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2}$

When solving problems with the combined gas law, temperatures must always be in Kelvin. The units for pressure and volume may be any appropriate units, but the units of pressure must be the same for $P_1$ and $P_2$ , and the units of volume for $V_1$ and $V_2$ must also be the same.

Example:

A sample of gas has a volume of $400$ . liters when its temperature is $20.^\circ C$ and its pressure is $300.\;\mathrm{mm \ of \ Hg}$ . What volume will the gas occupy at STP?

Solution:

Step 1: Assign known values to the appropriate variable.

$& P_1 = 300.\ \text{mm\ of\ Hg} & & P_2 = 760.\ \text{mm of Hg (standard pressure)}\\ & V_1 = 400.\ \text{liters} & & V_2 = x\ \text{(the unknown)}\\ & T_1 = 20.^\circ \text{C} + 273 = 293\ \text{K} & & T_2 = 0^\circ \text{C} + 273 = 273\ \text{K}$

Step 2: Solve the combined gas law for the unknown variable.

$\frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ V_2 = \frac {P_1V_1T_2} {P_2T_1}$

Step 3: Substitute the known values into the formula and solve for the unknown.

$V_2 = \frac {(300.\ \text{mm of Hg}) \cdot (400.\ \text{L}) \cdot (273\ \text{K})} {(760.\ \text{mm of Hg}) \cdot (293\ \text{K})} = 147 \ \text{liters}$

Example:

A sample of gas occupies $1.00\;\mathrm{liter}$ under standard conditions. What temperature would be required for this sample of gas to occupy $1.50\;\mathrm{liters}$ and exert a pressure of $2.00\;\mathrm{atm}$ ?

Step 1: Assign known values to the appropriate variable.

$& P_1 = 1.00\ \text{atm (standard pressure)} & & P_2 = 2.0\ \text{atm} \\ & V_1 = 1.00\ \text{liter} & & V_2 = 1.50\ \text{liters} \\ & T_1 = 273\ \text{K (standard temperature)} & & T_2 = x \ \text{(unknown)}$

Step 2: Solve the combined gas law for the unknown variable.

$\frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ T_2 = \frac {P_2V_2T_1} {P_1V_1}$

Step 3: Substitute the known values into the formula and solve for the unknown.

$T_2 = \frac {(2.00\ \text{atm}) \cdot (1.50\ \text{L}) \cdot (273\ \text{K})} {(1.00\ \text{atm}) \cdot (1.00\ \text{L})} = 819\ \text{K}$

Example:

A $1.00\;\mathrm{liter}$ sample of oxygen gas under standard conditions has a density of $1.43\;\mathrm{g/L}$ . What is the density of oxygen gas at $500.\;\mathrm{K}$ and $760.\;\mathrm{torr}$ ?

Solution:

You can find the mass of oxygen in the $1.00\;\mathrm{liter}$ sample by multiplying volume times density, which yields a mass of $1.43\;\mathrm{grams.}$ Changing the temperature and/or pressure of a sample of gas changes its volume and therefore its density, but it does not change the mass. Therefore, when the new volume of the gas is found, the mass of oxygen gas in it will still be $1.43\;\mathrm{grams.}$ The density under the new conditions can be found by dividing the mass by the volume the gas now occupies.

Step 1: Assign \known values to the appropriate variable.

$&P_1 = 760.\ \text{torr} & &P_2 = 760.\ \text{torr} \\ &V_1 = 1.00\ \text{L} & &V_2 = x \ \text{unknown} \\ &T_1 = 273\ \text{K}& &T_2 = 500.\ \text{K}$

Step 2: Solve the combined gas law for the unknown variable.

$V_2 = \frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ \frac {P_1V_1T_2} {P_2T_1}$

Step 3: Substitute the known values into the formula and solve for the unknown.

$\frac {(760.\ \text{torr}) \cdot (1.00\ \text{L}) \cdot (500.\ \text{K})} {(760.\ \text{torr}) \cdot (273\ \text{K})} = 1.83\ \text{liters}$

$D_{O_2@500\ \text{K}} = \frac {\text{mass}} {\text{volume}} = \frac {1.43\ \text{g}} {1.83\ \text{L}} = 0.781\ \text{g/L}$

Lesson Summary

Boyle's law states that for a gas at constant temperature, volume is inversely proportional to pressure.
Charles's law states that for a gas at constant pressure, volume in directly proportional to temperature.
Gay-Lussac's law states that for a gas at constant volume, pressure is directly proportional to temperature.
The volume of a mass of gas is dependent on the temperature and pressure. Therefore, these conditions must be given along with the volume of a gas.
Standard conditions for temperature and pressure are $0^\circ \text{C}$ and $1.0\;\mathrm{atm}$ .
The combined gas law relates the temperature, pressure, and volume of a gas.

Review Questions

When a sample of gas is placed in a larger container at the same temperature, what happens to the total force of the molecules hitting the walls?
When a sample of gas is placed in a larger container at the same temperature, what happens to the pressure exerted by the gas?
If $X$ and $Y$ are quantities that are related to each other by inverse proportion, what will the value of $Y$ become when the value of $X$ is increased by a factor of five?
Under what conditions will the value for the constant, $K$ , change in the equation for Boyle’s Law, $PV = K$ .
A sample of gas has a volume of $500.\ \mathrm{mL}$ under a pressure of $500.\ \mathrm{mm \ of \ Hg}$ . What will be the new volume of the gas if the pressure is reduced to $300.\ \mathrm{mm \ of \ Hg}$ at constant temperature?
A graph is made illustrating Charles’s Law. Which line would be appropriate assuming temperature is measured in Kelvin?
At constant pressure, the temperature of a sample of gas is decreased. Will the volume of the sample
1. increase
2. decrease
3. remain the same?
A sample of gas has its temperature increased from $-43^\circ C$ to $47^\circ C$ at constant pressure. If its volume at $-43^\circ C$ was $500.\;\mathrm{mL}$ , what is its volume at $47^\circ C$ ?
A gas is confined in a rigid container and exerts a pressure of $250.\ \mathrm{mm \ of \ Hg}$ at a temperature of $17^\circ \text{C}$ . To what temperature must this gas be cooled in order for its pressure to become $216\ \mathrm{mm \ of \ Hg}$ ? Express this temperature in $^\circ \text{C}$ .
What is the abbreviation used to indicate standard conditions for temperature and pressure?
A sample of gas has a volume of $500.\ \mathrm{mL}$ at standard conditions. Find its volume at $47^\circ \text{C}$ and $800.\ \mathrm{torr}$ .
A sample of gas has a volume of $100.\ \mathrm{L}$ at $17^\circ \text{C}$ and $800.\ \mathrm{torr}$ . To what temperature must the gas be cooled in order for its volume to become $50.0\ \mathrm{L}$ at a pressure of $600. \ \mathrm{torr}$ ?

Universal Gas Law

Lesson Objectives

The student will:

solve problems using the universal gas law, $PV = nRT$ .
state Avogadro’s law of equal molecules in equal volumes.
calculate molar mass (M) from $M = gRT/PV$ given mass, temperature, pressure, and volume.

Vocabulary

$&\mathbf{Avogadro's \ law} &\mathbf{universal \ gas \ law} & &\mathbf{universal \ gas \ law \ constant \ (R)}$

Introduction

The individual gas laws (Boyle's, Charles's, and Gay-Lussac's) and the combined gas law all require the quantity of gas remain constant. The universal gas law (sometimes called the ideal gas law) allows us to make calculations when there are different quantities of gas.

Avogadro’s Law

Avogadro’s law postulates that equal volumes of gas with the same temperature and pressure contain the same number of molecules. Mathematically, this would be written as: $V = k_4n$ , where $n$ represents the number of moles of molecules and $k_3$ represents a constant. This law was known as Avogadro’s hypothesis for the first century of its existence. Since Avogadro's hypothesis can now be demonstrated mathematically, it was decided that it should be called a law instead of a hypothesis. If we think about the definitions of the volumes, pressures, and temperatures for gases, we can also develop Avogadro’s conclusion.

Suppose we have a group of toy robots that are all identical in strength. They do not have the same size or look, but they all have exactly the same strength at an assigned task. We arrange a tug-of-war between groups of robots. We arrange the rope so that we can see one end of the rope, but the other end disappears behind a wall. On the visible end, we place eight robots to pull. On the other end, an unknown number of robots will pull. When we say “go,” both sides pull with maximum strength, but the rope does not move. How many robots are pulling on the hidden end of the rope? Since each robot pulls with the exact same strength, to balance the eight robots we know are on our end, there mustbe eight robots on the other end. We can think of molecules as robots by recognizing that molecules at the same temperature have exactly the same striking power upon collision. As a result, equal volumes of gas with equal pressures and temperatures must contain an equal number of molecules.

In the early 1800s, the first attempts to assign relative atomic weights to the atoms were accomplished by assigning hydrogen to have an atomic mass of $1.0.$ and by decomposing compounds to determine the mass ratios in the compounds. Some of the atomic weights found this way were accurate, but many were not. In the 1860s, Stanislao Cannizzaro refined the process of determining relative atomic weights by using Avogadro’s law. If gas $X$ and gas $Y$ were heated to the same temperature, placed in equal volume containers under the same pressure (gas would be released from one container until the two containers had the same pressure), Avogadro’s conditions would be present. As a result, Cannizzaro could conclude that the two containers had exactly the same number of molecules. The mass of each gas was then determined with a balance (subtracting the masses of the containers), and the relationship between the masses would be the same as the mass relationship of one molecule of $X$ to one molecule of $Y$ . That is, if the total mass of gas $X$ was $10.\ \mathrm{grams}$ and the total mass of gas $Y$ was $40.\ \mathrm{grams}$ , then Cannizzaro knew that one molecule of $Y$ must have four times as much mass as one molecule of $X$ . If an arbitrary value such as $1.0$ dalton was assigned as the mass of gas $X$ , then the mass of gas $Y$ on that same scale would be $4.0$ daltons.

The Universal Gas Law Constant

We have considered three laws that examine how the volume of a gas depends on pressure, temperature, and number of moles of gas (Gay-Lussac's law requires a constant volume).

Boyle’s law: $V = \frac {k_1} {P}$ at constant temperature and constant moles of gas, where $k_1$ is a constant.
Charles’s law: $V = k_2T$ at constant pressure and constant moles of gas, where $k_2$ is a constant.
Avogadro’s law: $V = k_4n$ at constant temperature and pressure, where $n$ is the number of moles of gas and $k_3$ is a constant.

These three relationships can be combined to form the expression $PV = nRT$ , where $R$ is the combination of the three constants. The equation is called the universal gas law(or the ideal gas law), and $R$ is called the universal gas law constant. When the pressure is expressed in atmospheres and the volume in liters, $R$ has the value $0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}$ . You can convert the value of $R$ into values for any set of units for pressure and volume. Moles, of course, always have the unit moles, while the temperature must always be Kelvin.

In our analysis of gas behavior, we have used a pair of assumptions that are not always true. We have assumed that the volume of the actual molecules in a gas is insignificant compared to the volume of the empty space between molecules. We have treated the molecules as if they were geometric points that took up no space. For most gases, this assumption will be true most of the time, and the gas laws work well. If, however, a gas is highly compressed (at very high pressure), the molecules will be pushed together very closely, with much of the empty space between molecules removed. Under such circumstances, the volume of the molecules becomes significant, and some calculations with the gas laws will be slightly off.

Another assumption that we have used is that the molecules are not attracted to each other so that every collision is a perfectly elastic collision. In other words, we have assumed that no energy is lost during the collision. This assumption also works well most of the time. Even though the molecules do have some attraction for each other, usually the temperature is high enough that the molecular motion readily overcomes any attraction and the molecules move around as if there were no attraction. If, however, we operate with gases at low temperatures (temperatures near the phase change temperature of the gas), the molecular attractions have enough effect to cause our calculations to be slightly off. If a gas follows the ideal gas laws, we say that the gas behaves ideally. Gases behave ideally at low pressure and high temperature. At low temperatures or high pressures, gas behavior may become non-ideal. As a point of interest, if you continue your study of chemistry, you will discover that there is yet another gas law equation for these non-ideal situations.

Example:

A sample of nitrogen gas $(\text{N}_2)$ has a volume of $5.56\ \mathrm{liters}$ at $0^\circ \text{C}$ and a pressure of $1.50\ \mathrm{atm}$ . How many moles of nitrogen are present in this sample?

Solution:

Step 1: Assign known values to the appropriate variable.

$& P = 1.50\ \mathrm{atm} \\ & V = 5.56\ \mathrm{L} \\ & n = \mathrm{unknown} \\ & T = 0^\circ \text{C} + 273 = 273\ \mathrm{K} \\ & R = 0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}$

Step 2: Solve the combined gas law for the unknown variable.

$PV = nRT \ \ \ \ \ \text{so} \ \ \ \ \ n = \frac {PV} {RT}$

Step 3: Substitute the known values into the formula and solve for the unknown.

$n = \frac {PV} {RT} = \frac {(1.50\ \text{atm}) \cdot (5.56 \ \text{L})} {(0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}) \cdot (273\ \text{K})} = 0.372\ \text{mol}$

Example:

$2.00\ \mathrm{moles}$ of methane gas $(\text{CH}_4)$ are placed in a rigid $5.00\ \mathrm{liter}$ container and heated to $100.^\circ \text{C}$ . What pressure will be exerted by the methane?

Solution:

Step 1: Assign known values to the appropriate variable.

$& P = \text{unknown} \\ & V = 5.00\ \text{liters} \\ & N = 2.00\ \text{moles} \\ & T = 100.^\circ\text{C} + 273 = 373\ \text{K} \\ & R = 0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}$

Step 2: Solve the combined gas law for the unknown variable.

$PV = nRT \ \ \ \ \ \text{so} \ \ \ \ \ n = \frac {PV} {RT}$

Step 3: Substitute the known values into the formula and solve for the unknown.

$P = \frac {nRT} {V} = \frac {(2.00\ \text{mol}) \cdot (0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}) \cdot (373\ \text{K})} {(5.00\ \text{L})} = 12.2\ \text{atm}$

Example:

A sample gas containing $0.300\ \mathrm{moles}$ of helium at a pressure of $900.\ \mathrm{torr}$ is cooled to $15^\circ \text{C}$ . What volume will the gas occupy under these conditions?

Note: If we are to use $0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}$ for the value of $R$ , then the pressure must be converted from torr to atm.

Solution: Step 1: Assign known values to the appropriate variable.

$& P = (900.\ \text{torr})\cdot(\frac{1.00 \ \text{atm}} {760.\ \text{torr}}) = 1.18\ \text{atm} \\ & V = \ \text{unknown} \\ & N = 0.300\ \text{mole} \\ & T = 15^\circ \text{C} + 273 = 288\ \text{K} \\ & R = 0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}$

Step 2: Solve the combined gas law for the unknown variable.

$PV = nRT \ \ \ \ \ \text{so} \ \ \ \ \ V = \frac {nRT} {P}$

Step 3: Substitute the known values into the formula and solve for the unknown.

$V = \frac {nRT} {P} = \frac {(0.300\ \text{mol})\cdot(0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}) \cdot (373\ \text{K})} {(1.18\ \text{atm})}= 7.08\ \text{liters}$

For a video example of solving an ideal gas law problem (4h), see http://www.youtube.com/watch?v=JEsfU7ogbVQ (4:44).

Molar Mass and the Universal Gas Law

The universal gas law can also be used to determine the molar mass of an unknown gas provided the pressure, volume, temperature, and mass are known. The number of moles of a gas, $n$ , can be expressed as grams/molar mass. If we substitute $g/M$ for $n$ in the universal gas law, we get $PV = (\frac {g} {M})RT$ which can be re-arranged to $\ \mathrm{M} = \frac{gRT} {PV}$ .

Example:

$20.0$ grams of an unknown gas occupy $2.00 \ \text{L}$ under standard conditions. What is the molar mass of the gas?

Solution:

$\text{molar mass} = \frac {gRT} {PV} = \frac {(20.0\ \text{g})(0.08206\ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \ \text{K}^{-1}) \cdot (273\ \text{K})} {(1.00\ \text{atm}) \cdot (2.00\ \text{L})} = 224\ \text{g/mol}$

Density and the Universal Gas Law

The density of a gas under a particular set of conditions is the mass of the sample of gas divided by the volume occupied at those conditions, $D = m/V$ . In the universal gas law, both the mass of the sample of gas and the volume it occupies are represented. We can substitute density for $g/V$ where it appears in the equation and produce an equation that contains density instead of mass and volume. For example, in the equation $M = \frac{gRT} {PV}$ , mass appears in the numerator and volume appears in the denominator. We can substitute $D$ for those two variables and obtain the equation $M = DRT/P$ .

Example:

The density of a gas was determined to be $1.95\ \mathrm{g/mL}$ at $1.50\ \mathrm{atm}$ and $27^\circ \text{C}$ . What is the molar mass of the gas?

Solution

$\text{molar mass} = \frac {DRT} {P} = \frac {(1.95\ \text{g/L}) \cdot (0.08206\ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \cdot (300.\ \text{K})} {(1.50\ \text{atm})} = 32.0\ \text{g/mol}$

Lesson Summary

Avogadro’s law states that equal volumes of gases under the same temperature and pressure contain equal numbers of molecules.
The universal gas law, also known as the ideal gas law, relates the pressure, volume, temperature, and number of moles of gas: $PV = nRT.$
The universal gas law assumes that the gas molecules are geometric points that take up no space and that they undergo perfectly elastic collision.
The universal gas law can also be used to determine the molar mass and the density of an unknown gas.

Review Questions

What conditions of temperature and pressure cause gases to deviate from ideal gas behavior?
What volume will $2.00\ \mathrm{moles}$ of hydrogen gas occupy at $2.62 \ \mathrm{atm}$ of pressure and $300.^\circ \text{C}$ ?
How many moles of gas are required to fill a volume of $8.00\ \mathrm{liters}$ at $2.00\ \mathrm{atm}$ and $273\ \mathrm{K}$ ?
What is the molar mass of a gas if its density is $1.30\ \mathrm{g/L}$ at STP?

University High School Chemistry

Monday, April 29, 2013

Review of Gas Laws

The Three States of Matter

Lesson Objectives

Vocabulary

Introduction

The Assumptions of the Kinetic Molecular Theory

Phases of Matter

Characteristics of Solids

Characteristics of Liquids

Characteristics of Gases

Lesson Summary

Further Reading / Supplemental Links

Review Questions

Gas Laws

Lesson Objectives

Vocabulary

Introduction

Boyle’s Law

Charles’s Law

Gay-Lussac’s Law

Standard Conditions for Temperature and Pressure (STP)

The Combined Gas Law

Lesson Summary

Further Reading / Supplemental Links

Review Questions

Universal Gas Law

Lesson Objectives

Vocabulary

Introduction

Avogadro’s Law

The Universal Gas Law Constant

Molar Mass and the Universal Gas Law

Density and the Universal Gas Law

Lesson Summary

Further Reading / Supplemental Links

Review Questions

No comments:

Post a Comment