14.1: The Ideal Gas Law
Objectives
The student will:
 Explain the Ideal Gas Law.
 Solve problems using the Ideal Gas Law.
Vocabulary

Avogadro’s number: The constant, \begin{align*}6.022 \times 10^{23}\end{align*}
6.022×1023 , representing the number of atoms in a gram atom, or the number of molecules in a gram molecule.  Boyle's Law: States that the absolute pressure and volume of a given mass of confined gas are inversely proportional, if the temperature remains unchanged within a closed system. Thus, it states that the product of pressure and volume is a constant for a given mass of confined gas as long as the temperature is constant, PV = k when T = k. The law was named after chemist and physicist Robert Boyle, who published the original law in 1662.
 Charles’s Law: An experimental gas law which describes how gases tend to expand when heated. A modern statement of Charles' Law is: At constant pressure, the volume of a given mass of an ideal gas increases or decreases by the same factor as its temperature on the absolute temperature scale (i.e. the gas expands as the temperature increases). \begin{align*} V = kT \end{align*}
 GuyLussac’s Law: The pressure of a gas of fixed mass and fixed volume is directly proportional to the absolute temperature of the gas. \begin{align*}\frac{P}{T}=k\end{align*}
 Ideal Gas Law: A law that describes the relationships between measurable properties of an ideal gas. The law states that P × V = n × (R) × T, where P is pressure, V is volume, n is the number of moles of molecules, T is the absolute temperature, and R is the gas constant (8.314 Joules per degree Kelvin or 1.985 calories per degree Celsius). A consequence of this law is that, under constant pressure and temperature conditions, the volume of a gas depends solely on the number of moles of its molecules, not on the type of gas. This is also called the Universal Gas Law.
 mole: A unit of measurement used in chemistry to express amounts of a chemical substance. A mole is defined as an amount of a substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as there are atoms in 12 grams of pure carbon12 (\begin{align*}^{12}C\end{align*}), which is the isotope of carbon with atomic weight 12. This corresponds to a value of \begin{align*}6.02214179 \times 10^{23}\end{align*} elementary entities of the substance. It is one of the base units in the International System of Units, and has the unit symbol mol.
 molecular mass: The mass of a molecule.
Introduction
In the chapter covering temperature and heat, we discussed the kinetic theory of temperature for an ideal gas. The kinetic theory stated that the temperature of such a gas is directly related to the kinetic energy of the molecules of the gas, which, in turn, is equal to the internal energy of the gas. The more we heat a gas, the faster the random motion of its molecules and the greater the increase in temperature and pressure. This seems perfectly reasonable if the gas is confined so that its volume cannot increase. Heat an empty glass bottle with a cork at the top for a sufficiently long time and the molecules of the air in the bottle gain enough kinetic energy to eventually “pop” the cork out of the bottle. In other words, the pressure (force per unit area) becomes great enough to force the cork out of the bottle. But what happens if we heat a gas and allow the volume to increase, or compress a gas while cooling it? In this chapter, we explore the relationships between the mass, volume, pressure, and temperature of ideal gases which are confined, compressed, and expanded.
Atomic Mass Units
Before we continue, it will be helpful to discuss how the masses of different atoms (and molecules) are compared to each other.
The idea that all matter is composed of tiny building blocks called atoms goes back to antiquity. It was only during the last two hundred years or so, however, that some measurements could be made suggesting the truth of the idea. Measurements made when different substances combined led to the conclusion that the ratios of atomic masses tended to be simple proportions and that hydrogen had the smallest mass of all atoms. Compared to a hydrogen atom, for example, a carbon atom had 12 times the mass, an oxygen atom16, and nitrogen14.
Today we define the mass of onetwelfth of a carbon atom as one unified atomic mass unit \begin{align*}(amu)\end{align*} or \begin{align*}(u)\end{align*}
\begin{align*} 1 \ u = 1.66 \times 10^{27} \ kg\end{align*}
Using this definition, a carbon atom has a mass of 12.0000 u, and a hydrogen atom has a mass of 1.0078 u.
The Mole
A basic unit in SI system is the mole (mol). The mole is defined as the amount of a substance that contains the same number of atoms or molecules as exactly 12 grams of carbon12. For example, it is found that 16 grams of oxygen contains the same number of atoms as does 12 grams of carbon12. Note that carbon12 is the most common isotope of carbon. We’ll have more to say about isotopes later.
Molar Mass
Molar mass is the mass of one mole of a substance.
Using the atomic mass of a compound, we can quickly determine the mass of one mole of the compound. For example, carbon monoxide (CO) is a chemical compound composed of one carbon atom and one oxygen atom. The unified atomic mass of carbon is 12 and the unified atomic mass of oxygen is 16. The total unified atomic mass number for carbon monoxide is therefore:
\begin{align*}12+16=28 \ u\end{align*}
The molar mass of carbon dioxide is numerically equal to the atomic mass number, but in grams. Therefore, carbon dioxide has a molar mass of 28 grams.
Another definition of the mole can be stated as: The number of grams of a substance that is equal to its molar mass.
Check Your Understanding
The atomic mass of aluminum oxide Al_{2}O_{3} (2 aluminum atoms and 3 oxygen atoms) is 102 u.
a. What is the mass of one mole of aluminum oxide?
Answer: 102 grams
b. What is the molar mass of aluminum oxide?
Answer: By definition, the mass of one mole of a substance is its molar mass: 102 grams
c. You are given 230 grams of aluminum oxide. How many moles \begin{align*}n\end{align*} of aluminum oxide do you have?
Answer: \begin{align*}n=\frac{230 \ g}{102 \frac{g}{mol}}=2.25 \ mol\end{align*}
d. What is the atomic mass of aluminum?
Answer: As cited above, one atom of oxygen has atomic mass 16 u.
Therefore, three oxygen atoms has atomic mass of
\begin{align*}3 \times 16 = 48 \ u\end{align*}
So, two atoms of aluminum must have anatomic mass of
\begin{align*}102 \ u  48 \ u = 54 \ u\end{align*}
Thus, one atom of aluminum has atomic mass of
\begin{align*}\frac{54 \ u}{2}= 27 \ u \end{align*}
Avogadro’s Number
In 1811, Amedeo Avogadro (17761856), in the Figure below, published a paper hypothesizing that equal volumes of different gases gas at equal pressure and temperature would have an equal number of molecules. Today we know that a volume of 22.4 liters of an ideal gas at standard atmospheric pressure and 20°C contains one mole of gas.
The number of molecules in one mole was not determined until the early 20th century. The number is known today as Avogadro’s Number \begin{align*}N_A=6.022 \times 10^{23}\end{align*} and is defined as one mole.
Amedeo Avogadro
http://www.youtube.com/watch?v=g_BelGwRxG8&feature=related
Check Your Understanding
1a. How many atoms of silver does one mole of silver contain?
Answer: One mole (abbreviated as mol), by definition, contains 6.022×10^{23} particles. In this case, it is 6.022×10^{23} atoms of silver.
1b. A United States silver dollar, Figure below, contains approximately 24.06 grams of silver. The molecular mass of silver is \begin{align*}107.87 \frac{g}{mol}\end{align*}. How many moles of silver are in a silver dollar?
Answer: \begin{align*}n=\frac{24.06 \ g}{107.87 \frac{g}{mol}}=0.22304 \rightarrow 0.2230 \ mol\end{align*}
United States silver dollar, late 19^{th} century
2a. The mass of a United States penny (primarily zinc) is 2.50 grams. What is the molar mass of one mole of pennies?
Answer: There are 6.022×10^{23} particles in every mole. Therefore, one mole of pennies is 6.022×10^{23} pennies.
\begin{align*}\left(2.50 \frac{g}{penny}\right)(6.022 \times 10^{23} \ pennies) = 1.5055 \times 10^{24} \rightarrow 1.51 \times 10^{24} \ grams \end{align*} or \begin{align*}1.51 \times 10^{21} \ kg\end{align*}. This is the mass of Pluto’s moon, Charon, see Figure below. Clearly, a mole is a very large number.
By comparison, the amount of zinc in about 23 pennies is equivalent to one mole of zinc atoms.
Clearly, atoms are very small.
Pluto’s largest moon Charon is 1,200 km in diameter
Three Laws
In the three laws that follow, we assume an ideal gas, as described by the Kinetic Theory. The mass of the gas is held constant, and that the pressure of the gas remains fairly low.
Boyle’s Law
Robert Boyle (16271691), in the Figure below, is regarded as the first modern chemist. In the early 1660s he stated what has become known as Boyle’s Law, which states that if the temperature of a gas is kept constant, then the volume \begin{align*}V\end{align*} and pressure \begin{align*}P\end{align*} form an inverse relationship.
\begin{align*}\rightarrow V \propto \frac{1}{P}, \ or \ PV = \text{constant}\end{align*}.
Figure below shows the relationship as a graph. Notice that as the pressure increases, the volume decreases, if the temperature remains constant. This may not seem obvious.
Robert Boyle
Let us try to understand why it happens. If the temperature does not increase, then the average velocity of the gas molecules does not increase. As we showed earlier, a change in momentum \begin{align*}(\Delta p=F \Delta t)\end{align*} is proportional to force. Recall that pressure is force per unit area. If the volume decreases, then more molecules impact a smaller surface area, thus there is more force per unit area.
The product remains constant;
In an 1802 paper, scientist Joseph Louis GayLussac, who we shall have more to say about shortly, credited Jacques Charles (17461823), Figure below, with what is now known as Charles’s Law. According to Charles’s Law, if the pressure of a gas is kept constant, the volume \begin{align*}V\end{align*} of a gas varies directly with the absolute temperature \begin{align*}T\end{align*} (expressed, as you may recall, in Kelvins), \begin{align*}V \propto T\end{align*} Figure below. Close to absolute zero, all gases will liquefy. The graph in Figure below is therefore extrapolated for temperatures close to absolute zero. We note that experimentally reducing the temperature of any substance to zero \begin{align*}K\end{align*} is not possible.
Jacques Charles
GuyLussac’s Law
Joseph Louis GayLussac (17781850), Figure below, is also credited with stating that when the volume of a gas is held constant, the pressure \begin{align*}P\end{align*} is directly related to the temperature \begin{align*}T\end{align*}(Kelvin), \begin{align*}P \propto T\end{align*} Figure below. This graph is also extrapolated for temperatures close to absolute zero.
GuyLussac
The Mass of a Gas
For the three laws above, we assumed that the mass of the gas was constant. But the mass of a gas can increase or decrease. Fill a tire with air (a gas) and there is an increase in mass. If the air flows out of the tire, the mass decreases. In both cases, the volume of the tire will change. Once a tire is close to its rated pressure, however, filling it with more air will hardly increase the volume. But the pressure can still increase a good deal. The increase in pressure is due to the increase in mass of the gas because there are more molecules hitting the inside of the tire. Both volume \begin{align*}V\end{align*} and pressure \begin{align*}P\end{align*}, therefore, increase with mass in direct proportion \begin{align*}m \rightarrow P \propto m\end{align*} and \begin{align*}V \propto m\end{align*}.
The Ideal Gas Law
The three gas laws we have previously discussed in addition to the relationship between volume, pressure, and mass, lead to a more general statement of proportionality \begin{align*}\rightarrow PV \propto mT\end{align*}. We can write this as an equation \begin{align*}\rightarrow PV = CmT\end{align*}, but we can show that if the mass \begin{align*}m\end{align*} is used in the equation, then the constant of proportionality \begin{align*}C\end{align*} depends upon the particular gas. If however, the mass is expressed using the number of moles \begin{align*}n\end{align*}, a constant of proportionality (known as the universal gas constant) \begin{align*}R=8.314 \frac{J}{molK}\end{align*}, is applicable for any gas.
Under these conditions the Ideal Gas Law can be expressed as \begin{align*}PV=nRT\end{align*}, where the temperature \begin{align*}T\end{align*} must be in Kelvins. No gas exactly follows the Ideal Gas Law. But if the assumptions made for the kinetic theory of gases are valid (that is, if the gas has high temperature and low pressure), then the Ideal Gas Law is applicable. Unless stated otherwise, throughout this chapter we will assume that we deal with an ideal gas.
http://www.youtube.com/watch?v=PjLA0Fmi44A
Illustrative Example 14.1.1
What is the volume of one mole of any gas at one atmosphere of pressure at room temperature \begin{align*}20.0^{\circ}C\end{align*}?
Answer:
Recall from earlier that \begin{align*} 1 \ atm = 1.01 \times 10^5 \frac{N}{m^2}\end{align*} and that the temperature in the Ideal Gas Law equation must be in Kelvin. Remember that
\begin{align*}& T_k = T_c+273 \rightarrow 20+273=293 \ K\\ & PV = nRT \rightarrow \left(1.01 \times 10^5 \frac{N}{m^2}\right)V = (1.00 \ mol)\left(8.314 \frac{J}{mol*K}\right)(293 \ K) \rightarrow \\ & V = \frac{(1.00 \ mol) \left(8.314 \frac{J}{mol*K})\right (293 \ K)}{\left(1.01 \times 10^5 \frac{N}{m^2}\right)}=0.0241=24.1 \times 10^{3} \frac{J*m^2}{N} \rightarrow \\ & 24.1 \times 10^{3} \frac{N*m^3}{N} \rightarrow 24.1 \times 10^{3} \ m^3\end{align*}
What would the volume be at \begin{align*}0.00^{\circ}C\end{align*}?
Answer: \begin{align*}22.4 \times 10^{3} m^3\end{align*}
Illustrative Example 14.1.2
An ideal gas is contained in a rigid vessel at a pressure of \begin{align*}1.15 \ atm \ at \ 29^{\circ}C\end{align*}. What is the pressure if the vessel is heated to a temperature of \begin{align*}90^{\circ}C\end{align*}? Express the answer in atm.
Answer:
The volume and mass remain constant, therefore \begin{align*}PV=nRT \rightarrow \frac{P}{T} = \frac{nR}{V}=\end{align*} constant. Before calculating, don’t forget to change the temperature to Kelvins!
\begin{align*}\frac{P_i}{T_i}=\frac{P_f}{T_f} \rightarrow \frac{1.15P_{atm}}{(29.0+273)K}=\frac{P_f}{(90.0+273)K} \rightarrow P_f = \frac{363K}{302K}(1.15 P_{atm})=1.382P_{atm} \rightarrow 1.38P_{atm}\end{align*}
Illustrative Example 14.1.3
A spherical helium balloon of radius 20.00 cm is initially inflated to a pressure \begin{align*}P_i\end{align*} of 1.05 atm at a temperature of \begin{align*}30.00^{\circ}C\end{align*}. The balloon remains in direct sunlight and the temperature of the helium increases to \begin{align*}37.00^{\circ}C\end{align*}. If the pressure inside the balloon increases by 1.00%, by what percent does the volume of the balloon increase?
Answer:
Since the number of moles remains unchanged, \begin{align*}\frac{P_iV_i}{T_i}=nR=\text{constant} \rightarrow\end{align*}
\begin{align*}\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f} \rightarrow V_f = \frac{T_f}{T_i} \frac{P_i}{P_f}V_i\end{align*}, however \begin{align*}P_f=P_i(1+0.01)=1.01 P_i \rightarrow \end{align*}
Substituting for \begin{align*}1.01 P_i\end{align*} for \begin{align*}P_f\end{align*}
\begin{align*}V_f=\frac{T_f}{T_i}\frac{P_i}{P_i(1.01)}V_i=\frac{(37.00+273)}{(30.00+273)}\frac{1}{1.01}V_i=1.013V_i \rightarrow V_f = 1.013 V_i \rightarrow\end{align*}
Percent increase in volume is \begin{align*}1.0131.000=0.013 \rightarrow 1.3 \%\end{align*}
Illustrative Example 14.1.4
Boltzmann’s constant \begin{align*}k\end{align*}, is defined as \begin{align*}k=\frac{R}{N_A}\end{align*}. Show that the internal energy of a gas can be stated as \begin{align*}U=\frac{3}{2}nRT\end{align*}.
Answer:
If \begin{align*}\frac{R}{N_A}\end{align*} is substituted for \begin{align*}k\end{align*} in \begin{align*}U=\frac{3}{2}kNT\end{align*}, then \begin{align*}U=\frac{3}{2}\frac{N}{N_A}RT\end{align*}.
But \begin{align*}\frac{N}{N_A}=n\end{align*}, since \begin{align*}N\end{align*} is the total number of particles divided by the number of particles in one mole \begin{align*}N_A\end{align*}. We can therefore state that \begin{align*}U=\frac{3}{2}nRT\end{align*}.
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Dec 06, 2012Last Modified:
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