College Physics for AP® Courses

13.3The Ideal Gas Law

College Physics for AP® Courses13.3 The Ideal Gas Law

Learning Objectives

By the end of this section, you will be able to:

• State the ideal gas law in terms of molecules and in terms of moles.
• Use the ideal gas law to calculate pressure change, temperature change, volume change, or the number of molecules or moles in a given volume.
• Use Avogadro’s number to convert between number of molecules and number of moles.
Figure 13.16 The air inside this hot air balloon flying over Putrajaya, Malaysia, is hotter than the ambient air. As a result, the balloon experiences a buoyant force pushing it upward. (credit: Kevin Poh, Flickr)

In this section, we continue to explore the thermal behavior of gases. In particular, we examine the characteristics of atoms and molecules that compose gases. (Most gases, for example nitrogen, $N2N2 size 12{N rSub { size 8{2} } } {}$, and oxygen, $O2O2 size 12{O rSub { size 8{2} } } {}$, are composed of two or more atoms. We will primarily use the term “molecule” in discussing a gas because the term can also be applied to monatomic gases, such as helium.)

Gases are easily compressed. We can see evidence of this in Table 13.2, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same $ββ size 12{β} {}$. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.

The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in Figure 13.17. Because atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them.

Figure 13.17 Atoms and molecules in a gas are typically widely separated, as shown. Because the forces between them are quite weak at these distances, the properties of a gas depend more on the number of atoms per unit volume and on temperature than on the type of atom.

To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into an initially deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If we continue to pump air into it, the pressure increases. The pressure will further increase when the car is driven and the tires move. Most manufacturers specify optimal tire pressure for cold tires. (See Figure 13.18.)

Figure 13.18 (a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure. (b) When the tire is filled to a certain point, the tire walls resist further expansion and the pressure increases with more air. (c) Once the tire is inflated, its pressure increases with temperature.

At room temperatures, collisions between atoms and molecules can be ignored. In this case, the gas is called an ideal gas, in which case the relationship between the pressure, volume, and temperature is given by the equation of state called the ideal gas law.

Ideal Gas Law

The ideal gas law states that

$PV=NkT,PV=NkT, size 12{ ital "PV"= ital "NkT"} {}$
13.18

where $PP size 12{P} {}$ is the absolute pressure of a gas, $VV size 12{V} {}$ is the volume it occupies, $NN size 12{N} {}$ is the number of atoms and molecules in the gas, and $TT size 12{T} {}$ is its absolute temperature. The constant $kk size 12{k} {}$ is called the Boltzmann constant in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value

$k=1.38×10−23 J/K.k=1.38×10−23 J/K. size 12{k=1 "." "38" times "10" rSup { size 8{ - "23"} } " J"/K} {}$
13.19

The ideal gas law can be derived from basic principles, but was originally deduced from experimental measurements of Charles’ law (that volume occupied by a gas is proportional to temperature at a fixed pressure) and from Boyle’s law (that for a fixed temperature, the product $PVPV size 12{ ital "PV"} {}$ is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of $VV size 12{V} {}$. The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that $NN size 12{N} {}$ is the total number of atoms and molecules, independent of the type of gas.)

Let us see how the ideal gas law is consistent with the behavior of filling the tire when it is pumped slowly and the temperature is constant. At first, the pressure $PP size 12{P} {}$ is essentially equal to atmospheric pressure, and the volume $VV size 12{V} {}$ increases in direct proportion to the number of atoms and molecules $NN size 12{N} {}$ put into the tire. Once the volume of the tire is constant, the equation $PV=NkTPV=NkT size 12{ ital "PV"= ital "NkT"} {}$ predicts that the pressure should increase in proportion to the number N of atoms and molecules.

Example 13.6

Calculating Pressure Changes Due to Temperature Changes: Tire Pressure

Suppose your bicycle tire is fully inflated, with an absolute pressure of $7.00×105 Pa7.00×105 Pa size 12{7 "." "00" times "10" rSup { size 8{5} } " Pa"} {}$ (a gauge pressure of just under $90.0lb/in290.0lb/in2 size 12{"90" "." 0"lb/in" rSup { size 8{2} } } {}$) at a temperature of $18.0ºC18.0ºC size 12{"18" "." 0°C} {}$. What is the pressure after its temperature has risen to $35.0ºC35.0ºC size 12{"35" "." 0°C} {}$? Assume that there are no appreciable leaks or changes in volume.

Strategy

The pressure in the tire is changing only because of changes in temperature. First we need to identify what we know and what we want to know, and then identify an equation to solve for the unknown.

We know the initial pressure $P0=7.00×105 PaP0=7.00×105 Pa$, the initial temperature $T0=18.0ºCT0=18.0ºC$, and the final temperature $Tf=35.0ºCTf=35.0ºC$. We must find the final pressure $PfPf$. How can we use the equation $PV=NkTPV=NkT$? At first, it may seem that not enough information is given, because the volume $VV$ and number of atoms $NN$ are not specified. What we can do is use the equation twice: $P0V0=NkT0P0V0=NkT0$ and $PfVf=NkTfPfVf=NkTf$. If we divide $PfVfPfVf$ by $P0V0P0V0$ we can come up with an equation that allows us to solve for $PfPf$.

$P f V f P 0 V 0 = N f kT f N 0 kT 0 P f V f P 0 V 0 = N f kT f N 0 kT 0$
13.20

Since the volume is constant, $VfVf size 12{V rSub { size 8{f} } } {}$ and $V0V0 size 12{V rSub { size 8{0} } } {}$ are the same and they cancel out. The same is true for $NfNf size 12{N rSub { size 8{f} } } {}$ and $N0N0 size 12{N rSub { size 8{0} } } {}$, and $kk size 12{k} {}$, which is a constant. Therefore,

$P f P 0 = T f T 0 . P f P 0 = T f T 0 . size 12{ { {P rSub { size 8{f} } } over {P rSub { size 8{0} } } } = { {T rSub { size 8{f} } } over {T rSub { size 8{0} } } } "." } {}$
13.21

We can then rearrange this to solve for $PfPf size 12{P rSub { size 8{f} } } {}$:

$P f = P 0 T f T 0 , P f = P 0 T f T 0 , size 12{P rSub { size 8{f} } =P rSub { size 8{0} } { {T rSub { size 8{f} } } over {T rSub { size 8{0} } } } ,} {}$
13.22

where the temperature must be in units of kelvins, because $T0T0 size 12{T rSub { size 8{0} } } {}$ and $TfTf size 12{T rSub { size 8{f} } } {}$ are absolute temperatures.

Solution

1. Convert temperatures from Celsius to Kelvin.

T 0 = 18 . 0 + 273 K = 291 K T f = 35 . 0 + 273 K = 308 K T 0 = 18 . 0 + 273 K = 291 K T f = 35 . 0 + 273 K = 308 K alignl { stack { size 12{T rSub { size 8{0} } = left ("18" "." 0+"273" right )" K"="291 K"} {} # T rSub { size 8{f} } = left ("35" "." 0+"273" right )" K"="308 K" {} } } {}
13.23

2. Substitute the known values into the equation.

$P f = P 0 T f T 0 = 7 . 00 × 10 5 Pa 308 K 291 K = 7 . 41 × 10 5 Pa P f = P 0 T f T 0 = 7 . 00 × 10 5 Pa 308 K 291 K = 7 . 41 × 10 5 Pa size 12{P rSub { size 8{f} } =P rSub { size 8{0} } { {T rSub { size 8{f} } } over {T rSub { size 8{0} } } } =7 "." "00" times "10" rSup { size 8{5} } " Pa" left ( { {"308 K"} over {"291 K"} } right )=7 "." "41" times "10" rSup { size 8{5} } "Pa"} {}$
13.24

Discussion

The final temperature is about 6% greater than the original temperature, so the final pressure is about 6% greater as well. Note that absolute pressure and absolute temperature must be used in the ideal gas law.

Making Connections: Take-Home Experiment—Refrigerating a Balloon

Inflate a balloon at room temperature. Leave the inflated balloon in the refrigerator overnight. What happens to the balloon, and why?

Example 13.7

Calculating the Number of Molecules in a Cubic Meter of Gas

How many molecules are in a typical object, such as gas in a tire or water in a drink? We can use the ideal gas law to give us an idea of how large $NN size 12{N} {}$ typically is.

Calculate the number of molecules in a cubic meter of gas at standard temperature and pressure (STP), which is defined to be $0ºC0ºC size 12{0°C} {}$ and atmospheric pressure.

Strategy

Because pressure, volume, and temperature are all specified, we can use the ideal gas law $PV=NkTPV=NkT size 12{ ital "PV"= ital "NkT"} {}$, to find $NN size 12{N} {}$.

Solution

1. Identify the knowns.

$T = 0 º C = 273 K P = 1 . 01 × 10 5 Pa V = 1 . 00 m 3 k = 1 . 38 × 10 − 23 J/K T = 0 º C = 273 K P = 1 . 01 × 10 5 Pa V = 1 . 00 m 3 k = 1 . 38 × 10 − 23 J/K$
13.25

2. Identify the unknown: number of molecules, $NN size 12{N} {}$.

3. Rearrange the ideal gas law to solve for $NN size 12{N} {}$.

PV = NkT N = PV kT PV = NkT N = PV kT alignl { stack { size 12{ ital "PV"= ital "NkT"} {} # size 12{N= { { ital "PV"} over { ital "kT"} } } {} } } {}
13.26

4. Substitute the known values into the equation and solve for $NN size 12{N} {}$.

$N = PV kT = 1 . 01 × 10 5 Pa 1 . 00 m 3 1 . 38 × 10 − 23 J/K 273 K = 2 . 68 × 10 25 molecules N = PV kT = 1 . 01 × 10 5 Pa 1 . 00 m 3 1 . 38 × 10 − 23 J/K 273 K = 2 . 68 × 10 25 molecules size 12{N= { { ital "PV"} over { ital "kT"} } = { { left (1 "." "01" times "10" rSup { size 8{5} } " Pa" right ) left (1 "." "00 m" rSup { size 8{3} } right )} over { left (1 "." "38" times "10" rSup { size 8{ - "23"} } " J/K" right ) left ("273 K" right )} } =2 "." "68" times "10" rSup { size 8{"25"} } "molecules"} {}$
13.27

Discussion

This number is undeniably large, considering that a gas is mostly empty space. $NN size 12{N} {}$ is huge, even in small volumes. For example, $1 cm31 cm3 size 12{1" cm" rSup { size 8{3} } } {}$ of a gas at STP has $2.68×10192.68×1019 size 12{2 "." "68"´"10" rSup { size 8{"19"} } } {}$ molecules in it. Once again, note that $NN size 12{N} {}$ is the same for all types or mixtures of gases.

It is sometimes convenient to work with a unit other than molecules when measuring the amount of substance. A mole (abbreviated mol) is defined to be the amount of a substance that contains as many atoms or molecules as there are atoms in exactly 12 grams (0.012 kg) of carbon-12. The actual number of atoms or molecules in one mole is called Avogadro’s number$(NA)(NA) size 12{ $$N rSub { size 8{A} }$$ } {}$, in recognition of Italian scientist Amedeo Avogadro (1776–1856). He developed the concept of the mole, based on the hypothesis that equal volumes of gas, at the same pressure and temperature, contain equal numbers of molecules. That is, the number is independent of the type of gas. This hypothesis has been confirmed, and the value of Avogadro’s number is

$N A = 6 . 02 × 10 23 mol − 1 . N A = 6 . 02 × 10 23 mol − 1 . size 12{N rSub { size 8{A} } =6 "." "02" times "10" rSup { size 8{"23"} } "mol" rSup { size 8{ - 1} } "." } {}$
13.28

One mole always contains $6.02×10236.02×1023 size 12{6 "." "02"´"10" rSup { size 8{"23"} } } {}$ particles (atoms or molecules), independent of the element or substance. A mole of any substance has a mass in grams equal to its molecular mass, which can be calculated from the atomic masses given in the periodic table of elements.

$N A = 6 . 02 × 10 23 mol − 1 N A = 6 . 02 × 10 23 mol − 1 size 12{N rSub { size 8{A} } =6 "." "02" times "10" rSup { size 8{"23"} } "mol" rSup { size 8{ - 1} } } {}$
13.29
Figure 13.19 How big is a mole? On a macroscopic level, one mole of table tennis balls would cover the Earth to a depth of about 40 km.

The active ingredient in a Tylenol pill is 325 mg of acetaminophen $(C8H9NO2)(C8H9NO2) size 12{ $$C rSub { size 8{8} } H rSub { size 8{9} } "NO" rSub { size 8{2} }$$ } {}$. Find the number of active molecules of acetaminophen in a single pill.

Example 13.8

Calculating Moles per Cubic Meter and Liters per Mole

Calculate: (a) the number of moles in $1.00 m31.00 m3 size 12{1 "." "00"" m" rSup { size 8{3} } } {}$ of gas at STP, and (b) the number of liters of gas per mole.

Strategy and Solution

(a) We are asked to find the number of moles per cubic meter, and we know from Example 13.7 that the number of molecules per cubic meter at STP is $2.68×10252.68×1025 size 12{2 "." "68"´"10" rSup { size 8{"25"} } } {}$. The number of moles can be found by dividing the number of molecules by Avogadro’s number. We let $nn size 12{n} {}$ stand for the number of moles,

$n mol/m 3 = N molecules/m 3 6 . 02 × 10 23 molecules/mol = 2 . 68 × 10 25 molecules/m 3 6 . 02 × 10 23 molecules/mol = 44 . 5 mol/m 3 . n mol/m 3 = N molecules/m 3 6 . 02 × 10 23 molecules/mol = 2 . 68 × 10 25 molecules/m 3 6 . 02 × 10 23 molecules/mol = 44 . 5 mol/m 3 . size 12{n"mol/m" rSup { size 8{3} } = { {N"molecules/m" rSup { size 8{3} } } over {6 "." "02" times "10" rSup { size 8{"23"} } "molecules/mol"} } = { {2 "." "68" times "10" rSup { size 8{"25"} } "molecules/m" rSup { size 8{3} } } over {6 "." "02" times "10" rSup { size 8{"23"} } "molecules/mol"} } ="44" "." 5"mol/m" rSup { size 8{3} } "." } {}$
13.33

(b) Using the value obtained for the number of moles in a cubic meter, and converting cubic meters to liters, we obtain

$10 3 L/m 3 44 . 5 mol/m 3 = 22 . 5 L/mol . 10 3 L/m 3 44 . 5 mol/m 3 = 22 . 5 L/mol . size 12{ { { left ("10" rSup { size 8{3} } "L/m" rSup { size 8{3} } right )} over {44 "." 5"mol/m" rSup { size 8{3} } } } ="22" "." 5"L/mol" "." } {}$
13.34

Discussion

This value is very close to the accepted value of 22.4 L/mol. The slight difference is due to rounding errors caused by using three-digit input. Again this number is the same for all gases. In other words, it is independent of the gas.

The (average) molar weight of air (approximately 80% $N2N2 size 12{N rSub { size 8{2} } } {}$ and 20% $O2O2 size 12{O rSub { size 8{2} } } {}$ is $M=28.8 g.M=28.8 g. size 12{M="28" "." 8" g" "." } {}$ Thus the mass of one cubic meter of air is 1.28 kg. If a living room has dimensions $5 m×5 m×3 m,5 m×5 m×3 m, size 12{5" m" times "5 m" times "3 m,"} {}$ the mass of air inside the room is 96 kg, which is the typical mass of a human.

The density of air at standard conditions $(P=1atm(P=1atm size 12{ $$P=1" atm"} {}$ and $T=20ºC)T=20ºC) size 12{T="20"°C$$ } {}$ is $1.28 kg/m31.28 kg/m3 size 12{1 "." "28"" kg/m" rSup { size 8{3} } } {}$. At what pressure is the density $0.64 kg/m30.64 kg/m3 size 12{0 "." "64 kg/m" rSup { size 8{3} } } {}$ if the temperature and number of molecules are kept constant?

The Ideal Gas Law Restated Using Moles

A very common expression of the ideal gas law uses the number of moles, $nn size 12{n} {}$, rather than the number of atoms and molecules, $NN size 12{N} {}$. We start from the ideal gas law,

$PV=NkT,PV=NkT, size 12{ ital "PV"= ital "NkT"} {}$
13.35

and multiply and divide the equation by Avogadro’s number $NANA size 12{N rSub { size 8{A} } } {}$. This gives

$PV = N N A N A kT . PV = N N A N A kT . size 12{ ital "PV"= { {N} over {N rSub { size 8{A} } } } N rSub { size 8{A} } ital "kT" "." } {}$
13.36

Note that $n=N/NAn=N/NA size 12{n=N/N rSub { size 8{A} } } {}$ is the number of moles. We define the universal gas constant $R=NAkR=NAk size 12{R=N rSub { size 8{A} } k} {}$, and obtain the ideal gas law in terms of moles.

Ideal Gas Law (in terms of moles)

The ideal gas law (in terms of moles) is

$PV=nRT.PV=nRT. size 12{ ital "PV"= ital "nRT"} {}$
13.37

The numerical value of $RR size 12{R} {}$ in SI units is

$R=NAk=6.02×1023mol−11.38×10−23J/K=8.31J/mol⋅K.R=NAk=6.02×1023mol−11.38×10−23J/K=8.31J/mol⋅K. size 12{R=N rSub { size 8{A} } k= left (6 "." "02" times "10" rSup { size 8{"23"} } "mol" rSup { size 8{ - 1} } right ) left (1 "." "38" times "10" rSup { size 8{ - "23"} } "J/K" right )=8 "." "31"J/"mol" cdot K} {}$
13.38

In other units,

R = 1 . 99 cal/mol ⋅ K R = 0 . 0821 L ⋅ atm/mol ⋅ K . R = 1 . 99 cal/mol ⋅ K R = 0 . 0821 L ⋅ atm/mol ⋅ K . alignl { stack { size 12{R=1 "." "99"" cal/mol" cdot K} {} # size 12{R"=0" "." "0821 L" cdot "atm/mol" cdot K "." } {} } } {}
13.39

You can use whichever value of $RR size 12{R} {}$ is most convenient for a particular problem.

Example 13.9

Calculating Number of Moles: Gas in a Bike Tire

How many moles of gas are in a bike tire with a volume of $2.00×10–3m3(2.00 L),2.00×10–3m3(2.00 L), size 12{2 "." "00"´"10" rSup { size 8{ +- 3} } " m" rSup { size 8{3} } $$2 "." "00 L"$$ ,} {}$ a pressure of $7.00×105Pa7.00×105Pa size 12{7 "." "00"´"10" rSup { size 8{5} } " Pa"} {}$ (a gauge pressure of just under $90.0lb/in290.0lb/in2 size 12{"90" "." 0" lb/in" rSup { size 8{2} } } {}$), and at a temperature of $18.0ºC18.0ºC size 12{"18" "." 0°C} {}$?

Strategy

Identify the knowns and unknowns, and choose an equation to solve for the unknown. In this case, we solve the ideal gas law, $PV=nRTPV=nRT size 12{ ital "PV"= ital "nRT"} {}$, for the number of moles $nn size 12{n} {}$.

Solution

1. Identify the knowns.

P = 7 . 00 × 10 5 Pa V = 2 . 00 × 10 − 3 m 3 T = 18 . 0 º C = 291 K R = 8 . 31 J/mol ⋅ K P = 7 . 00 × 10 5 Pa V = 2 . 00 × 10 − 3 m 3 T = 18 . 0 º C = 291 K R = 8 . 31 J/mol ⋅ K alignl { stack { size 12{P=7 "." "00" times "10" rSup { size 8{5} } " Pa"} {} # V=2 "." "00" times "10" rSup { size 8{ - 3} } " m" rSup { size 8{3} } {} # T="18" "." 0°C="291 K" {} # R=8 "." "31"" J/mol" cdot K {} } } {}
13.40

2. Rearrange the equation to solve for $nn size 12{n} {}$ and substitute known values.

n = PV RT = 7 . 00 × 10 5 Pa 2 . 00 × 10 − 3 m 3 8 . 31 J/mol ⋅ K 291 K = 0 . 579 mol n = PV RT = 7 . 00 × 10 5 Pa 2 . 00 × 10 − 3 m 3 8 . 31 J/mol ⋅ K 291 K = 0 . 579 mol alignl { stack { size 12{n= { { ital "PV"} over { ital "RT"} } = { { left (7 "." "00" times "10" rSup { size 8{5} } "Pa" right ) left (2 "." 00 times "10" rSup { size 8{ - 3} } m rSup { size 8{3} } right )} over { left (8 "." "31""J/mol" cdot K right ) left ("291"" K" right )} } } {} # " "=" 0" "." "579"`"mol" {} } } {}
13.41

Discussion

The most convenient choice for $RR size 12{R} {}$ in this case is $8.31 J/mol⋅K,8.31 J/mol⋅K, size 12{8 "." "31"" J/mol" cdot "K,"} {}$ because our known quantities are in SI units. The pressure and temperature are obtained from the initial conditions in Example 13.6, but we would get the same answer if we used the final values.

The ideal gas law can be considered to be another manifestation of the law of conservation of energy (see Conservation of Energy). Work done on a gas results in an increase in its energy, increasing pressure and/or temperature, or decreasing volume. This increased energy can also be viewed as increased internal kinetic energy, given the gas’s atoms and molecules.

The Ideal Gas Law and Energy

Let us now examine the role of energy in the behavior of gases. When you inflate a bike tire by hand, you do work by repeatedly exerting a force through a distance. This energy goes into increasing the pressure of air inside the tire and increasing the temperature of the pump and the air.

The ideal gas law is closely related to energy: the units on both sides are joules. The right-hand side of the ideal gas law in $PV=NkTPV=NkT size 12{ ital "PV"= ital "NkT"} {}$ is $NkTNkT size 12{ ital "NkT"} {}$. This term is roughly the amount of translational kinetic energy of $NN size 12{N} {}$ atoms or molecules at an absolute temperature $TT size 12{T} {}$, as we shall see formally in Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature. The left-hand side of the ideal gas law is $PVPV size 12{ ital "PV"} {}$, which also has the units of joules. We know from our study of fluids that pressure is one type of potential energy per unit volume, so pressure multiplied by volume is energy. The important point is that there is energy in a gas related to both its pressure and its volume. The energy can be changed when the gas is doing work as it expands—something we explore in Heat and Heat Transfer Methods—similar to what occurs in gasoline or steam engines and turbines.

Problem-Solving Strategy: The Ideal Gas Law

Step 1 Examine the situation to determine that an ideal gas is involved. Most gases are nearly ideal.

Step 2 Make a list of what quantities are given, or can be inferred from the problem as stated (identify the known quantities). Convert known values into proper SI units (K for temperature, Pa for pressure, $m3m3 size 12{m rSup { size 8{3} } } {}$ for volume, molecules for $NN size 12{N} {}$, and moles for $nn size 12{n} {}$).

Step 3 Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written list is useful.

Step 4 Determine whether the number of molecules or the number of moles is known, in order to decide which form of the ideal gas law to use. The first form is $PV=NkTPV=NkT size 12{ ital "PV"= ital "NkT"} {}$ and involves $NN size 12{N} {}$, the number of atoms or molecules. The second form is $PV=nRTPV=nRT size 12{ ital "PV"= ital "nRT"} {}$ and involves $nn size 12{n} {}$, the number of moles.

Step 5 Solve the ideal gas law for the quantity to be determined (the unknown quantity). You may need to take a ratio of final states to initial states to eliminate the unknown quantities that are kept fixed.

Step 6 Substitute the known quantities, along with their units, into the appropriate equation, and obtain numerical solutions complete with units. Be certain to use absolute temperature and absolute pressure.

Step 7 Check the answer to see if it is reasonable: Does it make sense?

Liquids and solids have densities about 1000 times greater than gases. Explain how this implies that the distances between atoms and molecules in gases are about 10 times greater than the size of their atoms and molecules.