The Ideal Gas Law

Chapter contents: Chapter 8: Thermal Physics

**Figure 8.3.1.** 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,

N_{2}

, and oxygen,

O_{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. 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

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 8.3.2. 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.

Spheres representing atoms and molecules; the spheres are relatively far apart and are distributed randomly. — **Figure 8.3.2.** 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 8.3.3.)

The figure has three parts, each part showing a pair of tires, and each tire connected to a pressure gauge. Each pair of tires represents the before and after images of a single tire, along with a change in pressure in that tire. In part a, the tire pressure is initially zero. After some air is added, represented by an arrow labeled Add air, the pressure rises to slightly above zero. In part b, the tire pressure is initially at the half-way mark. After some air is added, represented by an arrow labeled Add air, the pressure rises to the three-fourths mark. In part c, the tire pressure is initially at the three-fourths mark. After the temperature is raised, represented by an arrow labeled Increase temperature, the pressure rises to nearly the full mark. — **Figure 8.3.3.** (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,

(8.3.1)

where

P

is the absolute pressure of a gas,

V

is the volume it occupies,

N

is the number of atoms and molecules in the gas, and

T

is its absolute temperature. The constant

k

is called the Boltzmann constant in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value

k = 1 . 38 \times 10^{- 23} J / K .

(8.3.2)

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

PV

is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of

V

. The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that

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

P

is essentially equal to atmospheric pressure, and the volume

V

increases in direct proportion to the number of atoms and molecules

N

put into the tire. Once the volume of the tire is constant, the equation

PV = NkT

predicts that the pressure should increase in proportion to the number N of atoms and molecules.

Example 8.3.1

Calculating Pressure Changes Due to Temperature Changes: Tire Pressure

Suppose your bicycle tire is fully inflated, with an absolute pressure of

7 . 00 \times 10^{5} Pa

(a gauge pressure of just under

90.0 {lb/in}^{2}

) at a temperature of

18.0 º C

. What is the pressure after its temperature has risen to

35.0 º 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

P_{0} = 7 .00 \times 10^{5} Pa

, the initial temperature

T_{0} = 18 . 0ºC

, and the final temperature

T_{f} = 35 . 0ºC

. We must find the final pressure

P_{f}

. How can we use the equation

PV = NkT

? At first, it may seem that not enough information is given, because the volume

V

and number of atoms

N

are not specified. What we can do is use the equation twice:

P_{0} V_{0} = {NkT}_{0}

and

P_{f} V_{f} = {NkT}_{f}

. If we divide

P_{f} V_{f}

P_{0} V_{0}

we can come up with an equation that allows us to solve for

P_{f}

\frac{P_{f} V_{f}}{P_{0} V_{0}} = \frac{N_{f} {kT}_{f}}{N_{0} {kT}_{0}}

(8.3.3)

Since the volume is constant,

V_{f}

and

V_{0}

are the same and they cancel out. The same is true for

N_{f}

and

N_{0}

, and

k

, which is a constant. Therefore,

\frac{P_{f}}{P_{0}} = \frac{T_{f}}{T_{0}} .

(8.3.4)

We can then rearrange this to solve for

P_{f}

P_{f} = P_{0} \frac{T_{f}}{T_{0}},

(8.3.5)

where the temperature must be in units of kelvins, because

T_{0}

and

T_{f}

are absolute temperatures.

Solution

1. Convert temperatures from Celsius to Kelvin.

\begin{matrix} T_{0} = (18 . 0 + 273) K = 291 K \\ T_{f} = (35 . 0 + 273) K = 308 K \end{matrix}

(8.3.6)

2. Substitute the known values into the equation.

P_{f} = P_{0} \frac{T_{f}}{T_{0}} = 7 . 00 \times 10^{5} Pa (\frac{308 K}{291 K}) = 7 . 41 \times 10^{5} Pa

(8.3.7)

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 8.3.2

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

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

0 º C

and atmospheric pressure.

Strategy

Because pressure, volume, and temperature are all specified, we can use the ideal gas law

PV = NkT

, to find

N

Solution

1. Identify the knowns.

\begin{array}{l} T & = & 0 º C = 273 K \\ P & = & 1 . 01 \times 10^{5} Pa \\ V & = & 1.00 m^{3} \\ k & = & 1 . 38 \times 10^{- 23} J/K \end{array}

(8.3.8)

2. Identify the unknown: number of molecules,

N

3. Rearrange the ideal gas law to solve for

N

\begin{matrix} PV = NkT \\ N = \frac{PV}{kT} \end{matrix}

(8.3.9)

4. Substitute the known values into the equation and solve for

N

N = \frac{PV}{kT} = \frac{(1 . 01 \times 10^{5} Pa) (1 . {00 m}^{3})}{(1 . 38 \times 10^{- 23} J/K) (273 K)} = 2 . 68 \times 10^{25} molecules

(8.3.10)

Discussion

The calculated number,

2.68 \times 10^{25}

, is certainly very large. You might say that the volume of a cubic meter is also large (

1 m^{3} = 1000 L

), but even in a small volume of

1 {cm}^{3}

, which is about size of a thimble (

1 {cm}^{3} = 10^{−6} m^{3}

), a gas at STP has

2.68 \times 10^{19}

molecules in it (still a very large number). Once again, note that

N

is the same for all types or mixtures of gases.

Section Summary

The ideal gas law relates the pressure and volume of a gas to the number of gas molecules and the temperature of the gas.
The ideal gas law can be written in terms of the number of molecules of gas:
$PV = NkT,$
(8.3.11)
where $P$ is pressure, $V$ is volume, $T$ is temperature, $N$ is number of molecules, and $k$ is the Boltzmann constant
$k = 1 . 38 \times 10^{- 23} J/K .$
(8.3.12)
The ideal gas law is generally valid at temperatures well above the boiling temperature.

Conceptual Questions

Exercise 5

Under what circumstances would you expect a gas to behave significantly differently than predicted by the ideal gas law?

Exercise 6

A constant-volume gas thermometer contains a fixed amount of gas. What property of the gas is measured to indicate its temperature?

Problems & Exercises

Exercise 9

The gauge pressure in your car tires is

2.50 \times 10^{5} {N/m}^{2}

at a temperature of

35.0 º C

when you drive it onto a ferry boat to Alaska. What is their gauge pressure later, when their temperature has dropped to

- 40 . 0 º C

Solution

1.62 atm

Exercise 10

Convert an absolute pressure of

7 . 00 \times 10^{5} {N/m}^{2}

to gauge pressure in

{lb/in}^{2} .

(This value was stated to be just less than

90 . {0 lb/in}^{2}

in Example 8.3.1. Is it?)

Exercise 11

Suppose a gas-filled incandescent light bulb is manufactured so that the gas inside the bulb is at atmospheric pressure when the bulb has a temperature of

20.0 º C

. (a) Find the gauge pressure inside such a bulb when it is hot, assuming its average temperature is

60.0 º C

(an approximation) and neglecting any change in volume due to thermal expansion or gas leaks. (b) The actual final pressure for the light bulb will be less than calculated in part (a) because the glass bulb will expand. What will the actual final pressure be, taking this into account? Is this a negligible difference?

Solution

(a) 0.136 atm

(b) 0.135 atm. The difference between this value and the value from part (a) is negligible.

Exercise 12

Large helium-filled balloons are used to lift scientific equipment to high altitudes. (a) What is the pressure inside such a balloon if it starts out at sea level with a temperature of

10.0 º C

and rises to an altitude where its volume is twenty times the original volume and its temperature is

- 50 . 0 º C

? (b) What is the gauge pressure? (Assume atmospheric pressure is constant.)

Exercise 13

In the text, it was shown that

N / V = 2 . 68 \times 10^{25} m^{- 3}

for gas at STP. (a) Show that this quantity is equivalent to

N / V = 2 . 68 \times 10^{19} {cm}^{- 3},

as stated. (b) About how many atoms are there in one

{μm}^{3}

(a cubic micrometer) at STP? (c) What does your answer to part (b) imply about the separation of atoms and molecules?

Exercise 14

An airplane passenger has

100 {cm}^{3}

of air in his stomach just before the plane takes off from a sea-level airport. What volume will the air have at cruising altitude if cabin pressure drops to

7.50 \times 10^{4} {N/m}^{2} ?

Exercise 15

An expensive vacuum system can achieve a pressure as low as

1.00 \times 10^{- 7} {N/m}^{2}

20 º C

. How many atoms are there in a cubic centimeter at this pressure and temperature?

Exercise 16

The number density of gas atoms at a certain location in the space above our planet is about

1 . 00 \times 10^{11} m^{- 3},

and the pressure is

2.75 \times 10^{- 10} {N/m}^{2}

in this space. What is the temperature there?

Solution

- 73 . 9 º C

Exercise 17

A bicycle tire has a pressure of

7.00 \times 10^{5} {N/m}^{2}

at a temperature of

18.0 º C

and contains 2.00 L of gas. What will its pressure be if you let out an amount of air that has a volume of

100 {cm}^{3}

at atmospheric pressure? Assume tire temperature and volume remain constant.

Exercise 18

A high-pressure gas cylinder contains 50.0 L of toxic gas at a pressure of

1.40 \times 10^{7} {N/m}^{2}

and a temperature of

25.0 º C

. Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperature

(- 78.5 º C)

to reduce the leak rate and pressure so that it can be safely repaired. (a) What is the final pressure in the tank, assuming a negligible amount of gas leaks while being cooled and that there is no phase change? (b) What is the final pressure if one-tenth of the gas escapes? (c) To what temperature must the tank be cooled to reduce the pressure to 1.00 atm (assuming the gas does not change phase and that there is no leakage during cooling)? (d) Does cooling the tank appear to be a practical solution?

Solution

(a)

9 . 14 \times 10^{6} {N/m}^{2}

(b)

8 . 23 \times 10^{6} {N/m}^{2}

(d) No. The final temperature needed is much too low to be easily achieved for a large object.

Exercise 19

(a) What is the gauge pressure in a

25.0 º C

car tire containing 3.60 mol of gas in a 30.0 L volume? (b) What will its gauge pressure be if you add 1.00 L of gas originally at atmospheric pressure and

25.0 º C

? Assume the temperature returns to

25.0 º C

and the volume remains constant.

ideal gas law

the physical law that relates the pressure and volume of a gas to the number of gas molecules or number of moles of gas and the temperature of the gas

Boltzmann constant

k

, a physical constant that relates energy to temperature;

k = 1.38 \times 10^{–23} J/K