Lesson Explainer: Bulk Properties of an Ideal Gas Physics

In this explainer, we will learn how to calculate the relationship between the changes in the pressure, volume, and temperature of an ideal gas.

A gas is a collection of particles that have more space between them compared to other states of matter. When describing a gas, it is very difficult to account for every single molecule at once, as each one is moving in its own direction with its own speed.

Rather than futilely attempting to document every molecule, gases are instead assigned bulk properties. This means that when describing a gas, its entirety is looked at as a single bulk object, rather than what it really is, a collection of tiny objects.

The collection on the left is how we will be viewing the properties of a gas, rather than as individuals on the right. Bulk properties are the average properties of all the molecules in a collection. The three properties we will look at in this explainer are pressure, volume, and temperature.

Pressure is the force of the gas over the area of the container it is being held in. That is, pressureforcearea=.

In the case of a gas, the force is being exerted by the molecules colliding with the walls of the container, and the area is the area being measured. It is possible to have an uncontained gas, such as clouds, but these shapes are highly irregular and are difficult to measure.

For gases, pressure is usually measured in pascals (Pa), which are newtons per square metre (N/m2).

Volume is the total volume the gas takes up. Since gases expand to fill the container they are in, this is often simply the volume of the container. For gases, the unit is typically the cubic metre (m3).

Temperature is the average temperature of the gas. It must always be measured in kelvins for gases. Recall that kelvins can be obtained from degrees Celsius by adding 273.15: 𝑇=𝑇+273.15.KC

Whenever used in an equation, again, temperature must be in kelvins, or else the equations will not work.

These properties together make up the ideal gas equation, which relates these properties together.

Equation: Ideal Gas Equation

For an ideal gas, 𝑃𝑉=π‘˜π‘‡, where 𝑃 is pressure, 𝑉 is volume, 𝑇 is temperature in kelvins, and π‘˜ is the constant of proportionality.

The proportionality constant π‘˜ is dependent on the number of molecules and the type of gas. It remains constant so long as the mass of the gas does not change, that is, by adding or removing gas to an area.

It must be noted that this equation only works when used to describe an ideal gas. An ideal gas is one that has molecules of a negligible size that do not interact with each other. Normally, the molecules in a gas bump into each other, but it is easier mathematically to assume they do not, and the accuracy of the results are still useful.

Let’s look at an example.

Example 1: Bulk Property Relations of an Ideal Gas

Which of the following formulas most correctly represents the relation between the pressure, volume, and absolute temperature of an ideal gas?

  1. π‘‰π‘‡βˆπ‘ƒ
  2. π‘ƒπ‘‰βˆπ‘‡
  3. π‘ƒπ‘‰βˆπ‘‡
  4. π‘‰π‘ƒβˆπ‘‡
  5. π‘ƒπ‘‡βˆπ‘‰


The symbol ∝ is a shorthand way of saying β€œproportional to”. It is used to show that as one quantity increases, so does another.

Looking at the ideal gas equation, we see that 𝑃𝑉 is on one side, and π‘˜π‘‡ is on the other: 𝑃𝑉=π‘˜π‘‡.

So, we would usually say that 𝑃𝑉 is proportional to π‘˜π‘‡, as when one side increases, so too does the other to remain balanced. However, since π‘˜ remains the same unless the number of molecules in the gas change, it can be assumed to be a constant.

When looking at proportionality, constants are typically ignored, since we only care about how the variables change; we know the constant will not change. So, we could write out the ideal gas equation as a proportionality by ignoring π‘˜, like so: π‘ƒπ‘‰βˆπ‘‡.

Therefore, the correct answer is C, π‘ƒπ‘‰βˆπ‘‡.

Showing how this proportionality looks with some numbers would be helpful to see how they change.

Let’s look at an example.

Example 2: Proportional Changes in Bulk Properties

A volume of 40 m3 of an ideal gas is heated so that its temperature rises from 23∘C to 60∘C. By that time, its volume becomes 34 m3; then .

  1. its density has decreased
  2. its pressure has not changed
  3. its pressure has increased
  4. its density has not changed


Let’s look again at the ideal gas equation: 𝑃𝑉=π‘˜π‘‡.

The quantities that we are told have changed are volume, 𝑉, and temperature, 𝑇. Volume decreased from 40 m3 to 34 m3, and temperature increased from 23∘C to 60∘C. In the equation, the quantity of 𝑃 would be smaller, and that of 𝑇 would be larger: 𝑃𝑉=π‘˜π‘‡.

The constant, π‘˜, being a constant, is unaffected by any changes.

Normally, as 𝑇 increases, we would expect 𝑃𝑉 to increase as well. In this case, however, 𝑉 decreased. To compensate, the pressure, 𝑃, must increase to account not just for the temperature increase, but the volume decrease.

Therefore, the pressure, 𝑃, has increased. The correct answer is C.

Using the ideal gas equation, again, requires us to assume that the gas we are working with is ideal. None of the molecules interact with each other, and they are of a negligible size. Most problems (and real-life calculations) assume an ideal gas, unless they say explicitly otherwise.

Let’s look at the ideal gas equation using some changing numbers and how to solve it by looking at a compressed can of air.

This can of air in a cold room starts at a volume of 2 m3 and is compressed to 1 m3. During this compression, its temperature decreases from 2∘C to βˆ’60∘C. If the pressure started at 110 kPa, what is the pressure after the compression?

The ideal gas equation is 𝑃𝑉=π‘˜π‘‡.

Let’s start by making two versions of this equation: one version before the compression of the can, denoted by subscripts of 1, and the other version after the compression of the can, denoted by subscripts of 2: 𝑃𝑉=π‘˜π‘‡,𝑃𝑉=π‘˜π‘‡.

Note that the number of molecules of gas here stays the same, so π‘˜ is the same in both equations. This actually means that these equations can be related together. To see what we mean, let’s divide both equations by their temperatures to isolate π‘˜.

For the first equation, 𝑃𝑉𝑇=π‘˜π‘‡π‘‡.

π‘‡οŠ§ cancels to become 𝑃𝑉𝑇=π‘˜.

For the second equation, it is just the same: 𝑃𝑉𝑇=π‘˜π‘‡π‘‡π‘ƒπ‘‰π‘‡=π‘˜.

Since both of these equations are equal to π‘˜, we can then relate them to each other: π‘˜=π‘˜,𝑃𝑉𝑇=𝑃𝑉𝑇.

We wish to find the pressure of the can after it has been compressed, which is π‘ƒοŠ¨. To isolate this value, we can use the equation above. Let’s start by multiplying both sides by π‘‡οŠ¨: 𝑃𝑉𝑇×𝑇=𝑃𝑉𝑇×𝑇.

This cancels the π‘‡οŠ¨ on the right, making the equation 𝑃𝑉𝑇𝑇=𝑃𝑉.

Dividing both sides by π‘‰οŠ¨ to get rid of it on the right side gives 𝑃𝑉𝑇𝑇𝑉=𝑃𝑉𝑉.

Canceling gives us 𝑃𝑉𝑇𝑇𝑉=𝑃.

All that is left is to put in the values we know. π‘ƒοŠ§ is 110 kPa, π‘‰οŠ§ is 2 m3, π‘‡οŠ¨ is βˆ’6∘C, π‘‡οŠ§ is 2∘C, and π‘‰οŠ¨ is 1 m3.

This gives us (110)ο€Ή2(2)(βˆ’6)(1)=𝑃.kPamCCm∘∘

The units of cubic metres and degrees Celsius cancel, leaving behind only kilopascals. Crunching the numbers and rounding to the nearest kPa gives 𝑃=βˆ’73.kPa

This is not right though! Compressing a gas is supposed to increase pressure, not make it negative. What went wrong is that we calculated the temperature using degrees Celsius, instead of using kelvins. The ideal gas equation and its other forms require kelvins to be used exclusively.

To get the correct answer, we need to convert to kelvins. π‘‡οŠ§ is 2∘C, and π‘‡οŠ¨ is βˆ’6∘C. Adding 273.15 to each of these gives the values we can use in the ideal gas equation: 𝑇=2+273.15,𝑇=βˆ’6+273.15.KK

So, π‘‡οŠ§ is 275.15 K and π‘‡οŠ¨ is 267.15 K. Putting these values back into the equation gives us (110)ο€Ή2(275.15)(267.15)(1)=𝑃.kPamKKm

Crunching this now correct set of numbers and rounding gives 𝑃=227.kPa

Let’s look at some more examples.

Example 3: Cooling and Compression of a Cloud

A cloud has a volume of 3β€Žβ€‰β€Ž560 m3 when in air with a pressure of 106 kPa and a temperature of 290 K. The air temperature then drops to 275 K and the air pressure drops to 101 kPa, causing the cloud to compress, as shown in the diagram. What is the new volume of the cloud? Answer to the nearest cubic metre.


The mass of the gas remains the same, so π‘˜ is the same. Let’s recall the way we can relate the state of the cloud before and after compression using this, like so 𝑃𝑉𝑇=𝑃𝑉𝑇.

We know all of these values except for π‘‰οŠ¨, the volume of the cloud after compression. We need to find π‘‰οŠ¨, so let’s isolate it. Let’s start by multiplying both sides by π‘‡οŠ¨: 𝑃𝑉𝑇×𝑇=𝑃𝑉𝑇×𝑇, which cancels to 𝑃𝑉𝑇𝑇=𝑃𝑉.

We can then divide both sides by π‘ƒοŠ¨: 𝑃𝑉𝑇𝑇𝑃=𝑃𝑉𝑃.

Canceling gives us 𝑃𝑉𝑇𝑇𝑃=𝑉.

π‘ƒοŠ§ is 106 kPa, π‘‰οŠ§ is 3β€Žβ€‰β€Ž560 m3, and π‘‡οŠ§ is in kelvins, so there is no need to convert the value of 290 K. Similarly, π‘‡οŠ¨ is 275 K and π‘ƒοŠ¨ is 101 kPa. Putting all these values into the equation above gives (106)ο€Ή3560(275)(290)(101)=𝑉.kPamKKkPa

The units of kilopascals and kelvins cancel, leaving behind the units of volume, cubic metres. Multiplying and dividing through the whole equation gives 𝑉=3543.m

So, the total volume of the cloud has gone from 3β€Žβ€‰β€Ž560 m3 to 3β€Žβ€‰β€Ž543 m3.

Example 4: Balloon Air Transfer

A balloon holds 0.012 m3 of air at a pressure of 101 kPa and a temperature of 300 K. The air from this balloon is transferred to another balloon that is half the volume of the first one. 125 kPa of external pressure is required to transfer the air. What is the air temperature in the new balloon? Give your answer to the nearest kelvin.


The balloons contain the exact same air, which was just transferred, so π‘˜ is the same. This means we can use the equation that relates the old balloon (subscript 1) and the new balloon (subscript 2): 𝑃𝑉𝑇=𝑃𝑉𝑇.

Let’s first ensure we have all the values we need. The new balloon that the air is transferred to is half the volume of the first balloon. The volume of the first balloon, π‘‰οŠ§, is 0.012 m3, so half of this would be π‘‰οŠ¨: 0.0122=0.006.mm

So, π‘‰οŠ¨ is 0.0006 m3.

The question also mentions that 125 kPa is required to transfer the air, but this is not π‘ƒοŠ¨. It means that 125 kPa are needed to put all the air into this new balloon, so it is in addition to the old pressure.π‘ƒοŠ¨ is thus the two added together: 101+125=226.kPakPakPa

Thus, 226 kPa is the value of π‘ƒοŠ¨.

We are looking for the final temperature in the new balloon, π‘‡οŠ¨, so we should start by putting it on top. Let’s invert both sides of the equivalence equation: 𝑃𝑉𝑇=𝑃𝑉𝑇,𝑇𝑃𝑉=𝑇𝑃𝑉.

To get π‘‡οŠ¨ on one side, we should multiply both sides by (𝑃𝑉): 𝑇𝑃𝑉×(𝑃𝑉)=𝑇𝑃𝑉×(𝑃𝑉).

This cancels the (𝑃𝑉) on the right side, giving 𝑇𝑃𝑉𝑃𝑉=𝑇.

π‘‡οŠ§ is 300 K, π‘ƒοŠ¨ is 226 kPa, π‘‰οŠ¨ is 0.006 m3, π‘ƒοŠ§ is 101 kPa, and π‘‰οŠ§ is 0.012 m3. Putting all these values in the equation gives (300)(226)ο€Ή0.006(101)(0.012)=𝑇.KkPamkPam

The units of cubic metres and kilopascals cancel, leaving behind temperature in kelvins. Crunching the numbers yields 𝑇=336.K

So, the final temperature of the new balloon is 336 kelvins.

Let’s summarize what we’ve learned in this explainer.

Key Points

  • An ideal gas is one in which the molecules it consists of are of negligible size and do not interact with each other.
  • The bulk properties of a gas are pressure, volume, and temperature (which must be in kelvins).
  • The ideal gas law equation is 𝑃𝑉=π‘˜π‘‡, where 𝑃 is pressure, 𝑉 is volume, π‘˜ is the constant of proportionality, and 𝑇 is the temperature in kelvins.
  • When the mass of a gas does not change, π‘˜ is constant, and thus 𝑃𝑉𝑇=𝑃𝑉𝑇, where π‘ƒοŠ§, π‘‰οŠ§, and π‘‡οŠ§ are the pressure, volume, and temperature before a change and π‘ƒοŠ¨, π‘‰οŠ¨, and π‘‡οŠ¨ are the pressure, volume, and temperature after a change.

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