Lesson Explainer: The Ideal Gas Law in terms of Number of Moles Physics

In this explainer, we will learn how to calculate the relationship between the number of moles in an ideal gas and the values of its bulk properties.

The bulk properties of an ideal gas are

  • the volume occupied by the gas, 𝑉,
  • the temperature that the gas is at, 𝑇,
  • the pressure exerted by the gas, 𝑃.

It is important to understand that, in this sense, β€œa gas” does not refer to a substance (such as oxygen) but to a specific collection of particles of some substance (such as the oxygen molecules within a specific container).

The bulk properties of an ideal gas are related by the expression π‘ƒπ‘‰βˆπ‘‡.

This can be written as 𝑃𝑉=π‘˜π‘‡, where π‘˜ is a constant. The value of π‘˜ depends on the number of particles in the gas.

Let us consider a container of a constant volume containing a gas at a constant temperature. All the particles of the gas are identical.

As the temperature of the gas is constant, the average force on the container due to a gas particle colliding with a surface of the container is constant for any number of gas particles.

The constant temperature of the gas also means that the average speed at which gas particles travel between opposite surfaces of the container is constant for any number of gas particles.

As the volume of the container is also constant, the average time between the collisions of a gas particle and a surface of the container must be constant for any number of particles.

We can then see that if the number of particles is greater, the number of the collisions between the particles and a surface of the container in a given time is also greater, and therefore the pressure exerted by the gas on a surface of the container is greater if the number of particles in the gas is greater.

Instead of directly determining the number of particles in some object, it is often convenient to determine the number of moles of some substance that the object consists of. The number of moles of a gas is often used in place of the number of particles that the gas consists of.

The bulk properties of a given number of moles of an ideal gas can be related to each other by the molar form of the ideal gas law.

Formula: The Molar Form of the Ideal Gas Law

The molar form of the ideal gas law relates the pressure, 𝑃, volume, 𝑉, and temperature, 𝑇, of an ideal gas to the number of moles of the gas, 𝑛, by the equation 𝑃𝑉=𝑛𝑅𝑇, where 𝑅 is the molar gas constant, which has an approximate value of 8.31 J/Kβ‹…mol.

The unit J/Kβ‹…mol is written as m2β‹…kg/s2β‹…Kβ‹…mol in SI base units.

Let us look at an example of using the molar form of the ideal gas law.

Example 1: Determining the Pressure of an Ideal Gas Using the Molar Form of the Ideal Gas Law

A container of volume 0.225 m3 holds 2.24 moles of oxygen gas at a temperature of 320 K. Find the pressure on the container’s interior surfaces. Use a value of 8.31 m2β‹…kg/s2β‹…Kβ‹…mol for the value of the molar gas constant. Give your answer in kilopascals to one decimal place.

Answer

The molar form of the ideal gas law can be expressed by the equation 𝑃𝑉=𝑛𝑅𝑇, where 𝑃 is pressure, 𝑉 is volume, 𝑇 is temperature, 𝑛 is the number of moles, and 𝑅 is the molar gas constant.

The question asks us to find the pressure of the gas, so we must make 𝑃 the subject of the equation. We can do this by dividing both sides of the equation by 𝑉, as follows: 𝑃𝑉𝑉=𝑛𝑅𝑇𝑉𝑃𝑉𝑉=𝑃𝑃=𝑛𝑅𝑇𝑉.

Recalling that m2β‹…kg/s2β‹…Kβ‹…mol can be expressed as J/Kβ‹…mol, we can now substitute the known values of the quantities to obtain 𝑃=2.24Γ—8.31/β‹…Γ—3200.225.molJKmolKm

To one decimal place, 𝑃=26473.8.Pa

The question asks for the pressure in units of kPa, however. To convert the answer to the correct value, we must then convert the value in Pa to a value in kPa, as follows: 26473.8=26473.8100026473.8=26.4738.PakPaPakPa

To one decimal place, 𝑃=26.5.kPa

It is useful to note that the value of the mass of a particle of a gas is not required in order to determine the bulk properties of the gas using the molar form of the ideal gas law.

Let us now look at another example involving the molar form of the ideal gas law.

Example 2: Determining the Volume of an Ideal Gas Using the Molar Form of the Ideal Gas Law

A cloud of gas has a pressure of 220 kPa and a temperature of 440 K. The gas contains 8.2 moles of a particle with a molar mass of 10.5 g/mol. Find the volume of the cloud. Use 8.31 m2β‹…kg/s2β‹…Kβ‹…mol for the value of the molar gas constant. Give your answer to two decimal places.

Answer

The molar form of the ideal gas law can be expressed by the equation 𝑃𝑉=𝑛𝑅𝑇, where 𝑃 is pressure, 𝑉 is volume, 𝑇 is temperature, 𝑛 is the number of moles, and 𝑅 is the molar gas constant.

The question asks us to find the volume of the gas, so we must make 𝑉 the subject of the equation. We can do this by dividing both sides of the equation by 𝑃, as follows: 𝑃𝑉𝑃=𝑛𝑅𝑇𝑃𝑃𝑉𝑃=𝑉𝑉=𝑛𝑅𝑇𝑃.

We can now substitute the known values of the quantities.

To determine a volume in SI base units, m3, we must use SI base units for all the quantities in the equation. We must then convert 220 kPa to a value in Pa: 220=(220Γ—1000)220=2.2Γ—10.kPaPakPaPa

Substituting this and the other values into the molar form of the ideal gas law, and recalling that m2β‹…kg/s2β‹…Kβ‹…mol can be expressed as J/Kβ‹…mol, we obtain 𝑉=8.2Γ—8.31/β‹…Γ—4402.2Γ—10.molJKmolKPa

To two decimal places, 𝑉=0.14.m

Note that the molar mass of the particles of the gas was not required to determine 𝑉.

Let us now look at another such example.

Example 3: Determining the Temperature of an Ideal Gas Using the Molar Form of the Ideal Gas Law

A gas consisting of 25.6 moles of carbon fills a volume of 0.128 m3 and has a pressure of 135 kPa. Find the temperature of the gas. Use a value of 12.0107 g/mol for the molar mass of carbon and 8.31 m2β‹…kg/s2β‹…Kβ‹…mol for the value of the molar gas constant. Give your answer to the nearest kelvin.

Answer

The molar form of the ideal gas law can be expressed by the equation 𝑃𝑉=𝑛𝑅𝑇, where 𝑃 is pressure, 𝑉 is volume, 𝑇 is temperature, 𝑛 is the number of moles, and 𝑅 is the molar gas constant.

The question asks us to find the temperature of the gas, so we must make 𝑇 the subject of the equation. We can do this by dividing both sides of the equation by 𝑛𝑅, as follows: 𝑃𝑉𝑛𝑅=𝑛𝑅𝑇𝑛𝑅𝑛𝑅𝑇𝑛𝑅=𝑇𝑇=𝑃𝑉𝑛𝑅.

We can now substitute the known values of the quantities.

To determine a temperature in SI base units, K, we must use SI base units for all the quantities in the equation. We must then convert 135 kPa to a value in Pa: 135=(135Γ—1000)135=1.35Γ—10.kPaPakPaPa

Substituting this and the other values into the molar form of the ideal gas law, and recalling that m2β‹…kg/s2β‹…Kβ‹…mol can be expressed as J/Kβ‹…mol, we obtain 𝑇=1.35Γ—10Γ—0.12825.6Γ—8.31/β‹….PammolJKmol

To the nearest kelvin, 𝑇=81.K

Note that the molar mass of carbon was not required to determine 𝑇.

Let us now look at another such example.

Example 4: Determining the Number of Moles of an Ideal Gas Using the Molar Form of the Ideal Gas Law

A gas cylinder with a volume of 0.245 m3 contains a gas at a temperature of 350 K and a pressure of 120 kPa. Find the number of moles of the gas particles in the cylinder. Use 8.31 m2β‹…kg/s2β‹…Kβ‹…mol for the value of the molar gas constant. Give your answer to one decimal place.

Answer

The molar form of the ideal gas law can be expressed by the equation 𝑃𝑉=𝑛𝑅𝑇, where 𝑃 is pressure, 𝑉 is volume, 𝑇 is temperature, 𝑛 is the number of moles, and 𝑅 is the molar gas constant.

The question asks us to find the number of moles of the gas, so we must make 𝑛 the subject of the equation. We can do this by dividing both sides of the equation by 𝑅𝑇, as follows: 𝑃𝑉𝑅𝑇=𝑛𝑅𝑇𝑅𝑇𝑛𝑅𝑇𝑅𝑇=𝑛𝑛=𝑃𝑉𝑅𝑇.

We can now substitute the known values of the quantities.

To determine a number of moles in SI base units, mol, we must use SI base units for all the quantities in the equation. We must then convert 220 kPa to a value in Pa: 120=(120Γ—1000)120=1.2Γ—10.kPaPakPaPa

Substituting this and the other values into the molar form of the ideal gas law, and recalling that m2β‹…kg/s2β‹…Kβ‹…mol can be expressed as J/Kβ‹…mol, we obtain 𝑛=1.2Γ—10Γ—0.2458.31/β‹…Γ—350.PamJKmolK

To one decimal place, 𝑛=10.1.mol

Let us now look at an example in which the number of moles of a gas is not constant.

Example 5: Determining the Percent Change in the Number of Moles of an Ideal Gas Using the Molar Form of the Ideal Gas Law

A gas cylinder with a movable lid has an initial volume of 0.125 m3 and contains a gas at a temperature of 360 K and a pressure of 1.5000Γ—10 Pa. The lid of the container is not perfectly sealed; hence, gas can escape from the container when the lid is moved. The lid of the container is pushed downward, reducing the volume of the gas to 0.105 m3. The pressure of the gas after the lid is pushed down is 1.5496Γ—10 Pa and the temperature of the gas is 355 K. Find the percent of the moles of gas that escape the cylinder due to the lid being moved. Use 8.31 m2β‹…kg/s2β‹…Kβ‹…mol for the value of the molar gas constant. Give your answer to the nearest percent.

Answer

The molar form of the ideal gas law can be expressed by the equation 𝑃𝑉=𝑛𝑅𝑇, where 𝑃 is pressure, 𝑉 is volume, 𝑇 is temperature, 𝑛 is the number of moles, and 𝑅 is the molar gas constant.

In this question, the values of all the variables in the equation other than 𝑅 change, meaning that we have two equations, which we can write as 𝑃𝑉=π‘›π‘…π‘‡οŠ§οŠ§οŠ§οŠ§ and 𝑃𝑉=𝑛𝑅𝑇.

These two equations represent the gas in the cylinder before it is compressed and after a part of it exits the cylinder.

From there, we can separate the molar gas constant 𝑅 in each of the equations to have 𝑃𝑉𝑛𝑇=π‘…οŠ§οŠ§οŠ§οŠ§ and 𝑃𝑉𝑛𝑇=𝑅.

We can then see that 𝑅=𝑃𝑉𝑛𝑇=𝑃𝑉𝑛𝑇.

We can compare the number of moles of the gas before and after the change by finding the ratio of the number of moles of gas in the cylinder after the change, π‘›οŠ¨, to the number of moles of gas in the cylinder before the change, π‘›οŠ§, as follows: 𝑃𝑉𝑛𝑇=𝑃𝑉𝑛𝑇.

We then multiply both sides of the equation by π‘›οŠ¨: 𝑛𝑃𝑉𝑛𝑇=𝑛𝑃𝑉𝑛𝑇, and simplify the equation by canceling on the right-hand side: 𝑛𝑃𝑉𝑛𝑇=𝑃𝑉𝑇.

Placing the ratio of π‘›οŠ¨ to π‘›οŠ§ in brackets, we get 𝑛𝑛𝑃𝑉𝑇=𝑃𝑉𝑇.

Multiplying both sides of the equation by π‘‡οŠ§, we have 𝑛𝑛𝑃𝑉𝑇𝑇=𝑃𝑉𝑇𝑇.

Simplifying the equation by canceling on the left-hand side, 𝑛𝑛𝑃𝑉=𝑃𝑉𝑇𝑇.

Dividing both sides of the equation by π‘ƒπ‘‰οŠ§οŠ§, 𝑛𝑛𝑃𝑉𝑃𝑉=𝑃𝑉𝑇𝑇𝑃𝑉.

Simplifying the equation by canceling on the right-hand side, 𝑛𝑛=𝑃𝑉𝑇𝑇𝑃𝑉.

The known values can now be substituted into the rearranged equation, as follows: 𝑛𝑛=1.5496Γ—10Γ—0.105Γ—360355Γ—1.5000Γ—10Γ—0.125𝑛𝑛=0.879998.PamKKPam

To express this ratio as a percent, it must be multiplied by 100%, giving 87.9998%. Initially, the gas in the cylinder was, by definition, 100% of the gas, so the percent loss of gas is given by 100%βˆ’87.9998%.

To the nearest percent, this is 12%, which is the percent of the moles of the gas that escapes the cylinder when the lid is moved.

Let us now summarize what has been learned in this explainer.

Key Points

  • The molar form of the ideal gas law is expressed by the equation 𝑃𝑉=𝑛𝑅𝑇, where 𝑃 is pressure, 𝑉 is volume, 𝑇 is temperature, 𝑛 is the number of moles, and 𝑅 is the molar gas constant.
  • The molar gas constant has an approximate value of 8.31 J/Kβ‹…mol.
  • It is not necessary to know the mass of a particle of a gas to determine the bulk properties of an ideal gas when using the molar form of the ideal gas law.

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