Pressure of an Ideal Gas: Ideal Gas Law, Sample Questions

An ideal gas is theoretically a gas in which the particles move in random directions and there is no interparticle interaction between them. Ideal gases are of three types. The law for the ideal gas is known as the Ideal Gas Law and is a combination of various other laws. In this article, we will learn more about the ideal gas law and calculating the pressure of an ideal gas.

Keyterms: Gas, Ideal gas, Gas Law, Random motion, capacity, Maxwell-Boltzmann ideal gas, Quantum, Pressure

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What is an Ideal Gas?

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A gas whose particles have a random motion with no interparticle interaction is called an Ideal Gas. In simple terms, all the gases are primarily ideal gas if they are at standard pressure and temperature conditions where their gas molecules play the role of ideal particles. One mole of an ideal gas is said to have a capacity of 22.71 liters at standard temperature and pressure, as defined by IUPAC since 1982.

Ideal gases can also be divided into three basic types, namely:

• The Classical or Maxwell-Boltzmann ideal gas
• The ideal quantum Bose gas, composed of Bosons
• The ideal quantum Fermi gas comprises fermions.

Ideal Gas

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Ideal Gas Law

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The law given for Ideal Gas is called the Ideal Gas equation. This law is a combination of Boyle's law, Charle's law, Avogadro's law, and Gay-Lussac's law. 

The equation that describes the law of the ideal gas is:

PV= nRT

Where,

P = Pressure of the gas, which is measured in Pascal

V = volume of the gas, which is measured in Litres

n = number of moles of gas, which is measured in moles.

R = universal or ideal gas constant, which is 8.314JK^-1mol^-2

T = absolute temperature of the gas, which is measured in Kelvin

As mentioned earlier, the ideal gas law is a combination of various other laws like Boyle's Law, Charles's Law, Avogadro's Law.

These laws are individually written as:

• Boyle's law 

PV = k

• Charle's law 

V = kT

• Avogadro's law 

V = kn

Now, when these three Equations are combined, we get the equation for the ideal gas:

 V = RTn/P

=> PV = nRT 

Ideal Gas Law

There are various assumptions that were made while laying down the law for an ideal gas:

• The gas molecules are assumed to be so small that they cannot be distinguished from each other and are in the shape of hard spheres.
• All the laws of Newton are applicable to gas.
• The molecules keep moving in tandem and at different speeds but with constant movement.
• The average distance between the molecules is larger than the size of the molecule.
• There is no form of attraction or repulsion between the molecules.

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Calculating the pressure of an Ideal Gas

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For the calculation of the pressure of an ideal gas, let us consider an ideal gas filled in a cubical container. Here, one corner of the container is taken as the origin. The edges are considered as x, y, and z axes. Let A1 and A2 be the parallel faces of the cuboid which are perpendicular to the x-axis. Suppose, a molecule is moving with velocity 'v' in the container and the components of velocity along three axes are vx, vy, and vz. The collisions amongst the particles are assumed to be elastic so when this molecule collides with face A1 x component of velocity reverses while y and z components remain unchanged.

Calculating the pressure of an Ideal Gas

Change in the momentum of the molecule is given as

ΔP = -mvx -mvx = -2mvx. ………. (1)

The change in momentum of the wall is 2mvx as the momentum remains conserved.

After the collision, the molecule travels towards the face A2 along with the x component whose velocity is equal to −vx. 

Now, the distance traveled by a molecule from A1 to A2 = L

Therefore, time = L/vx

After a collision with A2, it again travels to A1. Hence, the time between two collisions= 2L/vx

So the number of collisions of molecule per unit time n = vx /2L……….(2)

From (1) and (2),

we get the Momentum imparted to the molecule by the wall per unit of time

ΔF=nΔP 

= m/L×vx2 

Therefore, the total force on wall A1 because of all the molecules is

F = Σm/L×vx²

F = m/LΣvx²

Σvx²=Σvy²=Σvz² (symmetry)

= 1/3Σv² 

Therefore, F=\(\frac{1}{3} * \frac{m}{L^3} *\)ΣV²

Now, the pressure becomes the force per unit area, hence

P=F/L²   

P=\(\frac{1}{3} * \frac{m}{L^3} * \)Σ\(\frac{V^2}{N}\)

P=\(\frac{1}{3}\)ρ∑\(\frac{V^2}{N}\)

Here, M=total mass of the gas

And ρ=density of the gas

Σv²/N is written as v². It is called mean square speed.

P= 1/3ρv²

So, this is what the pressure exerted by an ideal gas is.

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Things to Remember

• A gas whose particles have a random motion with no interparticle interaction is called an Ideal Gas.
• The Ideal Gas equation is a combination of Boyle's law, Charle's law, Avogadro's law, and Gay-Lussac's law.

PV= nRT

• Therefore, the total force on wall A1 because of all the molecules is

F = Σm/L×vx²

• The pressure exerted by an ideal gas: P= 1/3ρv²