Derivation of Ideal Gas Equation

ideal gas

Let this cube resemble an ideal gas, and one red molecule represents one of the monoatomic gas particles in the gas moving in constant random motion.

To simplify the problem, let’s consider one molecule of a monoatomic gas in a 1D plane,

2d

Let d be the length of one side of the cube.

As such,1.png

where v = the velocity of a particle moving in constant random motion,

and t = the time taken for a particle to travel along one side of the cube and back to its original position.

Let F be the force by the walls of the box on the particle that causes the particle to change direction.  By Newton’s third law, this is equal to the force of the particle on the wall.

By applying Newton’s second law of motion,

2.png

The pressure caused by the gas moving in a straight line and striking the wall perpendicularly is:

3

However, this resulting equation can only be applied in a 1D plane.

To apply this equation to 3D,

Since the velocity squared of the particle has an equal chance of being directed in any of the three dimensions, and it has an equal chance of hitting any of the three walls and causing pressure in any of the three dimensions, we multiply the equation by 1/3.

2.png

Screen Shot 2017-03-08 at 21.50.28

The pressure caused by N molecules of monoatomic gas can be given by multiplying the equation by N:

7Based on statistics (Maxwell-Boltzmann distribution curve),

9.jpg
Taken from: http://revisionworld.com/a2-level-level-revision/chemistry/energy-changes/maxwell-boltzmann-distribution

5

Assuming that all the particles act independently of each other, as no intermolecular forces are present, we can multiply the average kinetic energy, which is proportional to T, by N to find total kinetic energy.

Hence, 6.png

Finally, by putting the two equations together, we obtain the ideal gas equation:

8

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