Pressure of an Ideal Gas

Before we learn how to calculate the pressure of an ideal gas let us first know what exactly an ideal gas is. An ideal gas in simple words is a theoretical gas in which the gas particles move randomly and there is no interparticle interaction. An ideal gas doesn't exist in reality. It follows the ideal gas equation which is a simplified equation we will learn further and is susceptible to analysis under statistical mechanics. At standard pressure and temperature conditions, most gases are taken to behave as an ideal gas. As defined by IUPAC, 1 mole of an ideal gas has a capacity of 22.71 litres at standard temperature and pressure.

Failure of Ideal Gas Model

At lower temperatures and high pressure, when intermolecular forces and molecular size become important the ideal gas model tends to fail. For most of heavy gases such as refrigerants and gases with strong intermolecular forces, this model tends to fail. At high pressures, the volume of a real gas is often considerably larger than that of an ideal gas and at low temperatures, the pressure of a real gas is often considerably less than that of an ideal gas. At some point in low temperature and high-pressure real gases undergo phase transition which is not allowed in the ideal gas model. The deviation from the ideal gas model can be explained by a dimensionless quantity, called the compressibility factor (Z).

Ideal Gas Equation

Ideal gas law gives an equation known as the ideal gas equation which is followed by an ideal gas. It is a combination of the empirical Boyle's law, Charles's law, Avogadro's law, and Gay-Lussac's law. The ideal gas equation in empirical form is given as

PV=nRT

where P= pressure of the gas (pascal)

V= volume of gas (liters)

n= number of moles of gas (moles)

R= universal or ideal gas constant (\[=8.314JK^{-1}mol^{-1}\])

T= absolute temperature of the gas (Kelvin)

Ideal gas law is an extension of experimentally discovered gas laws. It is derived from Boyle's law, Charles law, Avogadro's law. When these three are combined, we get an ideal gas law.

Boyle's law =>  PV = k

Charle's law => V = kT

Avogadro's law => V = kn

Now, when we combine these three laws we use the proportionality constant 'R', which is the universal gas constant and we get the ideal gas equation as

V = RTn/P

=> PV = nRT 

Ideal Gas Model Assumptions

Various assumptions are made in the ideal gas model. They are as follows:

• Gas molecules are considered as indistinguishably very small and hard spheres.

• All motions are frictionless and the collisions are elastic, that is there is no energy loss in motion or collisions.

• All laws of Newton are applicable.

• The size of the molecules is much smaller than the average distance between them.

• There is a constant movement of molecules in random directions with distributed speeds.

• Molecules don't attract or repel each other apart from point-like collisions with the walls.

• No long-range forces exist between molecules of the gas and surroundings.

The Pressure of an Ideal Gas: Calculation

For the calculation let us consider an ideal gas filled in a container cubical in shape. One corner of the container is taken as the origin and the edges as x, y, and z axes. Let \[A_{1} and A_{2}\] 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 \[V_{x}, V_{y} and V_{z}\]. As we assume collisions to be elastic when this molecule collides with face \[A_{1}\] x component of velocity reverses while the y and z component remains unchanged.

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Change in the momentum of the molecule is 

\[\Delta P= -mv_{x} -mv_{x}  = -2mv_{x} ..... (1)\]

The change in momentum of the wall is \[2mv_{x}\] as the momentum remains conserved.

After the collision, the molecule travels towards the face \[A_{2}\] with the x component of the velocity equal to \[-v_{x}\] 

Now, the distance traveled by a molecule from \[A_{1}\] to \[A_{2}\] = L

Therefore, time = \[ \frac{L}{v_{x}}\]

After a collision with \[A_{2}\] it again travels to \[A_{1}\]. Hence, the time between two collisions= \[ \frac{2L}{v_{x}}\]

So the number of collisions of molecule per unit time \[n = \frac{v_{x}}{2L}\]……….(2)

From (1) and (2),

Momentum imparted to the molecule by the wall per unit time

∆F=n∆P

 \[ = \sum \frac{m}{L *  v_{x}^{2}}\]

Therefore, the total force on wall  \[A_{1}\] due to all the molecules is

\[F = \sum \frac{m}{L *  v_{x}^{2}}\]

\[F = \frac{m}{L * \sum v_{x}^{2}}\]

\[\sum v_{x}^{2} = \sum v_{y}^{2} = \sum v_{z}^{2} (symmetry)\]

= \[\frac{1}{3}\sum V^{2}\]

Therefore,  \[ F = \frac{m}{L*\frac{1}{3N}\sum V^{2}}\]

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

\[P = \frac{F}{L^{2}}\]

\[\frac{m}{L^{3}(\frac{1}{3N})\sum V^{2}}\]

\[P = \frac{3\rho }{v^{2}}\]

Here, M=total mass of the gas

And ρ=density of the gas

Now, \[\frac{\sum v^{2}}{N}\] is written as \[{v^{2}}\] and is called mean square speed.

\[P = \frac{3\rho }{v^{2}}\]

So, this is what the pressure exerted by gas.