Write expression for pressure exerted by an ideal gas in terms of density and root mean square speed

Expression for pressure exerted by a gas: Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.

The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.

The molecules of the gas exert pressure on the walls of the container due to the collision on it. During each collision, the molecules impart certain momentum to the wall. Due to the transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.

A molecule of mass m moving with a velocity `vec"v"` having components (vx, vy, vz) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with the same speed and its x-component are reversed. This is shown in the figure. The components of velocity of the molecule after the collision are (- vx, vy, vz).

The x-component of momentum of the molecule before collision = mvx

The x-component of momentum of the molecule after collision = – mvx

The change in momentum of the molecule in x-direction = Final momentum – initial momentum

= – mvx – mvx

= – 2mvx

According to the law of conservation of linear momentum, the change in momentum of the wall = 2mvx

The number of molecules hitting the right side wall in a small interval of time At.

The molecules within the distance of vx∆t from the right side wall and moving towards the right will hit the wall in the time interval ∆t. The number of molecules that will hit the right side wall in a time interval At is equal to the product of volume (Avx∆t) and number density of the molecules (n). Here A is an area of the wall and n is the number of molecules per unit volume `"N"/"V"`.
We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on average only half of the n molecules move to the right and the other half moves towards the left side.

The number of molecules that hit the right side wall in a time interval ∆t

= `"n"/2"` Avx∆t ..............(1)

In the same interval of time At, the total momentum transferred by the molecules

∆P = `"n"/2"Av"_"x"Δ"t" xx 2"mv"_"x" = "Av"_"x"^2"mn"Δ"t"` .......(2)

From Newton’s second law, the change in momentum in a small interval of time gives rise to force.

The force exerted by the molecules on the wall (in magnitude)

F = `(Δ"P")/(Δ"t") = "nmAv"_"x"^2` ...........(3)

Pressure, P = force divided by the area of the wall

P = `"F"/"A" = "nmv"_"x"^2` ..................(4)

Since all the molecules are moving completely in a random manner, they do not have the same speed. So we can replace the term `"v"_"x"^2` with the average `bar("v"_"x"^2)` in equation (4).

P = `"nm"bar("v"_"x"^2)` ........(5)

Since the gas is assumed to move in a random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has the same average speed in all three directions. So, `bar("v"_"x"^2) = bar("v"_"y"^2) = bar("v"_"z"^2)`. The mean square speed is written as

`bar("v"^2) = bar("v"_"x"^2) + bar("v"_"y"^2) + bar("v"_"z"^2) = 3bar("v"_"x"^2)`

`bar("v"_"x"^2) = 1/3 bar("v"^2)`

Using this in equation (5), we get

P = `1/3"nm"bar("v"^2)` or P = `1/3 "N"/"V" "m"bar("v"^2)` ...........(6)

as `["n" = "N"/"V"]`

In this article, we shall study to derive an expression for pressure exerted by gas on the walls of container. We shall also derivation of different gas laws using the kinetic theory of gases.

Expression for Pressure Exerted by a Gas Using Kinetic Theory of Gases:

Let us consider a gas enclosed in a cube whose each edge is of length ‘l’. Let A be the area of each face of the cube. So A = l². Let V be the volume of the cube (or the gas). So V = l³. Let ‘m’ be the mass of each molecule of the gas and ‘N’ be the total number of molecules of the gas and ‘M’ be the total mass of the gas. So M = mN.

Suppose that the three intersecting edges of the cube are along the rectangular co-ordinate axes X, Y and Z with the origin O at one of the corners of the cube.  By the kinetic theory of gases, we know that molecules of a gas are in a state of random motion so it can be imagined that on an average N/3 molecule are constantly moving parallel to each edge of the cube i.e. along the co-ordinate axes. Let velocities of N/3 molecules moving parallel to X-axis be C1, C2, C3, …. , CN/3 respectively.

Consider a molecule moving with the velocity C1in the positive direction of the X-axis. The initial momentum of the molecule,   

p1 = mC1,

It will collide normally with the wall ABCD and as the collision is perfectly elastic rebounds with the same velocity. Hence, momentum of molecule  after  collision, 

p2 =  – mC1

Change in the momentum of the molecule due to one collision with ABCD

Δp =  p2  –   p1

Δp = – mC1  – mC1     =  – 2 – mC1

Before the next collision with the wall ABCD the molecule will travel a distance 2l with velocity C1. So time interval between two successive collisions of the molecule with ABCD = 2l /C1

Number of collisions of the molecule per unit with wall ABCD =  C1/ 2l

Change in the momentum of the molecule per   unit   time

= –  2mC1  ×  C1/ 2l   = –  mC1² /l

But by Newton’s second law we know that the rate of change of momentum is equal to the impressed force.

So force exerted on the molecule by wall ABCD =- mC1² /l

From    Newton’s third law of motion, action and reaction are equal and opposite. So force exerted on the molecule by wall ABCD   = mC1² /l. Thus every molecule will exert a force on the wall ABCD. So total force exerted on the wall ABCD due to molecules moving in the positive X-axis direction is given by

Let P be the pressure of the gas. The pressure exerted on the wall ABCD is given by

The pressure exerted on each wall will be the same and i.e. equal to the pressure of the gas. By definition of r.m.s. velocity

This is an expression for pressure exerted by a gas on the walls of the container.

Boyle’s Law from Kinetic Theory of Gases:

Statement: 

The temperature remaining constant the pressure exerted by a certain mass of gas is inversely proportional to its volume.

Explanation: 

If P is the pressure and V is the volume of a certain mass of enclosed gas, then

P ∝1 / V      ∴ P V = constant

Proof:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

N = Number of molecules of a gas

m = Mass of each molecule of a gas.

But by assumptions of the kinetic theory of gases the average kinetic energy of a molecule is constant at a constant temperature. Thus the right-hand side of the equation is constant.

Thus,         PV = constant i.e.  P ∝ 1 / V .   This is Boyle’s Law.

Thus Boyle’s law is deduced from the kinetic theory of gases.

Relation Between r.m.s. Velocity of Gas Molecule and the Absolute Temperature of the Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

For one mole of a gas, the total mass of the gas can be taken as molecular weight

This is an expression for r.m.s. velocity of gas molecules in terms of its molecular weight. Now R is the universal gas constant, molecular mass M for a particular gas is constant.

Thus, the R.M.S. velocity of a gas is directly proportional to the square root of the absolute temperature of the gas.

Kinetic Energy of Gas:

Kinetic Energy Per Unit Volume of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

N = Number of molecules of a gas

m = Mass of each molecule of a gas.

This is an expression for the kinetic energy of gas molecules per unit volume of the gas.

Kinetic Energy of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

N = Number of molecules of a gas

m = Mass of each molecule of a gas.

This is an expression for the kinetic energy of gas molecules.

Average Kinetic Energy Per Mole of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

For one mole of a gas, the total mass of the gas can be taken as molecular weight

This is an expression for r.m.s. velocity of gas molecules in terms of its molecular weight. Now R is the universal gas constant, molecular mass M for a particular gas is constant.

This is an expression for kinetic energy per mole of a gas

Average Kinetic Energy Per Molecule of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

For one mole of a gas, the total mass of the gas can be taken as molecular weight

This is an expression for r.m.s. velocity of gas molecules in terms of its molecular weight. Now R is the universal gas constant, molecular mass M for a particular gas is constant.

This is an expression for kinetic energy per mole of a gas

Where N = Avogadro’s number

This is an expression for average kinetic energy per molecule of a gas.

Average Kinetic Energy Per Unit Mass of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

For one mole of a gas, the total mass of the gas can be taken as molecular weight

This is an expression for r.m.s. velocity of gas molecules in terms of its molecular weight. Now R is the universal gas constant, molecular mass M for a particular gas is constant.

This is an expression for kinetic energy per mole of a gas

This is an expression for average kinetic energy per unit mass of a gas.

Charle’s Law:

At constant pressure, the volume of a certain mass of enclosed gas is directly proportional to the absolute temperature of the gas.

Dalton’s Law of Partial Pressure:

At constant temperature, the pressure exerted by a mixture of two or more non-reacting gases in enclosed in a definite volume, is equal to the sum of the individual pressure which each gas exerts, if present alone in the same volume.

Explanation: Let P1, P2, P3, …. be the partial pressure of the mixture of non-reacting gases enclosed in a definite volume and at temperature T. Let P be the pressure of a mixture of these gases. Then by the law of partial pressure,

P = P1 + P2 + P3, ….

Maxwell Distribution of Molecular Speeds:

For a given mass of a gas, the velocities of all molecules are not the same, even when bulk parameters like pressure, volume and temperature are fixed. Collisions change the direction and speed of the molecules, but in a state of equilibrium, the distribution of speed is constant.

It is observed that the molecular speed distribution gives the number of molecules dN(v) between speeds v and v + dv, which is proportional to dv (difference in velocities) is called Maxwell distribution.

Following graph shows the distribution of velocity at different temperatures.

The fraction of the molecules with speed v and v + dv is equal to the area of the strip shown.

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