In this article, we shall study the concept of specific heat capacities or Specific heats of gases.

Solids and liquids have only one specific heat, while gases have two specific heats:

In the case of solids and liquids, a small change in temperature causes a negligible change in the volume and pressure, hence the external work performed is negligible. In such cases, all the heat supplied to the solid or liquid is used for raising the temperature. Thus there is only one value of specific heat for solids and liquids.

When gases are heated small change in temperature causes a considerable change in both the volume and the pressure. Due to which the specific heat of gas can have any value between 0 and ∞. Therefore to fix the values of specific heat of a gas, either the volume or pressure is kept constant. Hence it is necessary to define two specific heats of gases. viz. Specific heat at constant pressure and specific heat at constant volume.

Molar Specific Heat of Gas at Constant Volume:

The quantity of heat required to raise the temperature of one mole of gas through 1K (or 1 °C) when the volume is kept constant is called molar specific heat at constant volume. It is denoted by C_V. Its S.I. unit is J K^-1 mol^-1.

Molar Specific Heat of Gas at Constant Pressure:

The quantity of heat required to raise the temperature of one mole of gas through 1K (or 1 °C) when pressure is kept constant is called molar specific heat at constant pressure. It is denoted by C_P. Its S.I. unit is J K^-1 mol^-1.

Principal Specific Heat of Gas at Constant Volume:

The quantity of heat required to raise the temperature of a unit mass of gas through 1 K (or 1 °C) when its volume is kept constant, is called its principal specific heat at constant volume. It is denoted by c_V. Its S.I. unit is J K^-1 kg^-1.

Principal Specific Heat of Gas at Constant Pressure:

The quantity of heat required to raise the temperature of a unit mass of gas through 1 K (or 1 °C) when its pressure is kept constant, is called its principal specific heat at constant pressure. It is denoted by c_P. Its S.I. unit is J K^-1 kg^-1.

Relation Between Molar Specific Heats and Principal Specific Heats:

Molar specific heat at constant volume is Molecular mass times the principal specific heat at constant volume

C_V = M c_V

Molar specific heat at constant pressure is Molecular mass times the principal specific heat at constant pressure

C_P = M c_P

Explanation of C_P >C_V :

When a gas is heated at constant volume there is no expansion of gas thus external work done is zero. Hence the heat given to the gas is completely used for the increase in the internal energy of the gas.

Whereas when a gas is heated at constant pressure the heat given to the gas is used for two purposes a part of it is used for an increase in the internal energy of the gas and a part of it is used for doing external work.

Now in both cases of constant volume and constant pressure, the increase in internal energy is the same as the rise in temperature is the same for the same mass of the gas. Therefore, heat supplied at constant pressure is more than heat supplied at constant volume by an amount of heat which is used for doing external work. This explains why specific heat of a gas at constant pressure is greater than the specific heat at constant volume. i.e. C_P > C_V.

Further by Mayer’s relation C_P –  C_V = R . R is the universal gas constant and is positive. Hence C_P > C_V.

Mayer’s Relation:

Let us consider one mole of a perfect gas enclosed in a cylinder fitted with a frictionless weightless airtight movable piston. Let P, V and T be the pressure, volume and absolute temperature of the gas respectively. Let the gas be heated at a constant volume so that its temperature rises by dT.  Let dQ₁ be the heat given to the gas for this purpose. In this case, all the heat supplied to the gas is used for increasing the internal energy of the gas.

Now, dQ₁ =   1 × C_V × dT = dE   …………  (1)

Where C_V is the molar specific heat of the gas at constant volume.

Now let the gas be heated at constant pressure so that its temperature rises by dT.  Let dQ2 be the heat given to the gas for this purpose.

Now, dQ₂ =   1 × C_P ×   dT   = C_P × dT …………  (2)

Where C=   1 × C_P × dT is the molar specific heat of the gas at constant pressure.

In this case, the heat supplied to the gas is used for two purposes a part of it is used for an increase in the internal energy of the gas and a part of it is used for doing external work. Thus,

dQ₂ = dE   +   dW    …………  (3)

From equations (1) ,(2) and (3)

C_P ×   dT = C_V ×   dT +   dW …………  (4)

The gas is heated at constant pressure so that it expands and in the process, the piston moves upward through a distance dx. If A is the area of cross-section of the cylinder (area of the piston) and dV is the increase in the volume of the gas,

Then,       dV =   A dx.  ……… (5)

Let dW be the work done by the gas during the increase in the volume.

Now, Work   =   Force × Displacement

But,   Force = Pressure × Area

∴ Work =  Pressure  ×  Area  × Displacement

dW     =   P ×   A  × dx    ……… (6)

From equations (5) and (6) we have

dW    =   P  dV         ……… (7)

From equations (7) and (4)

C_P ×   dT = C_V ×   dT + P .dV…………  (8)

For one mole of a perfect gas

P V = R T

Where, R is the universal gas constant.

Differentiating both sides with respect to temperature.

P dV    = R dT.    ……………. (9)

From equations (8) and (9)

C_P ×   dT = C_V ×   dT +  R.dT

∴   C_P   = C_V   + R

∴   C_P – C_V = R

This relation is known as Mayor’s relation between the two molar specific heats of a gas.

Relation Between Principal Specific Heats of Gases:

By Mayer’s relation

C_P – C_V   =  R     ………..(1)

Let c_P and c_V be the principal heat capacities of the gas at constant pressure and constant volume.  Let M be the molecular weight of the gas.

Then, C_V = M c_V and C_P = M c_P

Substituting these values in equation (1)

M c_P – M c_V = R

∴ M (c_P – c_V) =   R

∴ (c_P – c_V) =   R/M

∴ (c_P – c_V) =   r

Where, r = gas constant for gas in consideration. It is different for different gases.

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Science > Physics > Kinetic Theory of Gases > Specific Heats of Gases

The post Specific Heats of Gases appeared first on The Fact Factor.

In this article, we shall study the concept of degree of freedom of gas molecules. Degree of freedom for different types of gases. We shall also find the ratio of specific heats for different types of gases.

Degree of Freedom of a Gas Molecule:

A molecule free to move in space needs three coordinates to specify its location. If a molecule is constrained to move along a line it requires one co-ordinate to locate it. Thus it has one degree of freedom for motion in a line. If a molecule is constrained to move in a plane it requires two coordinates to locate it. Thus it has two degrees of freedom for motion in a plane. If a molecule is free to move in a space it requires three coordinates to locate it. Thus it has three degrees of freedom for motion in a space.

Law of Equipartition of Energy:

Statement: In equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having average energy equal to

Explanation: A molecule has three types of kinetic energy a) Translational kinetic energy, b) Rotational kinetic energy, and c) Vibrational kinetic energy. Thus total energy of a molecule is given by

E_T = E_{Translational} + E_Rotational + E_Vibrational ……… (1)

For translational motion, the molecule has three degrees of freedom (along the x-axis, along the y-axis and along the z-axis). Hence,

For rotational motion, it has two degrees of freedom along its centre of mass (clockwise and anticlockwise).

For vibrational motion only one degree of freedom (to and fro).

Where k is the force constant and y is the vibrational coordinate.

Thus the total energy of the molecule is

It is to be noted that in the vibrational mode the energy has two components potential and kinetic and by the law of equipartition energy, each part is equal to

Thus the total vibrational component of the energy is

Specific Heats of Gases:

In the case of solids and liquids, small change in temperature causes a negligible change in the volume and pressure, hence the external work performed is negligible. In such cases, all the heat supplied to the solid or liquid is used for raising the temperature. Thus there is only one value of specific heat for solids and liquids.

When gases are heated small change in temperature causes a considerable change in both the volume and the pressure. Due to which the specific heat of gas can have any value between 0 and ∞. Therefore to fix the values of specific heat of a gas, either the volume or pressure is kept constant. Hence it is necessary to define two specific heats of gases. viz. Specific heat at constant pressure and specific heat at constant volume.

Molar Specific Heat of Gas at Constant Volume:

The quantity of heat required to raise the temperature of one mole of gas through 1K (or 1 °C) when the volume is kept constant is called molar specific heat at constant volume. It is denoted by C_V. Its S.I. unit is J K^-1 mol^-1.

Molar Specific Heat of Gas at Constant Pressure:

The quantity of heat required to raise the temperature of one mole of gas through 1K (or 1 °C) when pressure is kept constant is called molar specific heat at constant pressure. It is denoted by C_P. Its S.I. unit is J K^-1 mol^-1.

Relation Between C_P and C_V :

C_P – C_V = R

This relation is known as the Mayor’s relation between the two molar specific heats of a gas.

The ratio C_P/C_V is known as ratio of specific heats and is denoted by the letter ‘γ’

Ratio of Specific Heats for Different Gases:

Monoatomic Gases:

Helium and argon are monoatomic gases. The monoatomic gases have only translational motion, hence they have three translational degrees of freedom. The average energy of the molecule at temperature T is given by

By the law of equipartition of energy we have

Thus the energy per mole of the gas is given by

Thus the ratio of specific heat capacities of monoatomic gas is 1.67

Diatomic Gases:

Dinitrogen, dioxygen, and dihydrogen are diatomic gases. The diatomic gases have translational motion (three translational degrees of freedom) as well as rotational motion(rotational degree of freedom). The average energy of the molecule at temperature T is given by

By the law of equipartition of energy we have

Thus the energy per mole of the gas is given by

Thus the ratio of specific heat capacities of diatomic gas is 1.4

Triatomic Gas:

Trioxygen (ozone) and carbondioxide are triatomic gases. The triatomic gases have translational motion, rotational motion as well as vibrational motion, hence has three translational degrees of freedom and two rotational degrees of freedom. For non-rigid molecules, there is an additional vibrational motion.

The average energy of the molecule at temperature T is given by

By the law of equipartition of energy we have

Thus the energy per mole of the gas is given by

Polyatomic Gases:

A polyatomic molecule has 3 translational, 3 rotational and certain number (say f) of vibrational modes. By the law of equipartition of energy, one mole of such gas has

Thus the energy per mole of the gas is given by

This is an expression for the ratio of specific heats of polyatomic gases. Where f is the degree of freedom of vibration.

Polyatomic Gas having ‘f’ degree of Freedom:

By the law of equipartition of energy for ‘f’ degree of freedom we have

Thus the energy per mole of the gas is given by

The ratio of specific heats 1+ 2/f

The Expression for the Molar Specific Heat Capacity of Solid:

Consider a solid of N atoms, each vibrating about mean position. These atoms don’t have translational or rotational modes. The average energy of oscillation in one dimension is K_BT. Thus the average energy of oscillation is 3K_BT.

For one mole of Solid N = Avogadro’s number = N_A.

Thus total energy 

Since there is negligible change in the volume of a solid on heating, solids have only one specific heat.

∴ C = 3R

Molar Specific Heat Capacity of Water:

A water molecule has three atoms (2hydrogens and 1 oxygen). If we treat water like solid, the total energy of water molecules is three times the average energy of an atom of solid. Thus for Water

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Science > Physics > Kinetic Theory of Gases > Degree of Freedom of Gas Molecules

The post Degree of Freedom of Gas Molecules appeared first on The Fact Factor.

In this article, we shall study to find r.ms. speed of gas molecules, density of gas, pressure exerted by the gas using kinetic theory of gases.

Example 01:

Calculate the R.M.S. velocity of Hydrogen molecules at N.T.P. given: Density of Hydrogen at N.T.P. = ρ = 8.957 x 10^-2 kg/m³. Density of mercury = 13600 kg/m³, g = 9.8 m/s².

Given: Density of hydrogen = ρ = 8.957 x 10^-2 kg/m³, condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9. 8 N/m², Density of Mercury = 13600 kg/m³, g = 9.8 m/s².

To Find: r.m.s. speed = C =?

Solution:

Ans: r.m.s. velocity of hydrogen molecule is 1842 m/s 0r 1.842 km/s

Example 02:

Find the R.M.S. velocity of Nitrogen molecules at N.T.P. given that the density of Nitrogen at this temperature is 1.25 g/litre.

Given: Density = ρ = 1.25 g/litre = 1.25 x 1 = 1.25 kg/m³, condition N.T.P., P = 1.013 x 10⁵ N/m².

To Find: r.m.s. speed = C =?

Solution:

Ans: r.m.s. velocity of nitrogen molecule is 493.1 m/s

Example 03:

Calculate the R.M.S. velocity of Oxygen molecules at a pressure of 1.013 x 10⁵ N/m² (N.T.P.) given the density of Oxygen is 1.44 kg/m³.

Given: Density = ρ = 1.44 kg/m³, P = 1.013 x 10⁵ N/m².

To Find: r.m.s. speed = C =?

Solution:

Ans: r.m.s. velocity of oxygen molecule is 459.4 m/s

Example 04:

Calculate the density of He at N.T.P. given that R.M.S. velocity of He molecules at N.T.P. is 1300 m/s.

Given: Condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², Density r.m.s. speed = C = 1300 m/s.

To Find: Density = ρ =?

Solution:

Ans: density of He is 0.1798 kg/m³

Example 05:

Determine the pressure of oxygen at 0 °C if the density of oxygen at NTP is 1.44 kg/m³ and r.m.s speed of the molecule at NTP is 456.4 m/s.

Given: Density = ρ = 1.44 kg/m³, condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², r.m.s. speed = C = 456.4 m/s

To Find: Pressure of gas = P =?

Solution:

Ans: Pressure of oxygen = 10⁵ N/m²

Example 06:

Two gases are at temperatures of 77 ^oC and 27 ^oC. What is the ratio of the R.M.S. velocities of the molecules of the two gases?

Given: temperature of first gas = T₁ = 77 ^oC = 77 + 273 = 350 K, temperature of second gas = T₂ = 27 ^oC = 27 + 273 = 300 K

To Find: Ratio of r.m.s speeds =?

Solution:

Ans: The ratio of r.m.s. speed is 1.08:1

Example 07:

At what temperature will the R.M.S. speed of the molecules of gas be three times its value at N.T.P.?

Given: Condition = N.T.P., T₁ = 273 K, C₂ = 3C₁

To Find: Temperature = T₂ =?

Solution:

Ans: At a temperature of 2184 °C the r.m.s. speed of the molecules of a gas is three times its value at N.T.P.

Example 08:

At what temperature will the R.M.S. speed of the molecules of gas be four times its value at N.T.P.?

Given: Condition = N.T.P., T₁ = 273 K, C₂ = 4C₁

To Find: Temperature = T₂ =?

Solution:

Ans: At a temperature of 4095 °C the r.m.s. speed of the molecules of a gas is four times its value at N.T.P.

Example 09:

The R.M.S. velocity of Nitrogen molecules at N.T.P. is 497 m/s. Calculate the R.M.S. velocity of Hydrogen molecules at N.T.P. At what temperature will the R.M.S. velocity of Nitrogen molecules be 994 m/s?

Part – I:

Given: For nitrogen: Condition = N.T.P., T_N = 273 K, C_N = 497 m/s, P_N = 1.013 x 10⁵ N/m², for oxygen: Condition = N.T.P., T_N = 273 K, P_H = 1.013 x 10⁵ N/m².

To Find: r.m.s. speed of hydrogen = C_H =?

Solution:

Part – II:

Given: Condition = N.T.P., T₁ = 273 K, C₁ = 497 m/s, C₂= 994 m/s

To Find: Temperature = T₂ =?

Solution:

Ans: r.m.s. speed of hydrogen molecules at NTP1860 m/s, at a temperature of 819oC the speed of nitrogen molecules at NTP 994 m/s,

Example 10:

Calculate the R.M.S. velocity of Oxygen molecules at 27 °C. The density of Oxygen at N.T.P. 1.44 kg/m³.

Given: For nitrogen: Condition = N.T.P. T₁ = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², density of oxygen = ρ = 1.44 kg/m³, Temperature = T₂ = 27 ^oC= 27 = 273 = 300 K

To Find: r.m.s. speed = C₂ =?

Solution:

Now C₁ = 459.4 m/s, T₁ = 273 K, = T₂ = 27 ^oC= 27 = 273 = 300 K, C₂ = ?

Ans: The R.M.S. velocity of Oxygen molecules at 27 °C is 481.6 m/s

Example 11:

Compute the R.M.S. velocity of Oxygen molecules at 127 °C. Density of Oxygen at N.T.P. = 1.44 kg/m³.

Given: For nitrogen: Condition = N.T.P. T₁ = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², density of oxygen = r = 1.44 kg/m³, Temperature = T₂ = 127 ^oC= 127 = 273 = 400 K

To Find: r.m.s. speed = C₂ =?

Solution:

Now C₁ = 459.4 m/s, T₁ = 273 K, = T₂ = 127 ^oC= 127 = 273 = 400 K, C₂ =?

Ans: The R.M.S. velocity of Oxygen molecules at 127 °C is 556.1 m/s.

Example 12:

Calculate the R.M.S. velocity of Oxygen molecules at 225 °C. The density of oxygen at NTP is 1.42 kg/m⁵ and 1 atmosphere = 1.013 x 10⁵ N/m².

Given: For nitrogen: Condition = N.T.P. T₁ = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², density of oxygen = ρ = 1.44 kg/m³, Temperature = T₂ = 225 ^oC= 225 + 273 = 498 K

To Find: r.m.s. speed = C₂ =?

Solution:

Now C₁ = 459.4 m/s, T₁ = 273 K, = T₂ = 127 ^oC= 127 = 273 = 400 K, C₂ = ?

Ans: The R.M.S. velocity of Oxygen molecules at 127 °C is 624.8 m/s

Example 13: 

The density of a gas is 0.178 kg/m³ at N.T.P. Find the R.M.S. velocity of gas molecules. By what factor will the velocity of molecules increase at 200 °C?

Given: For nitrogen: Condition = N.T.P. T₁ = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², density of oxygen = ρ = 0.178 kg/m³, Temperature = T₂ = 200 ^oC= 200 + 273 = 473 K

To Find: r.m.s. speed = C₂/ C₁=?

Solution:

Now T₁ = 273 K, = T₂ = 200 ^oC= 200 + 273 = 473 K

Ans: The r.m.s. velocity of the gas molecule at NTP is 1.306 km/s. The r.m.s. velocity will increase by a factor of 1.316 at 200 °C.

Example 14: 

R M.S. velocity of oxygen molecules at 27 °C is 500 m/s. Calculate the R.M.S. and mean square velocities of oxygen molecules at 127 °C.

Given: C1 = 500 m/s at temperature T₁ = 27 ^oC = 27 + 273 = 300 K, Required speed at temperature = T₂ = 127 ^oC = 127 + 273 = 400 K,

To Find: C₂ =? and (C₂)²

Solution:

(C₂)² = (577.4)² = 3.33 x 10⁵ m²/s²

Ans: r.m.s. velocity is 577.4 m/s, and mean square velocity is 3.33 x 10⁵ m²/s²

Example 15:

Taking the R.M.S. velocity of Hydrogen molecules at N.T.P. as 1.84 km/s, calculate the R.M.S. velocity of Oxygen molecules at N.T.P. Molecular weights of Oxygen and Hydrogen are 32 and 2 respectively.

Given: r.m.s. velocity of hydrogen = C_H = 1.84 km/s, molecular mass M_O = 32, M_H = 2, temperature T_H = T_O = 273 K, pressure P_H = P_O = 1.013 x 10⁵ N/m².

To Find: r.m.s. velocity of oxygen molecule = C_O =?

Solution:

Ans: The R.M.S. velocity of Oxygen molecules at N.T.P. is 0.46 km/s

Example 16:

R.M.S. speed of Oxygen molecules is 493 m/s at a certain temperature. Calculate the R.M.S. speed of helium molecules at the same temperature. Molecular weights of Oxygen and Helium are 32 and 4 respectively.

Solution:

Given: r.m.s. velocity of oxygen = C_O = 493 m/s, molecular mass M_O = 32, M_He = 4,
temperature T_He = T_O, Pressure P_He = P_O.

To Find: r.m.s. velocity of helium molecule = C_He=?

Ans: The r.m.s. speed of helium molecule is 1394.2 m/s

Example 17:

R.M.S. speed of Oxygen molecules at N.T.P. is 459.3 m/s. Find the R.M.S. speed of Nitrogen molecules at 340 K. Molecular weights of Oxygen and Nitrogen are respectively 32 and 28.

Given: r.m.s. speed of oxygen = C_O1 = 459.3 m/s, T_O1 = 273 K, T_O2 = 340 K  = T_N , M_O = 32, M_N = 28

To Find: C_N =?

Solution:

Ans: r.m.s speed of nitrogen at 340 K is 548 m/s

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Science > Physics > Kinetic Theory of Gases > Numerical Problems on Kinetic Theory of Gases

The post Numerical Problems on Kinetic Theory of Gases appeared first on The Fact Factor.

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, …. , C_N/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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Science > Physics > Kinetic Theory of Gases > Pressure Exerted by Gas

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In this article, we shall study a very important concept called kinetic theory of gases, using which we can explain behaviour of the gases.

Ideal Gas:

A gas which obeys gas laws at all temperatures and pressures is called an ideal gas or Perfect gas e.g. an ideal gas does not liquefy even at low temperature but continues to obey Charle’s law and finally occupies no volume at –273 ^oC or 0 K.

Ideal gas consists of molecules in a continuous state of motion with neither attraction nor repulsion between them. Molecules of ideal gas collide with each other without any net loss of kinetic energy.

For an ideal gas, PV = nRT.

This relation is known as the ideal gas equation. Where, P = Pressure of the gas, V = Volume of the gas, n = number of moles of the gas, R = Universal gas constant, T = Absolute temperature of the gas.

In reality, there exists some negligible force of attraction and repulsion between the molecules of the gas. Besides, there is some loss of kinetic energy in the collision of the molecules of the gas. Hence ideal gas is an imaginary or hypothetical concept. Gas is said to non-ideal or real gas if it obeys gas laws only at low pressures and high temperatures.

Avogadro’s Hypothesis:

Equal volumes of all gases under the same conditions of temperature & pressure contain an equal number of molecules.

Explanation:

Let the pressure of the two gases be the same (P). By Avogadro’s law under such conditions of equal pressure, equal volume & equal temperature, the number of molecules of gas A in the container should be equal to the number of molecules of gas B in the container.

Importance of Avogadro’s Hypothesis:

• It differentiates between atoms and molecules of gasses.
• It modified Dalton’s atomic theory.
• It explains Gay-Lussac’s law of combining volume.
• It helps in the determination of the atomic mass of elements.
• It established that the number of molecules per unit volume is the same for all gases at a fixed temperature and pressure.
• It established that at N.T.P.one gram mole of any gas occupies 22.4 dm3 by volume. one mole of gas contains 6.023 X 1023 molecules of gas.
• It gives the relation between vapour density & molecular weight.

Molecular weight = 2 × vapour density.

Kinetic Theory of Gases:

Evidences of the Molecules of a Gas are Always in Constant Motion:

• Diffusion of gases.
• Indefinite expansion of gases.
• Gases exert pressure on the walls of the container.
• Brownian like motion in gases.

Assumptions of Kinetic Theory of Gases:

• A gas consists of a large number of extremely small molecules which are exactly identical in all respects.
• The molecules are rigid and perfectly elastic spheres of very small diameters.
• The intermolecular forces between gas molecules are negligible.
• The molecules are in a  state of random motion  i.e.  they move with all possible velocities in all possible directions.
• During their random motion the molecules collide with each other and with the walls of the container and these collisions are supposed to be perfectly elastic, (i.e. there is no loss of kinetic energy during these collisions).
• Between successive collisions, molecules move with uniform velocity in straight paths and these paths are called free paths.
• All free paths are not equal. Average of free paths is called mean free path.
• The actual volume of the molecules is negligible compared to the total volume of the gas. Therefore, molecules can be treated as geometrical points.
• The number of molecules per unit volume of a gas remains constant.
• At constant temperature, the average kinetic energy of the gas molecules remains constant. The average kinetic energy of the molecules of a gas depends only on the absolute temperature of the gas.
• The time of impact i.e. the time interval during which collision occurs is very small compared to the time interval between successive collisions.

Terminology of Kinetic Theory of Gases:

Mean Free Path of Gas Molecule:

The molecules of a gas are always moving in random motion i.e. in all possible directions with all possible velocities. This is also called as molecular chaos. Therefore they constantly collide with one another and with the walls of the container.

Between two successive collisions, a molecule travels in a straight line. The distance covered by a molecule between two successive collisions is called the free path.

All the free paths are not equal. Therefore their average value is considered. The average distance covered by a molecule between successive collisions is called its mean free path. It is denoted by ‘λ’. Its S.I. unit is m but the practical unit is angstrom.

If λ₁, λ₂,l₃, ….., λ_N are the free paths, then mean free path is given by

Where N is a number of collisions.

Mean or Average Velocity:

The mean velocity of a molecule of a gas is defined as the arithmetic mean of the velocities of the molecules of the gas at a given temperature.

Let C₁, C₂, C₃, …. , CN be the velocities of N molecules of a gas, then the mean velocity of molecules of a gas is given by

As the molecules of gases are in random motion, i.e. they can move in any direction with any possible velocity, by the probability theory the mean velocity of gas molecules should be zero. Its S.I. unit is m/s.

Mean Square Velocity of Gas Molecules:

The mean square velocity of a molecule of a gas is defined as the arithmetic mean of squares of the velocities of the molecules of the gas at a given temperature.

LetC₁, C₂, C₃, …. , CN be the velocities of N molecules of a gas, then the mean velocity of molecules of a gas is given by

Its S.I. unit is m²/s².

Root Mean Square Velocity of Gas Molecules:

The square root of the mean of the squares of the velocities of the molecules of a gas is called root mean square (r.m.s.) velocity of the molecules of a gas.

Let C₁, C₂, C₃, …. , CN be the velocities of   N   molecules of a gas, then the r.m.s. velocity of molecules of a gas is given by

It is clear that r.m.s.  Velocity cannot be zero.

Numerical Problems:

Example 01:

In the following table, n_i represents the number of molecules of a gas and C_i represents their speed in m/s Calculate the average and R.M.S. speeds of the molecules.

Ans: Average speed of molecule is 3.173 m/s and r.m.s. speed is 3.369 m/s

Example 02:

Find the average velocity, mean square velocity and root mean square velocity of six molecules having velocities 4, 5, 8, -6, -4, and 10 m/s respectively.

Ans: average speed = 2.83 m/s, mean square speed = 42.83 m²/s²; rms speed = 6.544 m/s

Example 03:

The velocities of seven molecules are 1, 2, 3, 4, 5, 6, 7 km/s respectively. Find the mean square velocity of the molecule.

Ans : Mean square velocity = 20 km²/s²

Example 04:

Find the r.m.s. velocity of three molecules having velocities 10, 20, 30 km/s.

Ans: r.m.s. speed = 21.60 km/s

Next Topic: Expression for Pressure Exerted by a Gas

Science > Physics > Kinetic Theory of Gases > Introduction to Kinetic Theory of Gases

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