PHY 103 General Physics III Lecture Notes PDF

# PHY 103: General Physics III ## DEPARTMENT OF PHYSICS FEDERAL UNIVERSITY LOKOJA LECTURE NOTES ## PHY 103: GENERAL PHYSICS Ⅲ (BEHAVIOUR OF MATTER) **LECTURE I** FIRST SEMESTER 2025 22ND JANUARY, 2025 **COURSE LECTURER:** PROF. Tajudeen O. AHMED ## COURSE OUTLINE 1. **KINETIC THEORY OF GASES** - MOLECULAR COLLISIONS & MEAN FREE PATH 2. **ELASTICITY** - HOOKE'S LAW - YOUNE'S MODULKI - SHEAR MODULUS - BULK MODULUS ## EMMENDED TEXTBOOKS - Advanced Level Physics by Nelkon & Parker - Collage Physics by Frederick J. Breche & Eugene Hea ## 1.0: KINETIC THEORY OF GASES - A gas is a Collection of large number of molecules which are in continuous motion. - The size of a molecule of gas (i.e., Molecular diameter) is of the order of 2x10⁻¹⁰m. - The distance between the gas molecules is of the order of 2x10⁻⁹m i.e., about 10 times as large as their size. - Therefore, the molecules can be considered to be moving freely with respect to each other. - However, during motion, when they come close to each other, they suffer a change in their velocities due to intermolecular force. - Thus, such changes in velocities of the molecules of gas are due to molecular collisions. - Rudolph Clausius (1822-1888) and James Clark Maxwell (1831-1875) developed the Kinetic theory of gases in order to explain the gas laws in terms of the motion of the gas molecules. - The theory is based on the following assumptions with regards to the motion of molecules and the nature of gases. 1. All gases consist of molecules. The molecules of gases are all alike and differ from those of other gases. 2. The molecules of a gas are very small in size as compared to the distance between them. 3. The molecules of a gas behave as perfect elastic spheres. 4. The molecules of a gas are always in random motion. They have velocities in all directions ranging from zero to infinity. 5. During their random motion, the molecules collide against one another and with the walls of the Containing vessel. The collisions of the molecules with one another & the walls of the Containing Vessel are perfectly elastic. 6. Between 2 collisions, a molecule moves along a straight line and the distance covered between 2 successive collisions is called the free-path of the molecule. 7. The collisions are almost instantaneous i.e., the time during which a collision takes place is negligible as compared to the time taken by the molecule to cover the free-path. 8. The molecules do not exert any force on each other except during collisions. ## 1.1: PRESSURE EXERTED BY A GAS ON THE WALLS OF THE CONTAINING VESSEL - Arising from the continuous collisions of the molecules against the walls of the containing vessel, pressure is exerted on its walls. - Consider the schematic representation of a containing vessel below: - **A- Area** - **L= Δx Δt** - **V** - Diagram with a box with a molecule moving along a straight line within the box. There are arrows showing the collision of the molecules. - Let n= number of molecules per unit volume inside the vessel - m = mass of each molecule - v = velocity of any gas molecule at any instant - If δx, δy and δz are the components of v along x, y and z axes respectively, then $v^2 = v_x^2 + v_y^2 + v_z^2$ - Suppose a molecule moves with momentum mv_x along the x-axis and strikes the face 'A'. - Since the collision is perfectly elastic, the molecule rebounds back with the same speed v_x. - Now, the momentum of the molecule after rebound becomes -mv_x. - Therefore, the change in momentum of the molecule along x-axis is: (-mv_x) - mv_x = -2mv_x - In time Δt, the length of the vessel is v_x Δt along x-axis. and the molecules lie in a volume A. v_x Δt. - Since the number of molecules per unit volume is n, then the number of molecules in this volume is n.A. v_x Δt. - In fact, on the average, half of this number is expected to move along -ve x-axis and the other half towards the face ‘AB’ along +ve x-axis. - Therefore, the number of molecules hitting the area A in time Δt along the x-axis is; $\frac{1}{2} n A v_x Δt$ - Thus, the total change in momentum in time Δt along x-axis is; $Δp_x = (-2mv_x) \times \frac{1}{2} n A v_x Δt = -mnA v_x^2 Δt$ - Since the molecules possess different velocities at different masses, v_x is represented with v_x^2 - Now, the force (rate of change of momentum) exerted by the walls of the vessel on the gas molecules along x-axis is $F_x = \frac {Δp_x} {Δt} = -mnA v_x^2$ - Therefore, the force exerted by these gas molecules on the walls of the containing vessel along x-axis is equal and opposite to (5). According to Newton's 3rd Law of motion; $F_x = - (-mnA v_x^2) = mnA v_x^2$ - Thus, the pressure exerted by the gas molecules along x-axis is then; $P_x = \frac {F_x} {A} = \frac{mnA v_x^2}{A}=mn v_x^2$ - If $P_y$ and $P_z$ are the pressures along y and z axes, then $P_y = mn v_y^2$ and $P_z = mn v_z^2$ - Where; - v̄2 = the mean square velocity along x, y, z - m = mass of the gas molecule - n = number of gas molecules per unit volume - F = force - A = area - Since gas molecules have the same properties in all directions, the choice of x-axis, y-axis and z-axis is arbitrary. In other words, pressure exerted by the gas molecules be the same in all directions i.e., $P_x = P_y = P_z =P$ - $\implies P = (P_x + P_y + P_z)$ - $\implies P = (\frac{1}{3}mn v_x^2 + \frac{1}{3}mn v_y^2 + \frac{1}{3}mn v_z^2)$ - $\implies P = \frac{1}{3}mn (v_x^2 + v_y^2 + v_z^2)$ - $\implies P = \frac{1}{3}mn \bar{v}^2$ - Since ρ is the density of the gas, eqn (5) takes the form: $P = \rho \bar{v}^2$ - From eqn (6), we have; $P = \frac{1}{3} \rho \bar{v}^2 = \frac{1}{3} n m \bar{v}^2$ - $\implies P = \frac{2}{3} n K.E.$ - If N is the total number of gas molecules in a volume V, then n = N/V and eqn (9) takes the form; $P = \frac{1}{3} (\frac{N}{V}) m \bar{v}^2$ - $\implies P = \frac{2}{3V} N \times (\frac{1}{2} m \bar{v}^2)$ - $\implies P = \frac{2}{3V} N * K.E$ - $\implies PV = \frac{2}{3} N * K.E$ - From (12), we could write; $P = \frac{2}{3} ( \frac{1}{2}m v^2) = \frac{1}{3} (\frac{2}{3} P V)$ - Where $\frac{1}{2} m \bar{v}^2$ = average K.E of the gas molecules per unit volume - .. $P = \frac{2}{3} \times$ average K.E per unit volume - .. $P = \times$ average K.E per unit volume ## 2: KINETIC ENERGY OF A MOLECULE AND OF 1 MOLE OF A GAS - The perfect gas equation is given as, $PV = RT$ for 1 mole gasses - From eqn (13), $P = \frac{1}{3} \rho \bar{v}^2 = \frac{1}{3} \frac{NM}{V} * \bar{v}^2$ - $\implies PV = \frac{1}{3} M \bar{v}^2$ - Comparing (13) and (14), we end up with; $\frac{1}{3} M \bar{v}^2 = RT$ - Therefore, average K.E of 1 mole of a gas is; $\frac{1}{2} M \bar{v}^2 = RT$ - Since 1 mole of a gas contains molecules equal to Avogadro’s number, N, then the mass of 1 molecule of a gas in relation to N is; $m = \frac{M}{N}$ - $\implies \frac{1}{2} (MN) \bar{v}^2 = RT$ - Therefore, average K.E of 1 molecule is; $\frac{1}{2} m \bar{v}^2 = \frac{3}{2} (\frac{R}{N}) T$ - $\frac{1}{2} m \bar{v}^2 = \frac{3}{2} k_B T$ $[K_B = \frac{R}{N}]$ $K_B = 1.38 \times 10^{-23} J mol^{-1} K^{-1}$ - From (17), it follows that $\frac{1}{2} m \bar{v}^2 \alpha T$, i.e., the higher the temperature, the more the average K.E. ## MEAN VELOCITY OF GAS MOLECULES - This is the arithmetic mean of the velocities of the gas molecules and it is denoted by $\bar{v}$ - $\bar{v} = \frac{v_1 + v_2 + v_3 +…. + v_n}{n} = \frac{\sum v_i}{n}$ - $\bar{v} = \sqrt{\frac{8k_B T}{\pi m}}$ - $\bar{v} = \sqrt{\frac{3P}{\rho}}$ - $\bar{v} = \sqrt{\frac{3RT}{M}}$ ## 1.4: ROOT-MEAN SQUARE VELOCITY OF GAS MOLECULES - This is defined as the square root of the mean of the squares of the velocities of the gas molecules. It is denoted by vrms and given by: $v_{rms} = \sqrt{\frac{v_1^2 + v_2^2 +... + v_n^2}{n}} = \sqrt{\bar{v^2}}$ - $v_{rms} = \sqrt{\frac{3P}{\rho}}$ - $v_{rms} = \sqrt{\frac{3RT}{M}}$ - $v_{rms} = \sqrt{\frac{3k_B T}{m}}$ ## 1.5: MEAN-FREE PATH - The molecules of a gas are in continuous random motion and they collide against one another during their random motion. - Between 2 collisions, a molecule moves along a straight line and the distance covered between 2 successive collisions is called the free-path of the molecule. - Due to the random nature of the motion of the gas molecules, the free-paths of a molecule during the collisions are bound to be of different lengths and zig-zag as shown; - Diagram of a zig-zag path of a molecule. - The mean free-path of a molecule is defined as the average distance travelled by the molecule between 2 successive collisions and is measured as the mean of a large number of free paths. - Consider that a molecule 'O' undergoes collisions at points A, B, C,..., so that the free paths of the molecule are l1, l2, l3 respectively. - Then, the mean free-path of the molecule is given by; $\bar{l} = \frac{l_1 + l_2 + l_3 +...}{total number of collisions}$ - To derive an expression for the mean free-path, the following assumptions are made: - #1: The gas molecules are perfectly spherical in shape. - #2: All the gas molecules except the one under consideration are at rest. - Consider a sample of gas which contains 'n' molecules per unit volume. - Let 'd' be the diameter of each gas molecule. - Consider a molecule ‘O’ moving with velocity V. The molecule ‘O’ will strike all those molecules whose centres lie within a distance ‘d’ from its path. - Obviously, all such molecules with which the molecule ‘O’ will strike in time t, lie inside a cylinder of length Vt and area of cross-section πd². - Thus, the number of collisions suffered by the molecule ‘O’ in time ‘t’ is; - **N= volume of cylinder of length Vt X area of cross-section πd² x number of molecules per unit volume** - **=Vt. πd².n** - .:. **N = πd²nVt** - Now, the mean free path of a gas molecule is; - **l = distance covered by the molecule in time t / no of collisions suffered in time t** - **= Vt / πd²nVt** - **= 1 / πd²n** - .. **$\bar{l} = \frac{1}{\pi d^2 n}$** - Taking the motion of all the gas molecules into consideration, the mean free path becomes. - .. **$\bar{l} = \frac{1}{\sqrt2 \pi d^2 n}$** - If m is the mass of each gas molecule, then density of the gas is ρ = mn or n = ρ/m. With this, eqn (23) becomes; - .. **$\bar{l} = \frac{1}{\sqrt2 \pi d^2 \rho / m}$** - From eqn (24), it follows that l is - #1. directly proportional to the mass of the gas molecule. - #2. inversely proportional to the density of the gas. - #3. inversely proportional to the square of the molecules diameter. ## TUTORIAL: Prove that l varies directly with the temperature and inversely proportional to the pressure of the gas. ## Q3. Calculate the I of the molecule, If the number of molecules per cm3 is 2x1019 and the diameter of each molecule is 2Å. - Soln: n = 3x1019 cm3 = 3x1025 m-3 - d = 2Å = 2x10⁻¹⁰m - $\implies \bar{l} = \frac{1}{\sqrt2 \pi d^2 n}$ - $\implies \bar{l} = \frac{1}{\sqrt2 \times \pi \times (2 \times 10^{-10})^2 \times 3 \times 10^{25}} = 1.876 \times 10^{-7} m$ ## Q4. How many collisions per second does each molecule of a gas make, when the average speed of a molecule is 500 m s-1 and the mean free-path is 2.66 x 10-7 m? - Soln: No of collisions per second = 500m s-1 / 2.66 x 10-7 m = 1.88 x 109 s-1 ## 3.Q ELASTICITY - Elasticity is a property of matter by virtue of which it regains its original shape and size when the deforming forces have been removed. - Deforming force is a force acting on a body instead of producing a change in its state of rest or uniform motion, produces a change in the shape of the body. - Elastic body is a body that returns to its original shape and size on the removal of the deforming force (when deformed within elastic limit). - Plastic body is a body that does not return to its original shape and size on the removal of deforming force, however small the magnitude of deforming force may be. ## HOOKE'S LAW - It states that the extension produced in a wire is directly proportional to the load attached to it. - That is; - extension ∝ load - However, this proportionality holds good upon a certain limit called the elastic limit. - Young (1807) pointed out that the load and the extension are more scientifically described in terms of stress and strain respectively. - Thus, Hooke’s law may be restated as stress is directly proportional to strain. - That is, - Stress ∝ Strain - Stress = Constant x Strain - $\frac{Stress}{Strain} = Constant$ - This Constant of proportionality is called Modulus of elasticity or coefficient of elasticity of the material. - Its value depends upon the nature of the material of the body and the manner in which the body is deformed. - There are 3 moduli of elasticity namely Young’s modules, (Y), bulk modulus (K) and modulus of rigidity (η) corresponding to the 2 types of the strain. ## STRESS - It is defined as the restoring force per unit area setup in the body when deformed by the external force. - That is; - Stress = restoring force / area - Since the restoring force is equal and opposite to the external deforming force, the stress may be defined as either - Stress = external force applied / area - Stress can be of the following 2 types: - (i) Normal stress: defined as the deforming force acting per unit area, normal to the surface of the body. - For example, when a wire is pulled by a force, the force acts along the length of the wire and normal to its cross-sectional area. - (ii) Tangential Stress: defined as the deforming force acting per unit area tangential to the surface. - For example, a body being sheared (when force applied parallel to the surface of the body) is under the tangential Stress. - Stress is measured in Nm-2 in SI units. - The dimensional formula is ML-1T-2. ## STRAIN - It is defined as the ratio of change in dimension of the body to its original dimension. - That is, - Strain = Change in length / original length = ΔL/L - Where L is the original length and ΔL the increase in length. - Strain can be of the following 2 types: - (i) Longitudinal Strain: defined as the increase in length per unit original length when deformed by the external force. It is also called Linear or tensile strain and expressed as the ratio of change in length to the original length. - (ii) Volumetric Strain: defined as the change in volume per unit original volume when deformed by the external force. - Volumetric Strain = Change in volume / Original volume = ΔV/V - Where V is the original volume and ΔV, change in volume. - (iii) Shear Strain: defined as the change in the shape of the body when deformed by the external force. - It is also defined as the angle (in radian) through which a line originally perpendicular to the fixed face gets turned on applying tangential deforming force. This angle through which the reference line turns, is called the angle of shear. ## YOUNG’S MODULUS - It is defined as the ratio of normal stress to the longitudinal strain. - It is denoted by Y and expressed as, - Y = normal stress / longitudinal strain - Y = $\frac{F/a}{L/L}$ - Y = $\frac{FL}{aL}$ - The units of Young's modulus are pascal (Pa) or Nm-2 ## Qy: A wire increases by 10% of its length, when a stress of 108 Nm-2 is applied on it. What is the Young’s modulus of the material of the wire? - Soln: Stress = 108 Nm-2 and Strain = 10-3 - Y = Stress/Strain = 108/10-3 = 1011 Nm-2 ## STRESS VERSUS STRAIN - The direct proportionality between stress and strain is found to be true only for small values of strain (about 0.3% of the original length). - If a wire is gradually loaded and strain produced is plotted against stress, then the graph obtained is of the shape shown in the figure below; - A graph of stress Vs strain with - **OP → proportional limit** - **DE → elastic limit** - **E → permanent set** - **F → yield point** - **G → breaking point** ## BULK MODULUS - It is defined as the ratio of the normal stress to the volumetric strain. - It is denoted by B and given mathematically as: - B = Normal stress / Volumetric Strain - Now consider a sphere of volume V and surface area a. - Diagram of a sphere with force F applied to it, and a change in volume from V to V - ΔV within the sphere. - Suppose that a force F which act uniformly over the whole surface of the sphere decreases its volume by ΔV, then; - normal stress = F/a - & Volumetric Strain = -ΔV/V - The –ve sign indicates that on increasing stress, the volume of the sphere decreases. - Therefore, $B = \frac{F/a}{-\Delta V/V} = -\frac{FV}{a\Delta V}$ - Since F/a is Pi then, - $B = -\frac{pV}{\Delta V}$ - The units of bulk modulus are Pa or Nm-2 in SI. ## Compressibility - The reciprocal of the bulk modulus of a material is called its compressibility. It is denoted by K and mathematically given as: - K = 1/B - The units are Pa-1 or Nm-2. ## Example: Find the bulk modulus of water from the following data. - Initial volume = 500.0 litre - Pressure increase = 100.0 atm - Final volume = 497.5 litre - Given that 1 atm = 1.013 x 105 Pa. - Soln: - p = 100.0 atm = 100.0 x 1.013 x 105 = 1.013 x 107 Pa (or Nm-2) - Initial volume = 500.0 litre = 500.0 x 10-3 m3 - Final volume = 497.5 litre = 497.5 x 10-3 m3 - Therefore, decrease in volume of Water, - ΔV = - 500.0 x 10-3 + 497.5 x 10-3 = -2.5 x 10-3 m3 - $\implies B = -\frac{pV}{\Delta V} = - \frac{1.013 \times 10^7 \times 500.0 \times 10^{-3}}{-2.5 \times 10^{-3}} = 2.026 \times 10^9 Pa$ ## SHEAR MODULUS (MODULUS OF RIGIDITY) - It is defined as the ratio of tangential stress to shear strain. - It is denoted by the Greek letter η (eta). Thus η = Tangential stress / Shear Strain - Consider a rectangular block where face ABCD is fixed and the upper face EFGH is subjected to tangential force F: - Diagram of a rectangular block with arrows showing the directions of force F. - Let a be the area of each of the face and AF = L be the distance between the 2 faces. - The tangential force will shear the rectangular block into a parallelopiped by displacing the upper face through a distance FF’ = X. - If ∠FAF’ = θ, then θ is the angle of shear. - $\implies$ tangential stress = F/a - $\implies$ & shear strain = θ = x/L - $\implies$ η = F/a / x/L - From ΔAFF’, we have tan θ = FF’/AF = X/L - In practice, for solids, the angle of shear is very small. Therefore tan θ ≈ θ = x/L - The distance x through which the upper face has been displaced is called lateral displacement. - Therefore, Shear strain may also be defined as the ratio of the lateral displacement to its distance from the fixed layer. - The unit is radian or Nm-2. ## Example: A cube of aluminium of each side 4 cm is subjected to a tangential (shearing) force. The top face of the cube is sheared through 0.02 cm w.r.t the bottom face. Find (i) Shearing strain (ii) Shearing stress (iii) Shearing force - Given that η = 2.08 x 1010 dyne cm-2 - η = 2.08 x 1010 x10-5 Nm-2 x 10-4 m-2 = 2.08 x 102 Nm-2 - Soln: - (i) Shearing strain θ = x/L = 0.012 cm / 4 cm = 0.003 - (ii) Shearing stress; - From η = F/a / θ = F/a / x/L - $\implies$ F/a = η θ - $\implies$ F = η θ a - $\implies$ F= 2.08 x 102 Nm-2 x 0.003 x (4 cm)2 = 6.24 x 10-2 Nm-2 = 6.24 x 10-2 N / 10-4 m2 - (iii) Shearing Force F = Shearing Stress x Area = 6.24 x 10-2 Nm-2 x (4 x 10-2 m)2 = 9.984 x 10-4 N

PHY 103 General Physics III Lecture Notes PDF

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