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1.Basic Maths (1) : Vectors
- Vector and Scalar, Representation of Vectors, Need for Co-ordinate System, Distance & Displacement
- Mathematics of Vectors, Triangle Law and Parallelogram Law
- Addition More than Two Vectors, Subtraction of Vectors- Displacement vector
- Elementary Maths
- Unit Vectors, Special Unit Vectors, Resolution of Vectors
- Addition & Subtract using Unit Vectors, 3 D Vectors, Product of Vectors
- Chapter Notes – Basic Maths (1) : Vectors
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2.Basic Maths (2) : Calculus
- Delta, Concept of Infinity, Time Instant Interval, Rate of Change, Position and Velocity
- Fundamental Idea of Differentiation- Constant Multiplication Rule, Sum/Difference Rule
- Trigonometric functions, Log function, Product Rule, Quotient Rule, Chain Rule
- Integration- Formulas of Integration, Use of Integration
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3.Unit and Measurement
- Unit, History of Unit of Length-Metre, Properties of a Good Unit
- Concept of Derived Units, Fundamental Physics Quantities and Prefix of Units
- Unit-less Derived Quantities, Supplementary Quantities, Systems of Unit, Unit Conversion
- Dimensional Analysis, Dimension and Unit, Dimensionless Quantities
- Principle of Homogeneity
- Dimensionally Correct/Incorrect Equations, Use of Dimensional Analysis
- More Units of Length and Measurement of Length
- Errors and Their Reasons
- Combination of Errors
- Round Off, Significant Figures, Exponent Form of Numbers/Scientific Notation
- Chapter Notes – Unit and Measurement
- NCERT Solutions – Unit and Measurement
- Revision Notes – Unit and Measurement
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4.Motion (1) : Straight Line Motion
- Meaning of Dimension; Position; Distance & Displacement
- Average Speed & Velocity; Instantaneous Speed & Velocity
- Photo Diagram; Acceleration- Direction of acceleration, Conceptual Examples
- Constant Acceleration; Equations of constant acceleration
- Average Velocity Examples and Concepts; Reaction Time
- Free Fall under Gravity
- Variable Acceleration; Derivation of Constant Acceleration Equations
- Chapter Notes – Motion (1) : Straight Line Motion
- NCERT Solutions – Straight Line Motion
- Revision Notes Straight Line Motion
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5.Motion (2) : Graphs
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6.Motion (3) : Two Dimensional Motion
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7.Motion (4) : Relative Motion
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8.Newton's Laws of Motion
- Force and Newton’s Laws
- Normal Reaction, Free Body Diagram(F.B.D), Normal on circular bodies, Mass and Weight
- Tension Force(Ideal Pulley, Clamp Force), Internal & External Force, Heavy Rope
- Spring Force(Sudden Change, Series and Parallel Cutting of Spring)
- Inertia and Non-Inertial Frames(Pseudo Force), Action-Reactin Pair, Monkey Problem
- Chapter Notes – Newton’s Laws of Motion
- NCERT Solutions – Laws of Motion
- Revision Notes Laws of Motion
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9.Constrain Motion
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10.Friction
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11.Circular Motion
- Ex. on Average Acc. and Angular Variables Theory and Ref. Frame
- Uniform Circular Motion and Centripetal Force
- Non-Uniform Center of Mass – Theory by Ex 2; Friction
- Centrifugal Force and Banking of Roads
- Radius of Curvature- Radius of Curvature; Axial Vector; Well of Death
- Chapter Notes – Circular Motion
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12.Work Energy Power
- Work & its calculation and Work-done on curved path
- Work-done by Different Forces
- Work Energy Theorem and W.E. th in Non-inertial frame, W.E. th and Time
- Work Energy Theorem for System
- Energy and Different Forms of Energy-and Energy of Chain; Potential Energy & Reference Frame
- Potential Energy Curve and Power
- Normal Reaction, Vertical Circular Motion, Motion in Co-Concentric Spheres
- Motion on Outer Surface of Sphere, Motion on Inner Surface of Fixed Sphere
- Motion on Rope, Motion on Rod
- VCM – 1
- VCM – 2
- VCM – 3
- Chapter Notes – Work Energy Power
- NCERT Solutions – Work Energy Power
- Revision Notes Work Energy Power
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13.Momentum
- Introduction and Conservation of Momentum
- Impulsive Force – Characteristics of Impulsive Force
- Momentum Conservation in Presence of External Force – Two Steps Problems
- Questions Involving Momentum & Work Energy Theorem
- Collision – Head – on Collision and Special Cases of Head – on Collision
- Oblique Collision
- Collision of Ball with Flat Surface
- Impulse and Average Force
- Advanced Questions
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14.Center of Mass
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15.Rotational Motion
- Rigid Body – Motion of Rigid Body; Axis of Rotation
- Vector Product/ Cross Product; Torque
- Couple and Principle of Moments
- Pseudo Force and Toppling – Overturning of Car
- Moment of Inertia
- Parallel Axis Theorem; Perpendicular Axis Theorem; Quantitative Analysis; Radius of Gyra
- Analogy b/w Transnational & Rotational Motion; Relation b/w Linear and Angular Velocity; Dynamics of Rotation
- Angular Momentum
- Angular Momentum of a Particle
- Rotational Collision
- Kinetic Energy, Work, Power; Potential Energy; Linear & Angular Acceleration; Hinge Force; Angular Impulse
- Chapter Notes – Rotational Motion and Rolling Motion
- NCERT Solutions – Rotational Motion
- Revision Notes Rotational Motion
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16.Rolling Motion
- Introduction to Rolling Motion
- Rolling Motion on Spool
- Friction
- Direction of Friction
- Rolling on Moving Platform and Motion of Touching Spheres
- Rope Based Questions
- Work-done by Friction in Rolling Motion, Kinetic Energy in Transnational + Rotational Motion
- Angular Momentum in Rotation + Translation
- Angular Collision
- Instantaneous Axis of Rotation
- De-Lambart’s Theorem
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17.Gravitation
- Gravitation force, Universal Law of Gravitation, Gravitational Force due to Hollow Sphere and Solid Sphere
- Acceleration due to Gravity and Rotation of Earth
- Potential Energy, Questions and Solutions
- Satellites, Circular Motion, Geostationary Satellites and Polar Satellites
- Polar Satellites, Weightlessness in Satellites, Trajectories and Kepler’s Laws
- Chapter Notes – Gravitation
- NCERT Solutions – Gravitation
- Revision Notes Gravitation
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18.Simple Harmonic Motion
- Oscillatory Motion – Horizontal Spring Block System, Qualitative Analysis of Horizontal Spring System
- Quantitative Analysis of Horizontal Spring System; Frequency and Angular Frequency; Velocity and Acceleration; Mechanical Energy
- Relating Uniform Circular Motion and SHM and Phasor Diagram
- Equation of SHM and Problem Solving using Phasor Diagram
- Questions
- More Oscillating Systems – Vertical Spring Block System
- Angular Oscillations – Simple Pendulum
- Compound / Physical Pendulum, Torsional Pendulum, Equilibrium of Angular SHM; Differentiation by Chain Rule
- Energy Method to find Time Period
- Finding Amplitude of SHM
- Block Over Block and Elastic Rope
- Superposition of Horizontal SHMs and Perpendicular
- Damped Oscillations
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19.Waves (Part-1)
- Wave, Plotting and Shifting of Curves, Meaning of y/t and y/x Graph, Wave is an Illusion!, 1D Wave on String
- Wave Equation, Analysis of Wave Equation and Wave Velocity
- Sinusoidal Wave (Harmonic Wave), Wave Equation for Sinusoidal Wave, Particle Velocity, Slope of Rope, Wave Velocity
- Superposition of Waves
- Reflection of Waves
- Standing Waves
- Tuning Fork, Sonometer and Equation of Standing Waves
- Energy in Waves
- Chapter Notes – Waves
- NCERT Solutions – Waves
- Revision Notes Waves
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20.Waves (Part-2)
- Waves, Propagation of Sound Wave and Wave Equation
- Sound as a Pressure Wave
- Speed of Sound, Laplace Correction and Intensity of Sound Waves
- Spherical and Cylindrical Sound Waves
- Addition of Sin Functions, Interference of Sound Waves of Same Frequency, Interference of Coherent Sources
- Quinke’s Apparatus
- Interference of Sound Waves of Slightly Different Frequencies (Beats)
- Reflection of Sound Waves, Standing Waves, End Correction
- Standing Waves in Terms of Pressure, Standing Waves on Rods, Kund’s Tube, Resonance Tube Experiment
- Doppler Effect, Reflection from Wall, Doppler Effect in 2 Dimension
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21.Mechanical Properties of Solids
- Rigid body,Strain, Stress,Hook’s Law
- Breaking Stress
- Shear Stress and Strain, Bulk Modulus, Elasticity and Plasticity, Stress-Strain Curve, Young’s Modulus
- Chapter Notes – Mechanical Properties of Solids
- NCERT Solutions – Mechanical Properties of Solids
- Revision Notes Mechanical Properties of Solids
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22.Thermal Expansion
- Linear Expansion; Second’s Pendulum; Bimetallic Strip; Expansion of Hole; Thermal Stress
- Areal/Superficial Expansion; Volume Expansion; Thermal Expansion of Liquid; Measurement of Temperature; Anomal
- Arial/Superficial Expansion; Volume Expansion; Thermal Expansion of Liquid; Measurement of Temperature
- Chapter Notes – Thermal Expansion
- NCERT Solutions – Thermal Expansion
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23.Heat and Calorimetry
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24.Heat Transfer
- Conduction; Comparison between Charge Flow & Heat Flow
- Equivalent Thermal Conductivity; Heat Transfer and Calorimetry; Use of Integration; Length Variation
- Convection; Radiation, Black Body, Prevost Theory, Emissive Power & Emissivity, Kirchoff’s Law, Stefan – Boltzman Law
- Newton’s Law of Cooling, Cooling Curve; Wien’s Displacement Law; Thermo Flask
- Chapter Notes – Heat Transfer
- Revision Notes Heat Transfer
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25.Kinetic Theory of Gases
- Model of Gas,Postulates of Kinetic Theory of Gases, Ideal Gas, Mean free Path, Maxwell’s speed Distribution
- Volume, Pressure of Gases, Kinetic Energy, Temperature, Ideal Gas Equation
- Gas Laws, Internal energy of Gas, Degree of Freedom, Degree of Freedom of Mono-atomic and Diatomic Gas
- Chapter Notes – Kinetic Theory of Gases
- NCERT Solutions – Kinetic Theory of Gases
- Revision Notes Kinetic Theory of Gases
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26.Thermodynamics
- State Equation; Thermodynamic Process; Process Equation & Graph; Work done by Gas
- Heat – Work Equivalence; 1st Law of Thermodynamics; Adiabatic Process
- Workdone in Adiabatic Process; Specific Molar Heat Capacity
- Poly-tropic Process, Bulk Modulus; Free Expansion; Mixture of Gases
- Heat Engine, Refrigerator or Heat Pump, Energy Conservation, Kelvin-Plank Statement, Clausius Statement
- Carnot Cycle, Reversible and Irreversible Process, Specific Heat Capacity of Solids and Water
- Chapter Notes – Thermodynamics
- NCERT Solutions – Thermodynamics
- Revision Notes Thermodynamics
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27.Fluids
- Introduction, Pressure of Liquid
- Manometer, Barometer
- Pascal Law, Hydraulic Lift
- Accelerated Liquid, Vertical and Horizontal Acceleration, Pressure Variation in Horizontally Accelerated Liquid
- Rotating Liquid, Rotating Liquid in U-Tube
- Archimedes’ Principle, Hollow Objects
- Apparent Weight, Variation of Liquid Force with Height
- Multiple Liquids
- Center of Bouyancy
- Fluid Dynamics, Equation of Continuity
- Magnus Effect
- Venturimeter, Pitot Tube
- Questions and Solutions
- Chapter Notes – Fluids
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28.Surface Tension and Viscosity
- Surface Tension, Surface Energy
- Force of Cohesion, Force of Adhesion, Angle of Contact, Radius of Meniscus, Capillary Rise
- Pressure Difference Across Meniscus, Variation of Surface tension with Temperature
- Viscous Force
- Terminal Velocity, Velocity Gradient, Renolds Number, Turbulent Flow, Streamline Flow
- Chapter Notes – Surface Tension and Viscosity
NCERT Solutions – Kinetic Theory of Gases
Answer
Diameter of an oxygen molecule, d = 3Å
Radius, r = d/2 = 3/2 = 1.5 Å = 1.5 × 10–8 cm
Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm3
Molecular volume of oxygen gas, V = (4/3)πr3N
Where, N is Avogadro’s number = 6.023 × 1023 molecules/mole
∴ V = (4/3) × 3.14 × (1.5 × 10-8)3 × 6.023 × 1023 = 8.51 cm3
Ratio of the molecular volume to the actual volume of oxygen = 8.51 / 22400
= 3.8 × 10-4.
13.2. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Answer
The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:
PV = nRT
Where,
R is the universal gas constant = 8.314 J mol–1 K–1
n = Number of moles = 1
T = Standard temperature = 273 K
P = Standard pressure = 1 atm = 1.013 × 105 Nm–2
∴ V = nRT / P
= 1 × 8.314 × 273 / (1.013 × 105)
= 0.0224 m3
= 22.4 litres
Hence, the molar volume of a gas at STP is 22.4 litres.
13.3. Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true: T1 > T2 or T1 < T2?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00 ×10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H2 = 2.02μ, of O2 = 32.0μ, R = 8.31 J mo1–1 K–1.)
Answer
(a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio PV/T is equal.
μR (μ is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas.
(b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature T1 is closer to the dotted plot than the curve of the gas at temperature T2. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, T1 > T2 is true for the given plot.
(c) The value of the ratio PV/T, where the two curves meet, is μR. This is because the ideal gas equation is given as:
PV = μRT
PV/T = μR
Where,
P is the pressure
T is the temperature
V is the volume
μ is the number of moles
R is the universal constant
Molecular mass of oxygen = 32.0 g
Mass of oxygen = 1 × 10–3 kg = 1 g
R = 8.314 J mole–1 K–1
∴ PV/T = (1/32) × 8.314
= 0.26 J K-1
Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is
0.26 J K–1.
(d) If we obtain similar plots for 1.00 × 10–3 kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).
We have:
PV/T = 0.26 J K-1
R = 8.314 J mole–1 K–1
Molecular mass (M) of H2 = 2.02 u
PV/T = μR at constant temperature
Where, μ = m/M
m = Mass of H2
∴ m = (PV/T) × (M/R)
= 0.26 × 2.02 / 8.31
= 6.3 × 10–2 g = 6.3 × 10–5 kg
Hence, 6.3 × 10–5 kg of H2 will yield the same value of PV/T.
Page No: 334
13.4. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32μ).
Answer
Volume of oxygen, V1 = 30 litres = 30 × 10–3 m3
Gauge pressure, P1 = 15 atm = 15 × 1.013 × 105 Pa
Temperature, T1 = 27°C = 300 K
Universal gas constant, R = 8.314 J mole–1 K–1
Let the initial number of moles of oxygen gas in the cylinder be n1.
The gas equation is given as:
P1V1 = n1RT1
∴ n1 = P1V1/ RT1
= (15.195 × 105 × 30 × 10-3) / (8.314 × 300) = 18.276
But n1 = m1 / M
Where,
m1 = Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
∴m1 = n1M = 18.276 × 32 = 584.84 g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume, V2 = 30 litres = 30 × 10–3 m3
Gauge pressure, P2 = 11 atm = 11 × 1.013 × 105 Pa
Temperature, T2 = 17°C = 290 K
Let n2 be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
P2V2 = n2RT2
∴ n2 = P2V2/ RT2
= (11.143 × 105 × 30 × 10-3) / (8.314 × 290) = 13.86
But n2 = m2 / M
Where,
m2 is the mass of oxygen remaining in the cylinder
∴ m2 = n2M = 13.86 × 32 = 453.1 g
The mass of oxygen taken out of the cylinder is given by the relation:
Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder
= m1 – m2
= 584.84 g – 453.1 g
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.
13.5. An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
Answer
Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10–6 m3
Bubble rises to height, d = 40 m
Temperature at a depth of 40 m, T1 = 12°C = 285 K
Temperature at the surface of the lake, T2 = 35°C = 308 K
The pressure on the surface of the lake:
P2 = 1 atm = 1 ×1.013 × 105 Pa
The pressure at the depth of 40 m:
P1 = 1 atm + dρg
Where,
ρ is the density of water = 103 kg/m3
g is the acceleration due to gravity = 9.8 m/s2
∴ P1 = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa
We have P1V1 / T1 = P2V2 / T2
Where, V2 is the volume of the air bubble when it reaches the surface
V2 = P1V1T2 / T1P2
= 493300 × 1 × 10-6 × 308 / (285 × 1.013 × 105)
= 5.263 × 10–6 m3 or 5.263 cm3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.
13.6. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.
Answer
Volume of the room, V = 25.0 m3
Temperature of the room, T = 27°C = 300 K
Pressure in the room, P = 1 atm = 1 × 1.013 × 105 Pa
The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as:
PV = kBNT
Where,
KB is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1
N is the number of air molecules in the room
∴ N = PV / kBT
= 1.013 × 105 × 25 / (1.38 × 10-23 × 300)
= 6.11 × 1026 molecules
Therefore, the total number of air molecules in the given room is 6.11 × 1026.
13.7. Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).
(i) At room temperature, T = 27°C = 300 K
Average thermal energy = (3/2)kT
Where k is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1
∴ (3/2)kT = (3/2) × 1.38 × 10-38 × 300
= 6.21 × 10–21J
Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10–21 J.
(ii) On the surface of the sun, T = 6000 K
Average thermal energy = (3/2)kT
= (3/2) × 1.38 × 10-38 × 6000
= 1.241 × 10-19 J
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19 J.
(iii) At temperature, T = 107 K
Average thermal energy = (3/2)kT
= (3/2) × 1.38 × 10-23 × 107
= 2.07 × 10-16 J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16 J.
13.8. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?
Answer
All the three vessels have the same capacity, they have the same volume.
Hence, each gas has the same pressure, volume, and temperature.
According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro’s number, N = 6.023 × 1023.
where,
k is Boltzmann constant
For the given gases, k and T are constants.
Hence vrms depends only on the mass of the atoms, i.e.,
vrms ∝ (1/m)1/2
Therefore, the root mean square speed of the molecules in the three cases is not the same. Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.
13.9. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer
Temperature of the helium atom, THe = –20°C= 253 K
Atomic mass of argon, MAr = 39.9 u
Atomic mass of helium, MHe = 4.0 u
Let, (vrms)Ar be the rms speed of argon.
Let (vrms)He be the rms speed of helium.
The rms speed of argon is given by:
= 2523.675 = 2.52 × 103 K
Therefore, the temperature of the argon atom is 2.52 × 103 K.
13.10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).
Answer
Mean free path = 1.11 × 10–7 m
Collision frequency = 4.58 × 109 s–1
Successive collision time ≈ 500 × (Collision time)
Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × 105 Pa
Temperature inside the cylinder, T = 17°C =290 K
Radius of a nitrogen molecule, r = 1.0 Å = 1 × 1010 m
Diameter, d = 2 × 1 × 1010 = 2 × 1010 m
Molecular mass of nitrogen, M = 28.0 g = 28 × 10–3 kg
The root mean square speed of nitrogen is given by the relation:
Collision frequency = vrms / l
= 508.26 / (1.11 × 10-7) = 4.58 × 109 s-1
Collision time is given as:
T = d / vrms
= 2 × 10-10 / 508.26 = 3.93 × 10-13 s
Time taken between successive collisions:
T ‘ = l / vrms
= 1.11 × 10-7 / 508.26 = 2.18 × 10-10 s
∴ T ‘ / T = 2.18 × 10-10 / (3.93 × 10-13) = 500
Hence, the time taken between successive collisions is 500 times the time taken for a collision.
Page No: 335
Additional Exercises
13.11. A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Answer
Length of the narrow bore, L = 1 m = 100 cm
Length of the mercury thread, l = 76 cm
Length of the air column between mercury and the closed end, la = 15 cm
Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is: 100 – (76 + 15) = 9 cm
Hence, the total length of the air column = 15 + 9 = 24 cm
Let h cm of mercury flow out as a result of atmospheric pressure.
∴Length of the air column in the bore = 24 + h cm
And, length of the mercury column = 76 – h cm
Initial pressure, P1 = 76 cm of mercury
Initial volume, V1 = 15 cm3
Final pressure, P2 = 76 – (76 – h) = h cm of mercury
Final volume, V2 = (24 + h) cm3
Temperature remains constant throughout the process.
∴ P1V1 = P2V2
76 × 15 = h (24 + h)
h2 + 24h – 1140 = 0
Height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm.
13.12. From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s–1. Identify the gas.
[Hint: Use Graham’s law of diffusion: R1/R2 = (M2/M1)1/2, where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]
Answer
Rate of diffusion of hydrogen, R1 = 28.7 cm3 s–1
Rate of diffusion of another gas, R2 = 7.2 cm3 s–1
According to Graham’s Law of diffusion, we have:
32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.
13.13. A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
n2 = n1 exp [-mg (h2 – h1)/ kBT]
Where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n2 = n1 exp [-mg NA(ρ – P′) (h2 –h1)/ (ρRT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [NA is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]
Answer
According to the law of atmospheres, we have:
n2 = n1 exp [-mg (h2 – h1) / kBT] … (i)
where,
n1 is the number density at height h1, and n2 is the number density at height h2
mg is the weight of the particle suspended in the gas column
Density of the medium = ρ‘
Density of the suspended particle = ρ
Mass of one suspended particle = m‘
Mass of the medium displaced = m
Volume of a suspended particle = V
According to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:
Weight of the medium displaced – Weight of the suspended particle
= mg – m‘g
= mg – V ρ’ g = mg – (m/ρ)ρ‘g
= mg(1 – (ρ‘/ρ) ) ….(ii)
Gas constant, R = kBN
kB = R / N ….(iii)
Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get:
n2 = n1 exp [-mg (h2 – h1) / kBT]
= n1 exp [-mg (1 – (ρ’/ρ) )(h2 – h1)(N/RT) ]
= n1 exp [-mg (ρ – ρ’)(h2 – h1)(N/RTρ) ]
13.14. Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].
∴ r = [ 3M / 4πρN]1/3
For Carbon,
M = 12.01 × 10-3 Kg
ρ = 2.22 × 103 Kg m-3
