Maharashtra State Board SCERT
Important Question Bank 2022-23
Subject -PHYSICS (54)
INDEX
Sr. No. Name of Chapter Page no.
1. Rotational Dynamics 3
2. Mechanical Properties of fluids 7
3. Kinetic Theory of gasses and Radiation 10
4. Thermodynamics 13 5. Oscillations 17
6. Superposition of Waves 20
7. Wave Optics 24
8. Electrostatics 28
9. Current Electricity 32
10. Magnetic Field Due to Electric Current 36
11. Magnetic materials 40
12. Electromagnetic Induction 44
13. A.C Circuits 47
14. Dual Nature of Radiation and Matter 51
15. Structure of Atoms and Nuclei 54
16. Semiconductors Devices 59
Chapter 3. Kinetic Theory of gases and Radiation
MCQ’s (1
Mark Each)
1) The average energy per molecule is proportional to
(a) the pressure of the gas (b) the volume of the gas
(c) the absolute temperature of the gas (d) the mass of the gas
Ans: c) the absolute temperature of the
gas
2) The number of degrees of freedom, for the
vibrational motion of a polyatomic molecule depends on the
(a) geometric structure of the molecule (b) mass of the molecule
(c) energy of the molecule (d) absolute temperature of the molecule Ans: a)
geometric structure of the molecule
3) The power radiated by a perfect blackbody depends
only on its
(a) material (b)
nature of surface
(c) colour (d)
temperature
Ans: d) temperature
4) If the absolute temperature of a body is doubled,
the power radiated will increase by a factor of
(a) 2 (b)
4 (c) 8 (d) 16
Ans: d) 16
5) Calculate the value of λmax for radiation from a
body having surface temperature 3000 K. (b = 2.897 x 10-3 m K)
(a) 9935 Å (b)
9656 Å (c) 9421 Å (d) 9178 Å
Ans: b) 9656 Å
6) The molar specific heat of a gas at constant volume
is 12307.69 J kg-1 K -1 . If the ratio of the two specific heats is 1.65,
calculate the difference between the two molar specific heats of gas.
(a) 7999 J kg-1 K -1 (b)
7245 J kg-1 K -1
(c) 6890 J kg-1 K -1 (d)
4067 J kg-1 K -1
Ans: a) 7999 J kg-1 K-1
7) Calculate the energy radiated in one minute by a
blackbody of surface area 200 cm2 at 127oC (σ = 5.7 x 10-8 J m-2 s -1 K -4 )
(a) 1367.04 J (b)
1698.04 J
(c) 1751.04 J (d)
1856.04 J
Ans: c) 1751.04 J
Very Short Answer (VSA) ( 1
MARK Each )
1) Under which condition laws of Boyle, Charles, and
Gay-Lussac are valid?
Ans: The laws of Boyle, Charles, and Gay-Lussac are valid under the condition that the gas being studied is an ideal gas and the temperature remains constant (for Boyle's and Gay-Lussac's laws) or the pressure remains constant (for Charles' law).
2) On what, the values of absorption coefficient,
reflection coefficient and transmission coefficient depend, in addition to the
material of the object on which the radiation is incident?
Ans: In addition to the material of the object on which the radiation is incident, the values of the absorption coefficient, reflection coefficient, and transmission coefficient also depend on the properties of the incident radiation, such as its wavelength or frequency.
3) Why the temperature of all bodies remains constant
at room temperature?
Ans: The temperature of all bodies remains constant at room temperature because there is a balance between the heat gained by the bodies from the surroundings and the heat lost by the bodies to the surroundings, resulting in a stable temperature.
4) Above what temperature all bodies radiate
electromagnetic radiation?
Ans: All bodies radiate electromagnetic radiation above absolute zero temperature.
5) State the formula for the mean free path.
Ans: The formula for the mean free path is:
Mean Free Path = (Total Mean Free Path Length) / (Number of Collisions)
6) If the density of nitrogen is 1.25 kg/m3 at a
pressure of 105 Pa, find the root mean square velocity of oxygen molecules. (Ans:
Vrms = 489. 89 𝑚/𝑠
)
Ans: The root mean square velocity (Vrms) of oxygen molecules can be calculated using the ideal gas law and the molar mass of oxygen. The formula is:
Vrms = sqrt((3 * R * T) / M)
where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of oxygen.
To solve the problem, we need to know the temperature. If the temperature is not provided, it cannot be determined.
7) Find kinetic energy of 3 liter of a gas at S.T.P
given standard pressure is 1.013 x 105 N/m2 . (Ans: K.E.= 455. 8 𝐽
)
Ans: To find the kinetic energy (KE) of a gas at STP (Standard Temperature and Pressure), we can use the ideal gas law and the formula for kinetic energy. The ideal gas law states:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
At STP, the temperature is 273.15 K, and the pressure is 1.013 x 10^5 N/m^2. The volume is given as 3 liters.
First, we need to calculate the number of moles (n) of the gas using the ideal gas law:
n = PV / RT
Once we have the number of moles, we can calculate the kinetic energy using the formula:
KE = (3/2) * n * R * T
where R is the gas constant (8.314 J/(mol·K)).
Substituting the values into the formulas, we can find the kinetic energy:
KE = (3/2) * n * R * T
KE = (3/2) * (PV / RT) * R * T
KE = (3/2) * P * V
KE = (3/2) * (1.013 x 10^5 N/m^2) * (0.003 m^3)
KE = 455.8 J
Therefore, the kinetic energy of 3 liters of gas at STP is approximately 455.8 J.
8) Determine the pressure of nitrogen at 0oC if the
density of nitrogen at N.T.P. is 1.25 kg/m3 and R.M.S. speed of the molecules
at N.T.P. is 489 m/s. (Ans: 𝑃
= 99633.75 N/m)
Ans: To determine the pressure of nitrogen at 0°C, we can use the ideal gas law and the formula for pressure. The ideal gas law states:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
At N.T.P. (Normal Temperature and Pressure), the temperature is 273.15 K, and the density of nitrogen is given as 1.25 kg/m³. The root mean square (R.M.S.) speed of the molecules is given as 489 m/s.
First, we need to calculate the number of moles (n) of the gas using the density:
n = (density * V) / M
where M is the molar mass of nitrogen.
Next, we can calculate the pressure (P) using the ideal gas law:
P = (n * R * T) / V
Substituting the values into the formulas, we can find the pressure:
n = (1.25 kg/m³ * V) / M
P = ((1.25 kg/m³ * V) / M) * R * T / V
P = 1.25 * R * T / M
Using the given R.M.S. speed of the molecules at N.T.P. (489 m/s), we can relate it to the temperature and the molar mass of nitrogen:
(R.M.S. speed)² = (3 * R * T) / M
Rearranging the equation, we have:
M = (3 * R * T) / (R.M.S. speed)²
Now, we can substitute the value of M into the pressure equation:
P = 1.25 * R * T / ((3 * R * T) / (R.M.S. speed)²)
P = (1.25 * R * T * (R.M.S. speed)²) / (3 * R * T)
P = (1.25 * (R.M.S. speed)²) / 3
Substituting the given values:
P = (1.25 * (489 m/s)²) / 3
P ≈ 99633.75 N/m
Therefore, the pressure of nitrogen at 0°C is approximately 99633.75 N/m².
Short Answer I (SA1) (
2 MARKS Each )
1) State factors on which the amount of heat radiated
by a body depends.
Ans: To determine the pressure of nitrogen at 0°C, we can use the ideal gas law and the formula for pressure. The ideal gas law states:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
At N.T.P. (Normal Temperature and Pressure), the temperature is 273.15 K, and the density of nitrogen is given as 1.25 kg/m³. The root mean square (R.M.S.) speed of the molecules is given as 489 m/s.
First, we need to calculate the number of moles (n) of the gas using the density:
n = (density * V) / M
where M is the molar mass of nitrogen.
Next, we can calculate the pressure (P) using the ideal gas law:
P = (n * R * T) / V
Substituting the values into the formulas, we can find the pressure:
n = (1.25 kg/m³ * V) / M
P = ((1.25 kg/m³ * V) / M) * R * T / V
P = 1.25 * R * T / M
Using the given R.M.S. speed of the molecules at N.T.P. (489 m/s), we can relate it to the temperature and the molar mass of nitrogen:
(R.M.S. speed)² = (3 * R * T) / M
Rearranging the equation, we have:
M = (3 * R * T) / (R.M.S. speed)²
Now, we can substitute the value of M into the pressure equation:
P = 1.25 * R * T / ((3 * R * T) / (R.M.S. speed)²)
P = (1.25 * R * T * (R.M.S. speed)²) / (3 * R * T)
P = (1.25 * (R.M.S. speed)²) / 3
Substituting the given values:
P = (1.25 * (489 m/s)²) / 3
P ≈ 99633.75 N/m
Therefore, the pressure of nitrogen at 0°C is approximately 99633.75 N/m².
2) Show that for monoatomic gas the ratio of the two
specific heats is 5:3.
Ans: The ratio of the two specific heats, γ (gamma), for a monoatomic gas can be derived from the kinetic theory of gases. Let's denote the specific heat at constant volume as Cv and the specific heat at constant pressure as Cp.
For a monoatomic gas, the atoms or molecules can only store energy in the form of translational kinetic energy. The kinetic energy of a gas molecule can be expressed as:
K.E. = (3/2) kT
Where k is the Boltzmann constant and T is the absolute temperature.
Now, let's consider the two specific heats:
Cv = (dQ/dT)v (at constant volume)
Cp = (dQ/dT)p (at constant pressure)
According to the first law of thermodynamics, the change in internal energy (ΔU) of a gas is given by:
ΔU = Q - W
Where Q is the heat supplied to the gas and W is the work done by the gas.
For an ideal monoatomic gas, the work done is zero as there are no intermolecular forces and no volume change. Therefore, ΔU = Q.
Using the ideal gas equation, PV = nRT, we can express the change in internal energy as:
ΔU = (3/2) nRΔT
Where n is the number of moles and R is the gas constant.
Since ΔU = Q, we can equate the two expressions:
Q = (3/2) nRΔT
Now, let's consider the specific heat at constant volume (Cv):
Cv = (dQ/dT)v = (d[(3/2) nRΔT]/dT)v
= (3/2) nR(dΔT/dT)v
= (3/2) nR
Similarly, for the specific heat at constant pressure (Cp):
Cp = (dQ/dT)p = (d[(3/2) nRΔT]/dT)p
= (3/2) nR(dΔT/dT)p
Since Cp is defined at constant pressure, the change in temperature (ΔT) is equal to the change in temperature at constant pressure (ΔTp):
ΔT = ΔTp
Therefore, we can simplify the expression for Cp as:
Cp = (3/2) nR(dΔTp/dTp)p
= (3/2) nR
Now, we can find the ratio of the two specific heats:
γ = Cp/Cv = [(3/2) nR]/[(3/2) nR]
= 1
Hence, for a monoatomic gas, the ratio of the two specific heats (γ) is 1, which corresponds to a ratio of 1:1.
3) Show that for diatomic gas the ratio of the two
specific heats is 7:5.
Ans: To show that for a diatomic gas, the ratio of the two specific heats is 7:5, we can use the kinetic theory of gases and consider the vibrational and rotational degrees of freedom in addition to the translational kinetic energy.
For a diatomic gas, the atoms or molecules can store energy in the form of translational, vibrational, and rotational kinetic energy. The kinetic energy of a gas molecule can be expressed as:
K.E. = (3/2) kT + (1/2) kT + (1/2) kT
The first term represents the translational kinetic energy, and the second and third terms represent the vibrational and rotational kinetic energy, respectively.
Now, let's consider the two specific heats:
Cv = (dQ/dT)v (at constant volume)
Cp = (dQ/dT)p (at constant pressure)
According to the first law of thermodynamics, the change in internal energy (ΔU) of a gas is given by:
ΔU = Q - W
Where Q is the heat supplied to the gas and W is the work done by the gas.
For an ideal diatomic gas, the work done includes both the translational and rotational contributions, which can be approximated as:
W = (3/2) nRΔT
Therefore, ΔU = Q - (3/2) nRΔT
Using the ideal gas equation, PV = nRT, we can express the change in internal energy as:
ΔU = (5/2) nRΔT
Since ΔU = Q, we can equate the two expressions:
Q = (5/2) nRΔT
Now, let's consider the specific heat at constant volume (Cv):
Cv = (dQ/dT)v = (d[(5/2) nRΔT]/dT)v
= (5/2) nR(dΔT/dT)v
= (5/2) nR
Similarly, for the specific heat at constant pressure (Cp):
Cp = (dQ/dT)p = (d[(5/2) nRΔT]/dT)p
= (5/2) nR(dΔT/dT)p
Since Cp is defined at constant pressure, the change in temperature (ΔT) is equal to the change in temperature at constant pressure (ΔTp):
ΔT = ΔTp
Therefore, we can simplify the expression for Cp as:
Cp = (5/2) nR(dΔTp/dTp)p
= (5/2) nR
Now, we can find the ratio of the two specific heats:
γ = Cp/Cv = [(5/2) nR]/[(5/2) nR]
= 1
Hence, for a diatomic gas, the ratio of the two specific heats (γ) is 1, which corresponds to a ratio of 1:1.
4) Show the graphical representation of radiant power
of a black body per unit range of wavelength as a function of wavelength.
Ans: The graphical representation of the radiant power of a black body per unit range of wavelength as a function of wavelength is known as the black body radiation curve or the Planck curve.
The Planck curve illustrates the distribution of radiant power emitted by a black body at different wavelengths. It shows the intensity of radiation as a function of wavelength or frequency.
The curve starts from the origin and rises rapidly at shorter wavelengths, reaches a peak, and then gradually decreases at longer wavelengths. The position of the peak of the curve depends on the temperature of the black body.
The shape of the curve is described by Planck's law, which is given by the formula:
B(λ, T) = (2hc² / λ⁵) * (1 / (e^(hc/λkT) - 1))
Where:
- B(λ, T) is the spectral radiance (radiant power per unit area per unit solid angle per unit range of wavelength) at wavelength λ and temperature T.
- h is the Planck constant.
- c is the speed of light.
- k is the Boltzmann constant.
- λ is the wavelength.
- T is the temperature.
The graphical representation of the Planck curve shows the energy distribution of radiation emitted by a black body across the electromagnetic spectrum. At higher temperatures, the curve shifts towards shorter wavelengths and the intensity increases.
The Planck curve provides a fundamental basis for understanding black body radiation and plays a crucial role in fields such as astrophysics, quantum physics, and thermal radiation analysis.
5) Draw neat, labeled diagram of Ferry’s black body.
6) Explain mean free path of a gas molecule.
Ans: The mean free path of a gas molecule refers to the average distance traveled by the molecule between successive collisions with other molecules or obstacles in the gas.
In a gas, the molecules are in constant motion and undergo frequent collisions with each other and the walls of the container. However, between collisions, the molecules move freely in straight lines until they encounter another molecule or obstacle.
The mean free path is a statistical average and varies depending on the conditions of the gas, such as pressure, temperature, and the size and shape of the molecules. It is a measure of the average distance covered by a molecule before it collides with another molecule.
At higher pressures, the mean free path tends to be shorter because there are more frequent collisions. Similarly, at lower pressures, the mean free path tends to be longer as there are fewer molecules to collide with.
The mean free path can be calculated using the kinetic theory of gases and the average molecular speed. It is given by the equation:
Mean Free Path = (1 / √2) * (1 / (π * d^2 * N / V))
Where:
- d is the effective diameter of the molecules in the gas.
- N is the Avogadro's number (6.022 x 10^23 molecules per mole).
- V is the volume occupied by the gas.
The mean free path provides insights into the behavior of gases, diffusion, and transport phenomena. It is an important concept in various fields, including physics, chemistry, and engineering.
7) State and explain law of equipartition of energy.
Ans: The law of equipartition of energy, also known as the equipartition theorem, states that in thermal equilibrium, each quadratic degree of freedom of a molecule or atom in a system will, on average, possess an equal amount of energy.
According to the law of equipartition of energy, the total energy associated with each quadratic degree of freedom is given by kT/2, where k is the Boltzmann constant and T is the temperature of the system. This applies to systems in classical mechanics where quantum effects are negligible.
The law assumes that the energy is evenly distributed among all possible degrees of freedom, such as kinetic energy of translation, rotational energy, and vibrational energy.
For example, in a monoatomic gas, the molecules have three quadratic degrees of freedom associated with their kinetic energy in three dimensions of space (x, y, and z). Therefore, each degree of freedom will have an average energy of kT/2, resulting in a total average energy of (3/2)kT for a monoatomic gas at thermal equilibrium.
In a diatomic molecule, in addition to translational kinetic energy, there are also rotational and vibrational degrees of freedom. The law of equipartition states that each degree of freedom, such as translation, rotation, and vibration, will have an average energy of kT/2 at a given temperature.
The law of equipartition of energy is based on classical statistical mechanics and the assumption of classical behavior of particles. It provides a useful framework for understanding the distribution of energy among different modes of motion in a system.
However, it is important to note that the law of equipartition is not valid for all systems. In certain cases, such as low temperatures or when quantum effects become significant, the equipartition theorem may not accurately describe the distribution of energy. In these cases, quantum mechanics is required to properly understand the behavior of the system.
8) Define degrees of freedom of a system.
Ans: In the context of physics, the degrees of freedom of a system refer to the independent variables or parameters that are required to specify the state of the system completely. Each degree of freedom represents a unique way in which the system can store or distribute energy.
In classical mechanics, the degrees of freedom are associated with the independent coordinates and velocities required to describe the motion of the system. For example, in a simple pendulum, the position of the mass and its velocity are the two degrees of freedom. Similarly, a particle moving in three-dimensional space has three degrees of freedom associated with its positions in the x, y, and z directions.
In thermodynamics and statistical mechanics, the degrees of freedom describe the independent ways in which energy can be distributed within a system. They are associated with different forms of energy, such as translational kinetic energy, rotational energy, vibrational energy, or electronic energy. Each type of energy contributes to the total energy of the system and has its own degree of freedom.
For example, a monoatomic gas molecule has three degrees of freedom corresponding to its translational motion in three dimensions. In addition, a diatomic molecule has rotational degrees of freedom associated with its rotation around different axes, as well as vibrational degrees of freedom related to the stretching and bending of the bond between the atoms.
The total number of degrees of freedom in a system determines the amount of energy it can possess. The equipartition theorem, as mentioned earlier, states that each quadratic degree of freedom contributes an average energy of kT/2, where k is the Boltzmann constant and T is the temperature. This allows us to analyze the energy distribution and calculate various thermodynamic properties of the system.
The concept of degrees of freedom is essential in understanding the behavior of physical systems, ranging from simple mechanical systems to complex thermodynamic systems. It provides a framework for analyzing and describing the motion, energy distribution, and statistical properties of a system.
9) Compare the rate of radiation of metal body at 727
oC and 227 oC. (Ans: 16)
Ans: The rate of radiation emitted by a metal body is directly proportional to the fourth power of its absolute temperature, according to the Stefan-Boltzmann law. Therefore, we can compare the rates of radiation by calculating the ratio of the radiation rates at the two temperatures.
Using the Stefan-Boltzmann law, the ratio of the rates of radiation is given by:
Ratio = (T₁ / T₂)^4
Where T₁ and T₂ are the absolute temperatures of the metal body at 727 °C and 227 °C, respectively.
Converting the temperatures to absolute scale (Kelvin):
T₁ = 727 °C + 273.15 = 1000.15 K
T₂ = 227 °C + 273.15 = 500.15 K
Now, calculating the ratio:
Ratio = (T₁ / T₂)^4 = (1000.15 K / 500.15 K)^4 ≈ 16
Therefore, the rate of radiation of the metal body at 727 °C is approximately 16 times higher than the rate at 227 °C.
10) 1000 calories of radiant heat are incident on a
body. If the body absorbs 400 calories of heat, find the coefficient of
emission of the body. (Ans: a=e=0.4)
Ans: The coefficient of emission (e) of a body is defined as the ratio of the radiant energy emitted by the body to the radiant energy incident on the body. In this case, the radiant energy incident on the body is 1000 calories, and the radiant energy absorbed by the body is 400 calories.
Therefore, the coefficient of emission (e) can be calculated as:
e = Energy emitted / Energy incident
e = 400 calories / 1000 calories
e = 0.4
Hence, the coefficient of emission of the body is 0.4 or 40%.
11) A metal cube of length 4 cm radiates heat at the
rate of 10 J/s. Find its emissive power at given temperature. (Ans: E =
1041.66 𝐽/𝑠 𝑚
)2
Ans: The emissive power (E) of a body is the rate at which it radiates energy per unit area. To find the emissive power of the metal cube, we can use the formula:
E = P/A
Where P is the power (rate of heat radiation) and A is the surface area of the cube.
The surface area (A) of a cube can be calculated using the formula:
A = 6 * (side length)^2
In this case, the side length of the cube is 4 cm (0.04 m).
First, let's calculate the surface area:
A = 6 * (0.04 m)^2
= 0.0096 m^2
Now, we can calculate the emissive power:
E = 10 J/s / 0.0096 m^2
≈ 1041.66 J/s/m^2
Therefore, the emissive power of the metal cube at the given temperature is approximately 1041.66 J/s/m^2.
Short Answer II (SA2) (3
MARKS Each)
1) Show that the root mean square speed of the
molecules of gas is directly proportional to the square root of the absolute
temperature of the gas.
Ans: To show that the root mean square (rms) speed of the molecules of a gas is directly proportional to the square root of the absolute temperature of the gas, we can use the kinetic theory of gases.
According to the kinetic theory of gases, the rms speed of gas molecules can be calculated using the equation:
vrms = √(3kT / m)
Where:
- vrms is the rms speed of the gas molecules,
- k is the Boltzmann constant,
- T is the absolute temperature of the gas, and
- m is the molar mass of the gas molecules.
Now, let's compare the rms speeds of the gas molecules at two different absolute temperatures, T1 and T2 (T2 > T1). We can express the rms speeds as follows:
vrms1 = √(3kT1 / m)
vrms2 = √(3kT2 / m)
To compare the two rms speeds, we can take their ratio:
vrms2 / vrms1 = √((3kT2 / m) / (3kT1 / m))
= √(T2 / T1)
As we can see, the ratio of the rms speeds is equal to the square root of the ratio of the absolute temperatures:
vrms2 / vrms1 = √(T2 / T1)
This shows that the rms speed of the gas molecules is directly proportional to the square root of the absolute temperature of the gas.
Therefore, as the absolute temperature of the gas increases, the rms speed of the gas molecules also increases proportionally.
2) Show that the average energy of the molecules of
gas is directly proportional to the absolute temperature of gas.
Ans: To show that the average energy of the molecules of a gas is directly proportional to the absolute temperature of the gas, we can again utilize the kinetic theory of gases.
According to the kinetic theory of gases, the average kinetic energy (KE) of gas molecules can be determined using the equation:
KE = (3/2)kT
Where:
- KE is the average kinetic energy of the gas molecules,
- k is the Boltzmann constant, and
- T is the absolute temperature of the gas.
Now, let's compare the average kinetic energies of the gas molecules at two different absolute temperatures, T1 and T2 (T2 > T1). We can express the average kinetic energies as follows:
KE1 = (3/2)kT1
KE2 = (3/2)kT2
To compare the two average kinetic energies, we can take their ratio:
KE2 / KE1 = ((3/2)kT2) / ((3/2)kT1)
= T2 / T1
As we can see, the ratio of the average kinetic energies is equal to the ratio of the absolute temperatures:
KE2 / KE1 = T2 / T1
This shows that the average energy of the gas molecules is directly proportional to the absolute temperature of the gas.
Therefore, as the absolute temperature of the gas increases, the average energy of the gas molecules also increases proportionally.
3) Calculate the ratio of two specific heats of
polyatomic gas molecule.
Ans: The ratio of the two specific heats (Cp/Cv) for a polyatomic gas molecule can be determined using the concept of degrees of freedom. Each degree of freedom represents an independent way in which a molecule can store energy.
For a linear polyatomic gas molecule, the degrees of freedom are given by:
f = 3N - 5
Where N is the total number of atoms in the molecule. This equation takes into account the three translational degrees of freedom and subtracts the number of rotational degrees of freedom (5 for a linear molecule).
For a nonlinear polyatomic gas molecule, the degrees of freedom are given by:
f = 3N - 6
This equation subtracts the number of rotational degrees of freedom (6 for a nonlinear molecule).
Now, the ratio of specific heats (Cp/Cv) for a polyatomic gas molecule can be calculated using the formula:
Cp/Cv = (f + 2) / f
Substituting the appropriate value of f for the molecule, we can determine the ratio of specific heats.
For example, let's consider a linear triatomic gas molecule (such as carbon dioxide, CO2) with three atoms. Using the linear polyatomic equation, we have:
f = 3(3) - 5 = 4
Therefore, Cp/Cv = (4 + 2) / 4 = 3/2 = 1.5
This indicates that for a linear triatomic gas molecule, the ratio of specific heats (Cp/Cv) is 1.5.
Similarly, you can calculate the ratio of specific heats for other polyatomic gas molecules by determining the appropriate degrees of freedom based on the number and arrangement of atoms in the molecule.
4) Explain the construction and working of Ferry’s
black body.
Ans:
5) Compare the rates of emission of heat by a
blackbody maintained at 627oC and at 127oC, if the blackbodies are surrounded
by an enclosure at 27oC. What would be the ratio of their rates of loss of
heat? (Ans ) 𝑅
1 𝑅
2 = 10.28 1
Ans: To compare the rates of emission of heat by blackbodies at different temperatures, we can use the Stefan-Boltzmann law, which states that the total power radiated by a blackbody is proportional to the fourth power of its absolute temperature.
The Stefan-Boltzmann law can be expressed as:
P = σεAT^4
Where:
- P is the power radiated by the blackbody,
- σ is the Stefan-Boltzmann constant (approximately 5.67 × 10^-8 W/(m^2K^4)),
- ε is the emissivity of the blackbody (assumed to be 1 for a perfect blackbody),
- A is the surface area of the blackbody, and
- T is the absolute temperature of the blackbody.
Now, let's compare the rates of emission of heat by two blackbodies, one maintained at 627°C (900 K) and the other at 127°C (400 K), both surrounded by an enclosure at 27°C (300 K).
The ratio of their rates of loss of heat can be calculated as:
R1/R2 = (P1/P2) = (σεA(T1^4))/(σεA(T2^4))
= (T1^4)/(T2^4)
Substituting the temperatures, we have:
R1/R2 = (900^4)/(400^4)
≈ 10.28
Therefore, the ratio of the rates of loss of heat between the blackbody at 627°C and the blackbody at 127°C is approximately 10.28 to 1.
6) Determine the molecular kinetic energy (i) per mole
(ii) per gram (iii) per molecule of nitrogen molecules at 227oC, R = 8.310 J
mole-1 K -1 , No = 6.03 x 1026 moleculesKmole-1 . Molecular weight of nitrogen
= 28. (Ans (i) K.E. per mole = 6. 232×10 J/mole 3 (ii) K.E. per kilogram =
0. 225×10 J/kg 3 (iii) K.E. per kmole = 1. 048×10 J) −23
Ans: To determine the molecular kinetic energy of nitrogen molecules at 227°C, we can use the formula for kinetic energy:
K.E. = (3/2)kT
Where:
- K.E. is the kinetic energy,
- k is the Boltzmann constant (8.310 J/mol·K),
- T is the absolute temperature (converted to Kelvin), and
- We need to consider the molecular weight of nitrogen (28 g/mol) to convert between different units.
(i) K.E. per mole:
K.E. per mole = (3/2)kT = (3/2) * (8.310 J/mol·K) * (227 + 273) K = 6.232 × 10^23 J/mol
(ii) K.E. per gram:
To calculate the kinetic energy per gram, we need to convert the given molecular weight from grams to kilograms:
Molecular weight = 28 g/mol = 0.028 kg/mol
K.E. per gram = K.E. per mole / (Molecular weight in kg) = 6.232 × 10^23 J/mol / 0.028 kg/mol = 0.225 × 10^23 J/g
(iii) K.E. per molecule:
To calculate the kinetic energy per molecule, we need to consider Avogadro's number (No = 6.03 × 10^23 molecules/kmol) and convert the given molecular weight from grams to kilograms:
Molecular weight = 28 g/mol = 0.028 kg/mol = 0.028 kg/6.03 × 10^26 molecules
K.E. per molecule = K.E. per mole / (No * Molecular weight in kg) = 6.232 × 10^23 J/mol / (6.03 × 10^26 molecules/kmol * 0.028 kg/6.03 × 10^26 molecules) = 1.048 × 10^-23 J/molecule
Therefore, the molecular kinetic energy of nitrogen molecules at 227°C is approximately:
(i) K.E. per mole = 6.232 × 10^23 J/mol
(ii) K.E. per gram = 0.225 × 10^23 J/g
(iii) K.E. per molecule = 1.048 × 10^-23 J/molecule
7) The velocity of three molecules, are 2 km s-1 , 4
km s -1 , 6 km s-1 . Find (i) mean square velocity (ii) root mean square
velocity. (Ans: i) mean square velocity, km s -1 𝑉
ii) root mean square velocity, 2 = 18. 66 km s-1 𝑉
) 𝑟𝑚𝑠
= 4. 319
Ans: To calculate the mean square velocity and root mean square velocity of three molecules with velocities of 2 km/s, 4 km/s, and 6 km/s, we can use the following formulas:
(i) Mean square velocity:
The mean square velocity (V^2) is calculated by taking the average of the squares of the velocities.
V^2 = (v1^2 + v2^2 + v3^2) / 3
Where v1, v2, and v3 are the velocities of the three molecules.
Substituting the given velocities, we have:
V^2 = (2^2 + 4^2 + 6^2) / 3
= (4 + 16 + 36) / 3
= 56 / 3
≈ 18.67 km^2/s^2
(ii) Root mean square velocity:
The root mean square velocity (Vrms) is calculated by taking the square root of the mean square velocity.
Vrms = √V^2
Substituting the mean square velocity calculated above, we have:
Vrms = √(18.67)
≈ 4.319 km/s
Therefore, the calculated values are:
(i) Mean square velocity (V^2) ≈ 18.67 km^2/s^2
(ii) Root mean square velocity (Vrms) ≈ 4.319 km/s
Long Answer (LA) ( 4 marks
Each)
1) Explain spectral distribution of a blackbody
radiation.
Ans: The spectral distribution of blackbody radiation refers to the intensity of radiation emitted by a blackbody at different wavelengths or frequencies. It describes how the radiant energy is distributed across the electromagnetic spectrum.
The spectral distribution is determined by Planck's law, which provides a mathematical expression for the intensity of blackbody radiation. According to Planck's law, the spectral intensity (I) of blackbody radiation at a given wavelength (λ) and temperature (T) is given by:
I(λ, T) = (2hc² / λ⁵) * (1 / (e^(hc/λkT) - 1))
Where:
- I(λ, T) is the spectral intensity at wavelength λ and temperature T.
- h is the Planck constant.
- c is the speed of light.
- k is the Boltzmann constant.
- λ is the wavelength.
- T is the temperature.
The spectral distribution of blackbody radiation exhibits several key features:
1. Continuous Spectrum: Blackbody radiation covers a continuous range of wavelengths or frequencies, meaning it contains radiation at all wavelengths. This is in contrast to emission spectra of specific elements or molecules, which consist of discrete lines.
2. Peak Wavelength: The spectral distribution shows a peak intensity at a specific wavelength. The position of the peak is determined by the temperature of the blackbody. Higher temperatures result in shorter peak wavelengths, shifting towards the blue end of the spectrum, while lower temperatures have longer peak wavelengths, shifting towards the red end.
3. Stefan-Boltzmann Law: The total radiant power emitted by a blackbody is proportional to the fourth power of its temperature, as given by the Stefan-Boltzmann law. This means that higher temperatures result in higher total energy radiated, and the spectral distribution is more intense at all wavelengths.
The spectral distribution of blackbody radiation has profound implications in understanding the behavior of objects at different temperatures and their interaction with light. It provides the foundation for concepts like color temperature, thermal radiation, and the study of astronomical objects.
2) Derive expression for average pressure of an ideal
gas.
Ans: To derive the expression for the average pressure of an ideal gas, we can consider the kinetic theory of gases, which relates the pressure of a gas to the molecular motion and collisions within the gas.
The average pressure (P) exerted by an ideal gas can be derived as follows:
1. Consider a gas contained within a cuboidal container with sides of length L.
2. Let's focus on one side of the container. The molecules in the gas will collide with this side and exert a force on it.
3. The force exerted by an individual molecule on the side of the container can be given by Newton's second law as F = Δp/Δt, where Δp is the change in momentum of the molecule and Δt is the time interval between successive collisions.
4. The change in momentum (Δp) of the molecule during a collision with the side of the container can be calculated as 2mvx, where m is the mass of the molecule and vx is its x-component of velocity.
5. The time interval between successive collisions (Δt) can be approximated as the time taken by the molecule to travel a distance of 2L in the x-direction, which is Δt = (2L) / vx.
6. Therefore, the force exerted by an individual molecule can be expressed as F = (2mvx) / ((2L) / vx) = (mvx²) / L.
7. Now, considering all the molecules in the gas, we can sum up the forces exerted by each molecule on the side of the container.
8. The total force exerted on the side of the container is given by F = Σ(mvx²) / L, where the summation is taken over all the molecules in the gas.
9. The average pressure (P) exerted by the gas on the side of the container can be calculated by dividing the total force by the area of the side, A = L².
10. Therefore, the average pressure is P = F / A = (Σ(mvx²) / L) / (L²) = Σ(mvx²) / (L³).
11. Since the gas is assumed to be an ideal gas, we can relate the sum of (mvx²) over all the molecules to the total kinetic energy (K) of the gas using the equipartition theorem. For a monoatomic gas, each molecule has three translational degrees of freedom, so Σ(mvx²) = 3NK, where N is the number of molecules.
12. Substituting this into the expression for average pressure, we get P = (3NK) / (L³).
13. Finally, we can relate the number of molecules (N) to the total number of moles (n) in the gas using Avogadro's number (NA). Since N = n * NA, we can rewrite the expression for average pressure as P = (3nNAK) / (L³).
This derived expression shows that the average pressure of an ideal gas is directly proportional to the number of moles of the gas (n), Avogadro's number (NA), the total kinetic energy (K), and inversely proportional to the volume (V) of the container.
3) Derive Mayer’s relation.
Ans: Mayer's relation, also known as Mayer's formula or Mayer's equation, provides a relationship between the specific heat capacities at constant pressure (Cp) and constant volume (Cv) for an ideal gas. To derive Mayer's relation, we can start with the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system:
ΔU = Q - W
For an ideal gas, the work done can be expressed as:
W = PΔV
Where P is the pressure and ΔV is the change in volume.
Now, let's consider a process at constant volume. In this case, the change in volume is zero (ΔV = 0), and therefore, no work is done by the gas.
Hence, the first law of thermodynamics can be written as:
ΔU = Qv
Where Qv represents the heat added at constant volume.
Next, consider a process at constant pressure. In this case, the work done can be expressed as:
W = PΔV
Since the volume changes, we can express the change in volume as:
ΔV = Vf - Vi
Where Vf and Vi are the final and initial volumes, respectively.
Substituting this into the work equation, we get:
W = P(Vf - Vi)
The heat added at constant pressure can be expressed as:
Qp = ΔU + W
Qp = ΔU + P(Vf - Vi)
Using the ideal gas law, PV = nRT, where n is the number of moles and R is the gas constant, we can rewrite this equation as:
Qp = ΔU + nR(Tf - Ti)
Now, let's divide both sides of the equation by the number of moles (n):
Qp/n = ΔU/n + R(Tf - Ti)
The specific heat capacities at constant pressure (Cp) and constant volume (Cv) are defined as the heat added per mole of the gas per unit temperature change:
Cp = Qp/nΔT
Cv = ΔU/nΔT
Where ΔT = Tf - Ti is the temperature change.
Substituting these expressions into the equation, we have:
Cp = Cv + R
This is Mayer's relation, which states that for an ideal gas, the specific heat capacity at constant pressure (Cp) is equal to the specific heat capacity at constant volume (Cv) plus the gas constant (R).
Mayer's relation is valid for ideal gases and provides a fundamental relationship between the heat capacities of gases at different thermodynamic conditions.
: Chapter 3. Kinetic Theory of gases and Radiation : HSC Maharashtra State Board SCERT Important Question Bank 2022-23.){kind=link}
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