VMC VIQ Scholarship Test
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In chapter 13, the Kinetic Theory of Gases is dealt with. In this chapter, the learners will come to know about the gases, their molecular movement, and how these phenomenons can be understood at both macroscopic and microscopic levels. This chapter can seem to be a bit complicated for the students in the beginning. But if one focuses on the terminology, the processes, and their applications step by step alongside NCERT exemplar Class 11 Physics solutions chapter 13 it will become easier to retain. In this chapter, the students are taught about the kinetic theory gases, gas laws of ideal gases, postulates, assumptions, and the non-ideal gas behavior. This chapter and NCERT exemplar Class 11 Physics chapter 13 solutions encapsulate all the topics related to gas molecule movement in a container along with the factors of temperature, pressure, and energy.
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Question:13.1
A cubic vessel (with face horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of in the vertical direction. The pressure of the gas inside the vessel as observed by us on the ground
(a) remains the same because is very much smaller than vrms of the gas.
(b) remains the same because the motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.
(c) will increase by a factor equal to where vrms was the original mean square velocity of the gas.
(d) will be different on the top wall and bottom wall of the vessel.
Answer:
The answer is the option (b)Question:13.2
1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K (figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time,
(a) the pressure on EFGH would be zero
(b) the pressure on all the faces will be equal
(c) the pressure of EFGH would be double the pressure on ABCD
(d) the pressure on EFGH would be half that on ABCD
Answer:
The answer is the option (d)Question:13.3
Boyle’s law is applicable for an
(a) adiabatic process
(b) isothermal process
(c) isobaric process
(d) isochoric process
Answer:
The answer is the option (b)Question:13.4
A cylinder containing an ideal gas is in a vertical position and has a piston of mass M that is able to move up or down without friction ( figure). If the temperature is increased
(a) both P and V of the gas will change
(b) the only P will increase according to Charles’ law
(c) V will change but not P
(d) P will change but not V
Answer:
The answer is the option (c)Question:13.5
Volume versus temperature graphs for a given mass of an ideal gas is shown in the figure. At two different values of constant pressure. What can be inferred about the relation between and ?
(a)
(b)
(c)
(d) Data is insufficient
Answer:
The answer is the option (a)Question:13.6
1 mole of gas is contained in a box of volume at T = 300 K. The gas is heated to a temperature of T= 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)
(a) same as the pressure initially
(b) 2 times the pressure initially
(c) 10 times the pressure initially
(d) 20 times the pressure initially
Answer:
The answer is the option (d)Question:13.7
A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole of oxygen (both considered as ideal). Let denotes the fraction of molecules with speed between v and (v + dv) with , similarly for oxygen. Then,
(a) obeys the Maxwell’s distribution law
(b) will obey the Maxwell’s distribution law separately
(c) neither will obey Maxwell’s distribution law
(d) will be the same
Answer:
The correct answer is the option (b)Question:13.8
An inflated rubber balloon contains one mole of an ideal gas, has a pressure P. volume V and temperature T. If the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will be
(a) 1.1 P
(b) P
(c) less than P
(d) between P and 1.1
Answer:
The correct answer is the option (d)Question:13.9
ABCDEFGH is a hollow cube made of an insulator (figure) face ABCD has a positive charge on it. Inside the cube, we have ionised hydrogen.
The usual kinetic theory expression for pressure
(a) will be valid
(b) will not be valid, since the ions would experience forces other than due to collisions with the walls
(c) will not be valid, since collisions with walls would not be elastic
(d) will not be valid because isotropy is lost
Answer:
The correct answer is the option (b) and (d)Question:13.10
Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory , E is
(a) the total energy per unit volume
(b) only the translational part of energy because rotational energy is very small compared to the translational energy
(c) only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum
(d) the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero
Answer:
The correct answer is the option (c)Question:13.11
In a diatomic molecule, the rotational energy at a given temperature
(a) obeys Maxwell’s distribution
(b) have the same value for all molecules
(c) equals the translational kinetic energy for each molecule
(d) is the translational kinetic energy for each molecule
Answer:
The correct answer is the option (a) and (d)Question:13.12
Which of the following diagrams (figure) depicts ideal gas behaviour?
Answer:
The correct answer is the option (a) and (c)Question:10.13
When an ideal gas is compressed adiabatically, its temperature rises the molecules on the average have more kinetic energy than before. The kinetic energy increases,
(a) because of collisions with moving parts of the wall only
(b) because of collisions with the entire wall
(c) because the molecules get accelerated in their motion inside the volume
(d) because of the redistribution of energy amongst the molecules
Answer:
The correct answer is the option (a)Question:13.14
Calculate the number of atoms in 39.4 g gold. The molar mass of gold is 197 g mole-1.
Answer:
Number of atoms on 197 g gold =Question:13.15
The volume of a given mass of a gas at , 1 atm is 100 cc. What will be its volume at ?
Answer:
Let, = 1 atm, = 1 atm, , = ? ,Question:13.18
Answer:
Degree of freedom is necessary to be known if we need to find the total energy of a particular molecule.Question:13.20
Answer:
According to the ideal gas situation, we have PV = nRTQuestion:13.22
Answer:
The volume of a sphere = the volume of one moleculeQuestion:13.23
Answer:
The volume of a gas is inversely proportional to the pressure of the gas at a constant temperature as stated by Boyle’s law. When air is continuously pumped inside the tyre, the number of moles of gas (i.e. the mass of the gas) increases. Since Boyle’s law is only applicable for a constant mass of gas, it will not hold true in this case.Question:13.24
A balloon has 5.0 mole of helium at . Calculate
(a) the number of atoms of helium in the balloon.
(b) the total internal energy of the system.
Answer:
Helium gas, n = 5 moles and T = 280 KQuestion:13.25
Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP.
Answer:
Degree of freedom of H2 molecule = 3 translational degree of freedom + 2 rotational degree of freedom = 5.Question:13.26
Answer:
Since the gas is monoatomic, only translational degree of freedom will be applicable. KE per molecule = . when the container which is insulated has suddenly stopped, a transfer of kinetic energy takes place to the gas molecules. Let be the slight increase in temperature and n be the number of moles of gas.Question:13.27
Explain why
(a) there is no atmosphere on the moon.
(b) there is a fall in temperature with altitude.
Answer:
a) The acceleration on the moon due to gravity is one-sixth that of EarthQuestion:13.28
Consider an ideal gas with the following distribution of speeds.
Speed | % of molecules |
200 | 10 |
400 | 20 |
600 | 40 |
800 | 20 |
1000 | 10 |
Answer:
(i)Question:13.3
Answer:
Let v1 = volume of box = 1 m3Question:13.31
Answer:
We assume ‘p’ to be the number of molecules per unit volume. Hence ‘p’ is the per unit volume molecular density. Let v be the velocity of gas molecules.NCERT questions are one of the most crucial ones if one wants to score high in their exams and also want to learn better for entrance exams. We have solved all the questions given in the main exercise as per the CBSE pattern and marking scheme. This will help the student learn better keeping in mind how to answer for better scoring. Our NCERT exemplar Class 11 Physics solutions chapter 13 pdf download are also downloadable so that one can access them offline in the PDF format. Simply, click on the given link and it is done.
Also, Read NCERT Solution subject wise -
Also, check NCERT Notes subject wise -
Kinetic theory of gases is the chapter that derives all its topics from the idea that in the gas, the molecules are smaller than the distance between the gas molecules. They move around the container and collide with each other on the walls. Now, this movement of the molecules with the effect of temperature, pressure, and other factors will help in describing the physical properties of the gas. The learner will learn about kinetic theory, its assumptions, and postulates. There are around 9 assumptions like the gas is made of particles, particles have zero volume, there is no interaction between particles, etc.
Students will also learn about the ideal gas’s gas laws like Avogadro’s law, Amonton’s law, Boyles’ law, Charles law, and Graham's law of diffusion. At the end of the chapter and NCERT exemplar Class 11 Physics chapter 13 solutions the learner will come across the Maxwell and Boltzmann energy and velocity distribution law, which has a lot of significance from an exam point of view.
· For a better understanding of the kinetic theory of gases, the first thing that one should pay attention to is the gas laws. 5 gas laws explain how an ideal gas behaves when one factor is kept constant. The formulas of these laws will help in solving numerical based questions in the exam.
· Secondly, a student should pay attention to non-ideal gas behaviour. In this topic, the real gases are discussed who do not follow the ideal gas behaviour under low pressure and high-temperature conditions as per NCERT exemplar Class 11 Physics solutions chapter 13.
· Lastly, learners should pay attention to the Maxwell and Boltzmann law of velocity and energy distribution. In this, the root means square and average velocity is discussed, of the gaseous particles which act under proportional kinetic energy and temperature.
Chapter 1 | Physical world |
Chapter 2 | Units and Measurement |
Chapter 3 | Motion in a straight line |
Chapter 4 | Motion in a Plane |
Chapter 5 | Laws of Motion |
Chapter 6 | Work, Energy and Power |
Chapter 7 | System of Particles and Rotational motion |
Chapter 8 | Gravitation |
Chapter 9 | Mechanical Properties of Solids |
Chapter 10 | Mechanical Properties of Fluids |
Chapter 11 | Thermal Properties of Matter |
Chapter 12 | Thermodynamics |
Chapter 13 | Kinetic Theory |
Chapter 14 | Oscillations |
Chapter 15 | Waves |
Our solutions are detailed and created most exhaustively, and this will help in understanding the topic, solving questions, and is also learning how to answer questions in the exam.
A student who is preparing for exams, for entrance exams like NEET or JEE or any other engineering and medical exam can make the most from these solutions.
Yes, these NCERT exemplar class 11 physics solutions chapter 13 are as per the CBSE pattern, as they are solved step by step and according to the marking scheme.
Application Date:09 September,2024 - 14 November,2024
Application Date:09 September,2024 - 14 November,2024
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