Important Questions Class 11 Physics Chapter 13

Important Questions Class 11 Physics Chapter 13

Important Questions for CBSE Class 11 Physics Chapter 13 – Kinetic Theory

Before the exams, students in Class 11 should review the Important Questions on Chapter 13. By doing so, they will be better able to comprehend the types of questions that may be asked in the Class 11 Physics exams. At Extramarks, these questions were created by subject matter experts for Class 11 Physics based on the questions that came most frequently in last year’s exams and sample question papers. The solutions have been written in a way that is simple and will aid the students in understanding the topics.

Chapter 13 of Class 11 Physics discusses Kinetic Theory. Some of the Important Questions for Class 11 Physics Chapter 13 are asked from atomic theory, the gas laws, the Boltzmann constant, the Avogadro number, the postulates of the kinetic theory, and specific heat capacities. Learning these topics is crucial because students may get several questions based on them.

CBSE Class 11 Physics Chapter 13 Important Questions

Study Important Questions for Class 11 Physics Chapter 13 – Kinetic Theory

Students can refer to some of the 1, 2, 3, 4, and 5-marks questions and answers discussed below. Alternatively, they can access more information on Class 11 Physics Chapter 13 Important Questions by clicking on the link provided here. 

1 Mark Questions and Answers

Q1. Explain Mean Free Path

Ans: The mean free path is the average distance a molecule travels between collisions. It is represented by lambda (). The units of measurement are in metres (m).

Q2. What variables affect the average kinetic energy of gas molecules?

Ans: The only factor directly proportional to average kinetic energy is absolute temperature.

Q3. Explain what an ideal gas is.

Ans: An ideal gas is one that complies with the laws or characteristics listed below.

1. A gas molecule has a size of zero.
2. Gas molecules do not have an attractive or repulsive force.

2 Marks  Questions and Answers

Q1. Make use of the kinetic theory of gases to derive  Boyle’s law.

Ans: Boyle’s law asserts that, the volume of a given amount of gas is inversely proportional to the pressure P, i.e., PV = constant, when the temperature is stable.

The kinetic theory of gases predicts the pressure that a gas will exert.

P = Pressure

V = Volume

V= Average Velocity

m = Mass of 1 molecule

N = Number of molecules

M = mN (Mass of gas)

P = 1mNV-23V

Pv = 13MV-2

Q2. Real gas acts like an ideal gas at high temperatures and very low pressures. Why?

Ans: The molecule volume of an ideal gas is zero, and there are no intermolecular forces present.

1. The amount of gas is so great at very low pressures that the volume of a molecule is negligible in relation to the amount of gas.
2. Intermolecular forces don’t matter since molecules have very high kinetic energies at very high temperatures.

As a result, real gases behave like ideal gases under low pressure.

Q3. What is Graham’s diffusion law and how was it derived?

Ans: Graham’s law of diffusion states that the rates of diffusion of two gases are inversely correlated to the square roots of their densities.

Imagine two gases colliding together at a specific pressure. Let SA and SB represent their densities.

As a result, we can infer that the root mean square velocities of the molecules of gases A and B will be,

VAr∙m∙s = 3PSA→1

VBr∙m∙s = 3PSB→2

Now, dividing equation 1 by 2,

VAr∙m∙sVBr∙m∙s= 3PSASB3P=SBSA→1

The rate of diffusion of a gas is now directly related to its molecules’ r.m.s. velocity. If rA is the diffusion rate of gas A and rB is the diffusion rate of gas B, then

rArB=VArm.sVBrm.s=SBSA

Therefore, Graham’s law can be derived as

rArB=SBSA

3 Marks Questions and Answers

Q1. State the assumptions of the Kinetic Theory of Gas.

Ans: The assumptions of the Kinetic Theory of Gas are as follows.

1. A gas is composed of a large number of molecules, each of which should be an identical elastic sphere.
2. The molecular motion of a gas is always swift and unpredictable.
3. Gas molecules are very small in relation to their distance from one another.
4. The molecules do not exhibit any attraction or repulsive forces.
5. Collisions between molecules and the vessel’s walls are perfectly elastic.

Q2. Nine particles have speeds of 5, 8, 12, 12, 12, 14, 14, 17, and 20 m/s. Calculate the following.

• Average Speed

Ans: To find the average speed, divide the total number of particles by the sum of their speeds.

Average Speed V5+8+12+12+12+14+14+17+209=12.7 m

• Most probable speed

Ans: The average value of the square of speeds is given by,

V2=52+82+122+122+122+142+142+172+2029

V2=25+64+144+144+144+196+196+289+4009

i.e., V2=178 m2/s2

r.m.s. speed, V r.m.s. = V-2=178=13.3m/s

• r.m.s. speed of the particles

Ans: Three particles have a speed of 12, two particles have a speed of 14, and the remaining particles have varying speeds. As a result, the most likely speed is

VmP = 12 m/s.

Q3. Establish the relation between Y=CPCV and degrees of freedom (n).

Ans: Now, Y=CPCV

Where, CP = Specific heat at constant pressure

CV = Specific heat at constant volume

n = Degrees of Freedom (the total number of co-ordinates or independent quantities required to fully describe the position and configuration of the system).

Suppose, a polyatomic gas molecule has ‘n’ degrees of freedom,

∴ Total energy associated with a gram molecule of gas (E) is

E = n×12KT×N=n2.

Here,

N = Total number of molecules

K = Boltzmann Constant

R = Universal Gas Constant

R = NK

So, we know that,

Specific heat at constant volume,

CV = dEdT

CV=ddTn2RT

CV=n2R

Now, Specific heat at constant pressure,

CP = n2R+R

CP=n2+1R

As, Y=CPCV

Y=n2+1Rn2R

Y=n2+12n

Y= 2/’/t’+1×2n

Hence, Y= 1+2n

4 Marks  Questions and Answers

Q1. Show that the molar volume is 22.4 litres at standard temperature and pressure (STP: 1 atmospheric pressure 0°C). One mole of any (ideal) gas occupies one mole of space under these conditions.

Ans: The relationship between Pressure (P), Volume (V), and absolute Temperature (T) is known as the ideal gas equation.

PV = nRT

where,

R = Universal Gas Constant

n = Number of moles = 1

T = Standard temperature = 273 K

P = Standard pressure = 1 atm = 1.013 × 105 Nm-2

Formula is,

∴V= nRTP

Substituting values,

= 1×8.314×2731.013×105=0.0224m3

i.e., V = 22.4 litres

Therefore, a gas’s molar volume at STP is 22.4 litres.

Q2. Gases are present in three equal-capacity vessels at the same pressure and temperature. The first container has monoatomic neon, the second container holds diatomic chlorine, and the third container holds polyatomic uranium hexafluoride. Do the vessels hold the same number of molecules? Are molecules moving at the same root mean square speed in all three scenarios? If not, which scenario represents the largest?

Ans: Yes. They all contain the same amount of molecules.

The highest root mean square speed is neon.

The three jars’ capacities are equal, hence they have the same volume.

As a result, each gas has the same pressure, volume, and temperature.

Avogadro’s law states that an equal number of the matching molecules will be present in each of the three vessels. This value corresponds to Avogadro’s number.

N=6.023×1023

The root mean square speed (vru) of a gas of mass m, and temperature T, is given as,

vr=3kTm

Here,

k is the Boltzmann constant.

As we know, the given gases, k and T are constants,

Hence, vr14 depends only on the mass of the atoms, i.e.,

vrne1m

As a result, the molecules’ root mean square speed varies between the three conditions. Of the three elements—neon, chlorine, and uranium hexafluoride, neon has the smallest mass. Therefore, neon has the fastest root mean square speed among the gases.

Q3. Calculate the average thermal energy of a helium atom at the following.

• Room temperature (27оC)

Ans: At room temperature, T = 27оC = 300K

Average thermal energy = 32kT

k is the Boltzmann constant = 1.38 × 10-23m2kg s-2K-1

∴32kT=32×1.38×10-98×300

=6.2110-21J

• The temperature on the Sun’s surface (6000K)

Ans: T = 6000K

Average thermal energy = 32kT

32×1.38×10-38×6000

= 1.24110-18J

• The temperature at the core of a star (10 million Kelvin)

Ans: T = 107K

Average thermal energy = 32kT

=32×1.38×10-38107

= 2.07 10-16 J

5 Marks  Questions and Answers

Q1. Explain the derivation of the ideal gas equation.

Ans: Consider the pressure exerted by the gas to be p

The volume of the gas to be v

The temperature be at T

The number of moles of gas as n

The Universal Gas Constant as R

According to Boyle’s Law,

The volume of a gas is inversely proportional to the pressure exerted under constant n and T

i.e., v∝1p1

According to Charles’ Law,

The volume of a gas is directly proportional to the temperature under constant p and n.

i.e., v∝n…(2)

According to Avogadro’s Law,

Under constant p and t, the volume of a gas is directly proportional to the number of moles in the gas.

i.e., v∝n…(3)

Combining all  three equations, we derive,

v∝nTp

Or

pv=nRT

Q2. Calculate a nitrogen molecule mean free path and collision frequency in a cylinder filled with nitrogen at a pressure of 2.0 atm and a temperature of 17℃. Assume that a nitrogen molecule has a radius of about 1.0oA. (Molecular mass of N2 = 28.0μ) Compare the collision time to the amount of free time the molecule has in between two collisions.

Ans: We know that, 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

Let temperature inside the cylinder T = οA 17οC-290K

Radius of a nitrogen molecule, r = 1.0 – 1 1010m

Diameter, d = 2 × 1 × 1010 = 2 × 1010m

Molecular mass of nitrogen, M = 28.0 g = 28 × 10-3kg

The relationship gives the nitrogen root mean square speed,

vnse=3RTM

Where,

R is the universal gas constant = 8.314 J mole-1K-1

∴vme=38.3142902810-4=508.26 m/s

The mean free path is given by the relation,

Now, l=1.3810-229023.14(210-12)22.026105

= 1.1110-7m

Collision frequency = vrecl

= 508.261.1110-7=4.58109s-1

Collision time is given as,

T=dvnsL

= 210-10508.26=3.9310-13s

The amount of time that passes between subsequent collisions,

T’ = 1vrm

= 1.1110-7508.26 m/s=2.1810-10s

∴T’T=2.1810-103.9310-11=500

As a result, the period between subsequent collisions is 500 times that of a single collision.

Conclusion

The question patterns in exams keep  changing, and students must be prepared to answer all types of questions. Important Questions helps students become familiar with current question formats and frequently asked questions. Having a good understanding of the types of questions that can be asked in exams greatly reduces stress and exam fear.

Students can directly access the Class 11 Physics Chapter 13 Important Questions on the Extramarks website at their convenience. This set of questions will help them  quickly revise and improve their answers. What’s more? These Class 11 Physics Chapter 13 Important Questions are prepared by subject matter experts as per the revised evaluation scheme and syllabus.