NCERT Solutions for class 11 Physics Chapter 13: Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 13: Kinetic Theory deals with the behaviour of gas molecules and ideal gas. An ideal gas is a gas that follows Boyle's law, Charles' law, Gay Lussac's law, and Avogadro’s law.

Class 11 Physics Chapter 13 Kinetic Theory belongs to Unit 9 Behaviour of Perfect Gases and Kinetic Theory of Gases. Along with Unit 7 and Unit 8, Unit 9 has a weightage of 20 marks. The NCERT Solutions for Chapter 13 deals with the molecular nature of matter and specific heat capacity.

Download PDF: NCERT Solutions for Class 11 Physics Chapter 13


NCERT Solutions for Class 11 Physics Chapter 13

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Class 11 Physics Chapter 13 – Concepts Covered

  • Boyle’s Law: According to this law, the volume (V) of a fixed mass of a gas is inversely proportional to the pressure (P) of that gas, given that the temperature of the gas is kept constant.
V ∝ 1/P or PV = constant
  • Charle’s Law: The volume (V) of a given mass of a gas is directly proportional to the temperature of the gas, given that the pressure of the gas remains constant.
V ∝ T or VT = constant
V1T= V2T2
  • Gay Lussac’s Law: The pressure P of a given mass of gas is directly proportional to its absolute temperature T, given that the volume V of the gas remains constant.
P ∝ T or PT = constant
P1T= P2T2
  • Equation of State of An Ideal Gas: The relation between pressure, volume, and absolute temperature of a gas is known as its equation of state. 

PV = nRT

n is the number of moles of the gas and R is the molar gas constant which is equal to 8.315 JK-1mol-1

  • Dalton’s Law of Partial Pressures states that the net pressure applied by a mixture of non-interacting gases is equivalent to the sum of their pressures.

P = P1 + P2 + ------- + Pn

CBSE CLASS XII Related Questions

  • 1.
    The energy of an electron in an orbit in hydrogen atom is \( -3.4 \, \text{eV} \). Its angular momentum in the orbit will be:

      • \( \dfrac{3h}{2\pi} \)
      • \( \dfrac{2h}{\pi} \)
      • \( \dfrac{h}{\pi} \)
      • \( \dfrac{h}{2\pi} \)

    • 2.
      A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.


        • 3.
          Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

            • attract with a force \( \frac{F}{2} \)
            • repel with a force \( \frac{F}{2} \)
            • repel with a force \( F \)
            • attract with a force \( F \)

          • 4.
            A circular coil of 100 turns and radius \( \left(\frac{10}{\sqrt{\pi}}\right) \, \text{cm}\) carrying current of \( 5.0 \, \text{A} \) is suspended vertically in a uniform horizontal magnetic field of \( 2.0 \, \text{T} \). The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate:
            the magnetic dipole moment of the coil, and
            the magnitude of the counter torque that must be applied to prevent the coil from turning.


              • 5.
                Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

                  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
                  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
                  • Assertion (A) is true, but Reason (R) is false.
                  • Both Assertion (A) and Reason (R) are false.

                • 6.
                  Four long straight thin wires are held vertically at the corners A, B, C and D of a square of side \( a \), kept on a table and carry equal current \( I \). The wire at A carries current in upward direction whereas the current in the remaining wires flows in downward direction. The net magnetic field at the centre of the square will have the magnitude:

                    • \( \dfrac{\mu_0 I}{\pi a} \) and directed along OC
                    • \( \dfrac{\mu_0 I}{\pi a \sqrt{2}} \) and directed along OD
                    • \( \dfrac{\mu_0 I \sqrt{2}}{\pi a} \) and directed along OB
                    • \( \dfrac{2\mu_0 I}{\pi a} \) and directed along OA
                  CBSE CLASS XII Previous Year Papers

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