NCERT Solutions For Class 11 Physics Chapter 12: Thermodynamics

NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics are given in this article. Thermodynamics is a branch of physics that deals with the study of the transformation of heat energy into several other forms of energy. For example, when we rub our palms together, the work done in rubbing produces heat. This process of rubbing hands denotes a Thermodynamic process.

Class 11 Physics Chapter 12 Thermodynamics is of Unit 8 which has a weightage of 20 marks along with unit 7 and Unit 9. Class 11 NCERT Solutions for Chapter 12 cover important concepts of laws of thermodynamicsCarnot Cycle, Born Haber Cycleand Enthalpy.

Download PDF: NCERT Solutions for Class 11 Physics Chapter 12


NCERT Solutions for Class 11 Physics Chapter 12

NCERT SolutionsNCERT SolutionsNCERT SolutionsNCERT SolutionsNCERT SolutionsNCERT SolutionsNCERT SolutionsNCERT SolutionsNCERT SolutionsNCERT Solutions

Class 11 Physics Chapter 12 – Concepts Covered

  • Thermodynamics is the branch of physics that deals with the study of the transformation of heat into other forms of energy and vice-versa.
A collection of a large number of atoms or molecules confined within certain boundaries in a way that it has specific values of pressure (P), volume (V) and temperature (T) is known as a thermodynamic system.
  • Zeroth Law of Thermodynamics identifies thermal equilibrium and introduces the concept of temperature as a tool for identifying equilibrium. 
If two systems are in thermal equilibrium with a third system then those two systems themselves are in equilibrium with each other.
  • Work is said to be done if a system moves a certain distance in the direction of the applied force.
It is given as dW = PdV
where P refers to the pressure of the gas in the cylinder.

The total change in heat energy in any system is the sum of the work done and the change in internal energy.

  • Second Law of Thermodynamics is the principle that disallows certain phenomena that are consistent with the First law of thermodynamics.
The two statements of second law of thermodynamics are as follows – 
- Kelvin-Planck Statement: It is impossible to form an engine, operating in a cycle, to extract heat from a hot body and then convert it completely into work without leaving change anywhere i.e., 100% conversion of heat into work is not possible.
- Clausius Statement: It is impossible for a self acting machine, operating in a cycle, without the help of any external energy to transfer heat from a cold body to a hot body. Heat can not flow itself from a colder body to a hotter body.

CBSE CLASS XII Related Questions

  • 1.
    Write the mathematical forms of three postulates of Bohr’s theory of the hydrogen atom. Using them prove that, for an electron revolving in the \( n \)-th orbit,
    (a) the radius of the orbit is proportional to \( n^2 \), and
    (b) the total energy of the atom is proportional to \( \frac{1}{n^2} \).


      • 2.
        Answer the following giving reason:
        (a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
        (b) The saturation current in case (a) is different for different intensity.
        (c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.


          • 3.
            (a) Consider the so-called ‘D-T reaction’ (Deuterium-Tritium reaction).
            In a thermonuclear fusion reactor, the following nuclear reaction occurs: \[ \ ^{2}_1 \text{H} + \ ^{3}_1 \text{H} \longrightarrow \ ^{4}_2 \text{He} + \ ^{1}_0 \text{n} + Q \] Find the amount of energy released in the reaction.
            % Given data Given:
            \( m\left(^{2}_1 \text{H}\right) = 2.014102 \, \text{u} \)
            \( m\left(^{3}_1 \text{H}\right) = 3.016049 \, \text{u} \)
            \( m\left(^{4}_2 \text{He}\right) = 4.002603 \, \text{u} \)
            \( m\left(^{1}_0 \text{n}\right) = 1.008665 \, \text{u} \)
            \( 1 \, \text{u} = 931 \, \text{MeV}/c^2 \)


              • 4.
                A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:
                vertically held bar magnet is dropped along the axis of a copper ring

                  • zero
                  • less than \( g \)
                  • \( g \)
                  • greater than \( g \)

                • 5.
                  The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

                    • \( \frac{\alpha}{r} \)
                    • \( \frac{\alpha r^2}{2} \)
                    • \( \frac{\alpha}{2r^2} \)
                    • \( -\frac{\alpha}{r} \)

                  • 6.
                    A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases. parallel plate capacitor

                      CBSE CLASS XII Previous Year Papers

                      Comments


                      No Comments To Show