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# S.S. Krotov Heat and Molecular Physics Solutions (Chapter 2)

SS Krotov Science for Everyone – Problems in Physics Solutions for Chapter 2 ‘Heat and Molecular Physics’ is a crucial resource for you to get well-versed with the concepts of Molecular Physics for IIT JEE. In this chapter of SS Krotov Physics solutions book, you will be introduced to the various concepts of heat. You will get to learn about heat transfers (Radiation, Convection, Advection), heat energy,  thermal conductivity, conduction of heat in a conducting rod, and also several process i.e. Isothermal, Adiabatic, Isobaric, and Cyclic process. Further, you will also come across Calorimetry principle and latent heat of fusion and sublimation.

SS Krotov Molecular Physics solutions for Heat and Molecular Physics contain 42 questions. With these solutions, you will be able to understand the molecular nature of matters, the behaviour of gases, kinetic theory of an ideal gas, gas laws, law of equipartition of energy, specific heat capacity and mean free path. You also require the knowledge of moles, assumptions of ideal gases, specific heats, and capillary tubes to tackle these questions.

SS Krotov Science for Everyone Heat and Molecular Physics Solutions are prepared by our subject experts keeping the latest syllabus of IIT JEE Physics in mind. These solutions are presented in a layman’s language so that you may not find difficulties in understanding the concepts of the chapter. They are prepared in accordance with the latest NEET and JEE Physics syllabus and will help you get familiarized with the latest exam patterns of JEE Mains, JEE Advanced and NEET.

## Important Topics for SS Krotov Science for Everyone – Problems in Physics Solutions Chapter 2: Heat and Molecular Physics

Heat can be defined as a continuous flow of energy in a system. Usually, energy flows from a hotter body to a cooler body. The sun is the most prominent example of heat energy in our solar system. We have three methods of heat transmission-

• Conduction- In this process, heat is transferred from a body with a higher temperature to the body with a cooler temperature. Usually, there is no involvement of the motion of particles in this mode of heat transfer.
• Convection- If we talk about the heat transmission in the fluids, then the process is called convection.

Thermal Conductivity-

It is nothing, it’s the ability or power of a material to transfer heat. Generally, denoted by an alphabet K  and also λ. Materials which have a high value of K can be used in heat sinks whereas low thermal conductivity materials are used as thermal Insulator.

Isothermal Process-

In the Isothermal process, the temperature of the surroundings remains constant.

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In this process, the amount of heat energy is zero. The vertical flow of air in the atmosphere is a common example of this process.

We know that-

Q = m C.  Δt

Q = 0

Isobaric Process-

In this system of the process, the pressure is kept constant that is, ΔP = 0.

Isochoric Process-

In a thermal process, the volume of the closed system remains constant.

Gas Laws-

Assuming permanent gases to be ideal, we have several gases laws-

Boyle’s Law-

Boyle’s told in his gas law that when the temperature of a gas is constant, then the volume of a gas of given mass will be inversely proportional to its pressure(p).

V ∝ 1/p

From this, we can deduce that-

P .V = constant

P1 .V1 = P2. V2

Charles’s Law-

In this law, the volume of a gas is directly proportional to the absolute temperature, when the pressure of a gas is kept constant.

V ∝ T

V / T   = constant

So, for a given gas-

V1 / T1   =  V2 / T2

We know, at constant pressure, the volume may be increased or decreased by 1 / 273.15 of its volume at 0°C for each 1°C rise or fall in temperature. So, volume at t-degree centigrade-

Vt = V0 (1 + t/273.15)

Where V0  is the initial volume of a gas.

Gay Lussac’s Law-

Gay Lussac told in his law that when a volume of a gas is kept constant, the pressure of a gas is directly proportional to the temperature of a gas.

p ∝ T

P / T = constant

P1 / T1 = P2 / T2

Avagadro stated that under similar conditions of temperature and pressure, the volume of all the gases will contain equal no. of molecules. Let see his hypothesis-

He gave the Avogadro number- NA = 6.023 X 1023 per gram mole ( the no of molecules present in a gram of gas). He said that at STP and NTP conditions,  each gas will contain  6.023 X 1023  molecules. He also said that 22,4 L volume of gas contains 1 mole.

Perfect Gas Equation-

Standard or perfect gas equations are those types of gas equations, which obeys all the gas laws in all the conditions of pressure and temperature.

Here we have a standard gas equation-

p V = n RT

Where

P = pressure

T= temperature

V= volume

R=  Universal Gas constant

n  = no. of moles

R = 8.31 J mol-1K-1

Van der Waals’ Gas Equation-

It is given by:

(p + a/V2) (V – b) = RT

Where a and b are van der waals’ constant.

Degree Of Freedom-

Degree of freedom describes how many directions the particles can move freely. It is denoted by f,

f  = 3 A – R

Where A = No. of particles in the system, R= No. of Independent relations

Let’s see the degree of freedom of monatomic, diatomic, and triatomic gases –

f  ( Monoatomic)  =  3

f  ( Diatomic )       = 5

f  ( Triatomic )      = 6

Specific Heat –

Specific heat is equal to the amount of heat divided by a mass of 1 kg substance when a temperature is raised by 1-degree celsius. It is denoted by C. Here’s the formula for specific heat-

Q = m C.  Δt

C =   Q  /  m . Δt

When m = 1 kg and Δt = 1 degree celsius-

C  =  Q

And,  At Constant volume and Constant pressure, Specific heat is denoted as Cv and Cp respectively.

Cv = f/ 2 . R

Cp =  (f/2 +1). R

Cp / Cv  = 1 + 2 / f

### Exercise Discussion for SS Krotov Science for Everyone – Problems in Physics Solutions Chapter 2: Heat and Molecular Physics

• There are around 20 questions based on Heat and related topic. This includes 4 questions on the cyclic process. You have to find the work done by a gas during the cycle.
• There are questions about heat capacity, specific heat, thermal conductivity, etc. You will also be asked to find the latent heat of sublimation. In question no. 32, the latent heat of fusion of ice and latent heat of vaporization of water is given, you will have to answer the latent heat of sublimation.
• Also, some questions are from the calorimetric principle. You need to go through once to solve questions related to it. You will also be given some questions based on the Carnot heat engine.
• Then, there are around 22 questions about Molecular Physics. There are questions about Insulating vessels. In question no. 26, a thermally insulated vessel is given that is full with two liquids, having specific heats, and the temperature conditions are given. All you need to determine the ratio of masses of liquids.
• There are questions about boiling points. In question no. 31, you are asked to find the density of saturated water vapour at the boiling point.
• Some questions on capillary tubes and pistons as well in this chapter.

### Why Use SS Krotov Science for Everyone – Problems in Physics Solutions Chapter 2: Heat and Molecular Physics by Instasolv?

• The answer to the questions of SS Krotov Science for Everyone – Problems in SS Krotov Physics solutions for Heat and Molecular Physics by Instasolv are indubitably the best study material for those who have issues in solving them.
• Practising with our solutions for SS Krotov Science for Everyone book also helps analyze your weaknesses regarding concepts, and with that, you can work on your mistakes, and you will always be prepared for your exams.
• Solutions of SS Krotov Science for Everyone are provided after thorough research of each and every aspect of the chapter. They can be very beneficial for you if you are preparing Molecular physics for JEE and NEET.
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