## Concepts of Thermodynamics for JEE Main Physics

This chapter gives us an answer about “what is thermodynamics”. Thermodynamics is a branch of physics in which we deal with the study of the transformation of heat energy into other forms of energy and vice versa. It also explains the various thermodynamics laws and their applications.

The chapter also includes the following concepts:

Thermal Equilibrium: A thermodynamics system is said to be in thermal equilibrium when macroscopic variables like temperature, pressure, volume, mass etc. that characterize the system don’t change with time.

Zeroth Law: According to this law, when a thermodynamics system A and B are separately in thermal equilibrium with a third thermodynamics system C, then systems A and B are in thermal equilibrium with each other.

Heat: Heat is a type of energy that moves from higher to lower temperatures.

Internal Energy: Internal energy of the system is the total energy possessed by the system due to molecular motion and molecular configuration.

Work: Work done by a system is energy transferred from the system to its surroundings by a process that allows the system to exert macroscopic forces on its surroundings spontaneously.

Now, let's move on to the important concepts and formulae related to JEE Main 2023 exams along with a few solved examples.

## JEE Main Physics Chapters 2024

### Important Topics of Thermodynamics Chapter

Thermodynamics Process and Example of Thermodynamics

Work done in Isothermal and Adiabatic Expansion

First Law of thermodynamics and its Limitations

Cyclic Process and Non-Cyclic Process

Heat Engine

Refrigerator or Heat Pump

Second Law of thermodynamics

Reversible and Irreversible processes

Carnot Engine

Efficiency of Carnot Engine

### Thermodynamics Important Concepts for JEE Main

Significance, Applications and Limitations of First and Second Law of Thermodynamics

### Heat Transfer:

Heat transfer is the transfer of thermal energy from one object to another. It is a fundamental process in physics and thermodynamics.

There are Three Main Modes of Heat Transfer:

Conduction: Conduction is the transfer of heat energy through direct contact between two objects. When two objects are in contact, the heat energy from the hotter object flows to the cooler object. The rate of heat transfer by conduction depends on the temperature difference between the two objects, the surface area of contact, and the thermal conductivity of the materials involved.

Convection: Convection is the transfer of heat energy through the movement of fluids. When a fluid is heated, it expands and rises. The cooler fluid then sinks to take its place. This creates a convection current, which transfers heat energy from the hotter fluid to the cooler fluid.

Radiation: Radiation is the transfer of heat energy through electromagnetic waves. All objects emit electromagnetic radiation, but the amount of radiation emitted depends on the temperature of the object. The hotter the object, the more radiation it emits.

### How do Engines work?

Engines are machines that convert heat energy into mechanical energy. They do this by using a working fluid, such as air or water, to perform a cycle of processes.

### The Basic Cycle of an Engine

The basic cycle of an engine consists of four processes:

Intake: The working fluid is drawn into the engine.

Compression: The working fluid is compressed, which increases its temperature and pressure.

Combustion: The working fluid is ignited, which causes it to expand rapidly.

Exhaust: The expanded working fluid is expelled from the engine.

### The Four-Stroke Cycle Engine:

The four-stroke cycle engine is the most common type of engine used in cars and trucks. It operates on a four-stroke cycle, which consists of the following steps:

Intake Stroke: The intake valve opens and the working fluid is drawn into the cylinder.

Compression Stroke: The intake valve closes and the piston compresses the working fluid.

Power Stroke: The spark plug ignites the working fluid, which causes it to expand rapidly and drive the piston down the cylinder.

Exhaust Stroke: The exhaust valve opens and the expanded working fluid is expelled from the cylinder.

### Two-Stroke Cycle Engines:

Two-stroke cycle engines are less common than four-stroke cycle engines, but they are often used in smaller engines, such as those found in motorcycles and lawnmowers. Two-stroke cycle engines operate on a two-stroke cycle, which consists of the following steps:

Intake and Compression Stroke: The piston rises in the cylinder, compressing the working fluid and closing the exhaust valve. The intake port opens and the working fluid is drawn into the cylinder.

Power and Exhaust Stroke: The spark plug ignites the working fluid, which causes it to expand rapidly and drive the piston down the cylinder. The exhaust port opens and the expanded working fluid is expelled from the cylinder.

### Efficiency of Engines:

The efficiency of an engine is a measure of how much of the heat energy that is used to power the engine is converted into mechanical energy. The efficiency of an engine is calculated using the following equation:

Efficiency = (Work output / Heat input) * 100%

The efficiency of an engine is limited by the second law of thermodynamics. No engine can be perfectly efficient.

### Thermodynamics Formula for JEE Main Exam

Understanding thermodynamics all formulas is essential for mastering the principles governing energy transfer, work, heat, and the behavior of gases. This thermodynamics formula sheet encompasses crucial equations used to analyze and solve problems in various thermodynamic processes, aiding in comprehensive preparation for the JEE Main Exam.

### JEE Main Thermodynamics Solved Examples

When the source temperature is $220^\circ C$, the adiabatic compression ratio in a Carnot reversible cycle is 8. Determine the temperature of the sink. (Given $\gamma = 1.5$)

Sol:

Given that, Source temperature $T_1=220+273=493K$

Adiabatic compression ratio, $\dfrac{V_2}{V_1}=8$

Temperature of the sink, $T_2=?$

To solve this problem, we have to apply the concept of adiabatic change. Now, according to this concept we are able to write,

$T_2V_2^{\gamma-1}=T_1V_1^{\gamma-1}$

On further simplification,

$T_2=T_1(\dfrac{V_1}{V_2})^{\gamma-1}$

$T_2=493(\dfrac{1}{8})^{1.5-1}$

$T_2=493(0.125)^{0.5}$

$T_2=493\times 0.353$

$T_2= 174.30 K$

$T_2= 174.30-273=-98.7^\circ C$

Hence, the temperature of the sink is $-98.7^\circ C$.

Key point: The concept of adiabatic change is important to solve this problem.

At $500^\circ C$, a reversible engine absorbs heat from a reservoir and sends it to the sink at $120^\circ C$. To perform usable mechanical work at a rate of 700 watt, how many calories per second must be extracted from the reservoir? (Given, 1 cal.= 4.2 J)

Sol:

Given that,

Temperature of the source, $T_1=500^\circ C=(500+273)K=773K$

Temperature of the sink, $T_2=120^\circ C=(120+273)K=393K$

Work done, $W= 700 \text{ watt}=700\text{ joule/sec}=\dfrac{700}{4.2} \text{ cal/sec}=166.66\text{ cal/sec}$

Heat extracted from reservoir, $Q_1=?$

To solve this problem we need to apply the concept of efficiency incase of a reversible engine. According to it, we can write efficiency as,

$\eta=1-\dfrac{T_2}{T_1}$

After putting the values of the variables, we get:

$\eta=1-\dfrac{393}{773}=1-0.508=0.492$

The efficiency of the reversible is also given as,

$\eta=\dfrac{W}{Q_1}$

Or, we can also write it as,

$Q_1=\dfrac{W}{\eta}$

After putting the values, we get:

$Q_1=\dfrac{166.66}{0.492}=338.73\text{ cal/sec}$

Hence, the heat extracted from the reservoir is $338.73\text{ cal/sec}$.

Trick: Use the quantities that are given in the question and after that apply the formula of efficiency for a reversible engine.

### Previous Years Questions from JEE Paper

A sample of gas with $\gamma= 1.5$ is taken through an adiabatic process in which the volume is compressed from $1200\,cm^3$ to $300\,cm^3$. If the initial pressure is 200 kPa. The absolute value of the work done by the gas in the process = _____________ J. (JEE Main 2021)

Sol:

Given that the gas is compressed adiabatically from volume $V_1=1200\,cm^3$ to volume $V_2=300\,cm^3$.

Initial pressure, $P_1=200\,kPa$ and $\gamma= 1.5$.

To solve this problem, we first use the concept of adiabatic process by applying its condition to find the final pressure and then use the formula of work done to get the answer.

The condition of adiabatic process is,

$P_1 V_1^\gamma=P_2 V_2^\gamma$

We can also write the above equation as;

$P_2=P_1(\dfrac{ V_1}{V_2})^\gamma$

After putting the values of known quantities, we get value final pressure as;

$P_2=200(\dfrac{ 300}{1200})^{1.5}$

$P_2=200(\dfrac{1}{4})^{\dfrac{3}{2}}$

$P_2=1600\,kPa$

Now, the formula of work done under adiabatic process is given as,

$W=\dfrac{P_2V_2-P_1V_1}{\gamma-1}$

After putting the values, we get;

$W=\dfrac{480-240}{1.5-1}=\dfrac{240}{0.5}$

$W=480\,J$

Hence, the absolute value of work done by the gas in the process is $480\,J$.

Trick: The concept of adiabatic process and the formula of work done under this process is essential to solve this problem.

A heat engine operates between a cold reservoir at temperature $T_2 = 400\,K$ and a hot reservoir at temperature $T_1$. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ______________ K. (JEE Main 2021)

Sol:

It is given that the heat given by a hot reservoir is $Q_1=300\,J$.

The heat taken by the cold reservoir, $Q_2=240\,J$

Temperature of the cold reservoir, $T_2=400\,K$

To find, the temperature of the hot reservoir i.e, $T_1$.

In order to solve this problem, we have to apply the formula of work done and the efficiency of the heat engine to find the temperature of the hot reservoir.

The work done formula in case of a heat engine is,

$W=Q_1-Q_2$

After putting the values of $Q_1$ and $Q_2$ , we get;

$W=300-240=60\,J$

Now the efficiency of heat is given as,

$\eta=\dfrac{W}{Q_1}$

Putting the values of $W$ and $Q_1$, we get;

$\eta=\dfrac{60}{300}=\dfrac{1}{5}$..........(1)

The efficiency formula in terms of $T_1$ and $T_2$ is,

$\eta=1-\dfrac{T_2}{T_1}$

After putting the value of $\eta$ using eq.(1) and $T_2$ according to the question, we get;

$\dfrac{1}{5}=1-\dfrac{400}{T_1}$

$\dfrac{400}{T_1}=\dfrac{4}{5}$

On simplification we get;

$T_1=500\,K$

Hence, the minimum temperature of the hot reservoir has to be 500K.

Trick: Use the quantities given in the questions i.e, values of heat, temperature and work done and apply the formula of efficiency and work done for a heat engine.

### Practice Questions

In a Carnot engine operating between $100^\circ C$ and $30^\circ C$, five moles of an ideal gas are taken. A single cycle produces 420 J of productive work. Calculate the ratio of gas volume at the beginning and end of an isothermal expansion. Take: R=8.4 J/mol k (Ans: 1.153)

When a diatomic gas ($\gamma=1.4$) is expanded isobarically, it produces 200 J of work. Determine the amount of heat delivered to the gas during this process. (Ans: 700 J)

## JEE Main Physics Thermodynamics Study Materials

Here, you'll find a comprehensive collection of study resources for Thermodynamics designed to help you excel in your JEE Main preparation. These materials cover various topics, providing you with a range of valuable content to support your studies. Simply click on the links below to access the study materials of Thermodynamics and enhance your preparation for this challenging exam.

## JEE Main Physics Study and Practice Materials

Explore an array of resources in the JEE Main Physics Study and Practice Materials section. Our practice materials offer a wide variety of questions, comprehensive solutions, and a realistic test experience to elevate your preparation for the JEE Main exam. These tools are indispensable for self-assessment, boosting confidence, and refining problem-solving abilities, guaranteeing your readiness for the test. Explore the links below to enrich your Physics preparation.

### Conclusion

In this article, we delve into the world of Thermodynamics, a fundamental physics chapter for JEE Main. We'll unravel key concepts, problem-solving strategies, and practical applications. You'll grasp the laws of energy and heat transfer, exploring how engines work and understanding entropy. Everything you need to ace your exam is right here. Our PDF resources offer detailed explanations, free to download. By studying these, you'll boost your exam readiness and succeed in your Thermodynamics journey.

## FAQs on Thermodynamics Chapter - Physics JEE Main

1. What is the weightage of the thermodynamics chapter in JEE?

Every year approximately two questions are asked from this chapter in JEE which ultimately lead to the weightage of 1-2% in the exam.

2. Is thermodynamics necessary for JEE?

Thermodynamics is a required chapter for both JEE Advanced and JEE Mains. This is an essential chapter that is available in both Chemistry and Physics. This idea is more comprehensive and logical in physics, and it encompasses a wide range of minor subjects. Therefore, it is advisable to study this chapter as it not only helps you in physics but in chemistry as well.

3. What is the level of difficulty of the questions from the thermodynamics chapter?

As this chapter includes both exams like in JEE Mains and JEE Advanced, so the level of questions asked in this chapter is from moderate to difficult level. Therefore, it is suggested that you must practice previous years' questions from basic to advanced level so that you are able to solve the difficult questions in the exam in an easy manner.