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Thermodynamics Chapter - Physics JEE Main

Last updated date: 15th Sep 2024
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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

 Name of the Concept Key Points of the Concepts Thermal Equilibrium Thermal equilibrium is a fundamental concept in thermodynamics. It occurs when two or more objects within a closed system have the same temperature. When objects are in thermal equilibrium, there is no net transfer of heat between them. Mechanical Equilibrium Mechanical equilibrium is a state in which an object is not moving and there is no net force acting on it. This means that the object is not accelerating. Thermodynamic equilibrium Thermodynamic equilibrium is a state in which a system is not changing and there is no net heat or work transfer between the system and its surroundings. This means that the system is not gaining or losing energy. Zeroth Law of Thermodynamics The Zeroth Law of Thermodynamics serves as the foundation for temperature measurement and thermal equilibrium. It states that if two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. This law enables the definition of temperature and the development of temperature scales, such as Celsius and Kelvin. The Concept of Temperature Temperature is a measure of the average kinetic energy of the particles in a system. In the context of thermodynamics, it plays a pivotal role in defining the direction of heat flow and determining the behavior of solids and liquids at different temperatures. The concept of absolute zero (0 Kelvin) is crucial in understanding temperature scales Heat, Work, and Internal Energy Heat, work, and internal energy are the three main forms of energy exchange in thermodynamics. Heat refers to the transfer of thermal energy between systems due to temperature differences. Work is done when a force acts over a distance, and internal energy is the energy stored within a system due to the motion and interaction of its particles. These concepts are vital in the First Law of Thermodynamics.Relationship Between Heat, Work, and Internal EnergyThe relationship between heat, work, and internal energy is described by the first law of thermodynamics:ΔU = Q - Wwhere:ΔU is the change in internal energy of the systemQ is the heat added to the systemW is the work done by the system First Law of Thermodynamics The first law of thermodynamics states that the quantity of the heat absorbed, when some amount of heat is given to a system that is capable of doing external work, is equal to the sum of the increase in internal energy of the system due to a rise in temperature and external work done during expansion.The first law of thermodynamics is generally represented by the equation-$\Delta U = Q -W$Where $\Delta U$ = change in internal energy of the thermodynamic system$\Delta Q$ = heat given to the system$\Delta W$ = work done on the systemDifferential form of the first law of thermodynamics equation-dU - dQ - dWThe first law of Thermodynamics is also called ‘Law of Conservation of Energy’. The law of conservation of energy states that “Energy can neither be destroyed nor be created, it can only be transferred from one form to another”. Second Law of Thermodynamics The Second Law of Thermodynamics introduces the concept of entropy and the direction of natural processes. It states that heat flows spontaneously from a hotter body to a colder one and that any isolated system will tend to increase in entropy over time. It leads to the classification of processes as reversible and irreversible, with irreversible processes leading to an increase in total entropy. Reversible and Irreversible processes A reversible process is a process that can be perfectly reversed, such that the system and its surroundings return to their original states. However, reversible processes are only theoretical and cannot exist in the real world.An irreversible process is a process that cannot be perfectly reversed, such as heat transfer from a hot object to a cold object. Irreversible processes are all spontaneous processes, and they are the only type of process that can occur in the real world. Carnot Engine and Its Efficiency The Carnot engine, named after Sadi Carnot, is an idealized heat engine that operates between two temperature reservoirs. It serves as a benchmark for the maximum efficiency of any heat engine. The Carnot efficiency formula is η = 1 - (Tc/Th), Where,η is efficiencyTc - absolute temperature of the cold reservoirTh - absolute temperature of the hot reservoir. Working Principle of a Carnot Engine A Carnot engine operates on a cycle that consists of four processes:Isothermal expansion: The working fluid is heated at a constant temperature. This causes the fluid to expand and do work.Adiabatic expansion: The working fluid expands without any heat transfer. This causes the fluid to cool down and do more work.Isothermal compression: The working fluid is cooled at a constant temperature. This causes the fluid to compress and require work to be done on it.Adiabatic compression: The working fluid is compressed without any heat transfer. This causes the fluid to heat up.

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

 First Law of Thermodynamics Second Law of Thermodynamics Significance The relation between heat and work is established by the first law of thermodynamics.Both Work and Heat are equivalent to each other.The exact equivalent amount of energy of the surrounding will be lost or gained, if any system gains or loses energy.Applied heat is always equal to the sum of work done and change in internal energy.The energy is constant for an isolated system. It defines the natural direction of energy flow, emphasizing that heat flows from hot to cold objects.Sets a maximum limit on the efficiency of heat engines, making it essential for engine design.Introduces the concept of entropy, which is central in understanding disorder and spontaneity in systems.Provides a benchmark for the maximum efficiency in the Carnot engine. Applications The first law of thermodynamics is commonly used in heat engines.Refrigerators is another example where the first law of thermodynamics is used.Sweating is a great example of the first law of thermodynamics since the heat of the body is transferred to sweat.When an ice cube is put in a drink, the ice cubes absorb the heat of the drink which makes it cool. Design and analysis of heat engines.Refrigeration and air conditioning systems.Predicting the spontaneity of chemical reactions.Environmental science applications, including resource utilization and energy transfer. Limitations The first law of thermodynamics does not state anything about the heat flow direction.The process is not reversible.It is difficult to distinguish whether the process is spontaneous or not. Based on idealized scenarios and not always representative of real-world processes.Offers qualitative insights but not always precise quantitative predictions.Some scenarios may not adhere to the second law, especially at the quantum level.Rooted in statistical mechanics, making it challenging to apply in certain cases.Complex interactions with quantum mechanics in microscopic systems.

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.

Description

Q = mc$\Delta$ T

Heat transfer formula, where Q is heat, m is mass, c is specific heat, and $\Delta$ T is temperature change.

W = -P$\Delta$ V

Work done in an isobaric process, where W is work, P is pressure, and $\Delta$ V is volume change.

$\Delta$ U = Q - W

First Law of Thermodynamics equation, representing change in internal energy.

PV = nRT

Ideal Gas Law equation, relating pressure, volume, moles, gas constant (R), and temperature.

$\frac{Q}{T}=\Delta$ S

Relation between heat, temperature, and entropy change.

$\Delta$ G = $\Delta$ H - T$\Delta$ S

Gibbs Free Energy equation, combining enthalpy, entropy, and temperature.

$\frac{dQ}{dt}$ = kA$(T_h - T_c)$

Rate of heat transfer through conduction, with k as thermal conductivity, A as area, $T_h$ as hot side temperature, and $T_c$ as cold side temperature.

$PV^\gamma = \text{constant}$

Adiabatic process equation for ideal gases, where \gamma is the heat capacity ratio.

$\eta = \frac{Q_{\text{in}}}{W_{\text{net}}} \times 100\%$

Efficiency formula for heat engines.

$W_{\text{cycle}} = Q_{\text{in}} - Q_{\text{out}}$

Work done in a thermodynamic cycle.

JEE Main Thermodynamics Solved Examples

1. 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.

1. 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

1. 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.

1. 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

1. 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)

1. 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

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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.

4. What are the key topics covered in the Thermodynamics chapter for JEE Main Physics?

The Thermodynamics chapter in JEE Main Physics encompasses concepts like laws of thermodynamics, heat transfer, work done, internal energy, and specific heat capacities.

5. How can I effectively prepare for Thermodynamics in JEE Main Physics?

To excel in Thermodynamics, focus on understanding fundamental principles, practice solving diverse problems, and consider using online platforms like Vedantu for expert guidance and interactive learning sessions.

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