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# JEE Important Chapter - Thermodynamics

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## Introduction of Thermodynamics for JEE

Last updated date: 21st Sep 2023
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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 2022 exams along with a few solved examples.

## JEE Main Physics Chapter-wise Solutions 2022-23

### Important Thermodynamics Topics

• 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 Crucial Concepts for JEE

 Name of the Concept Key Points of the Concepts 1. Thermodynamics Process and Example of Thermodynamics When certain changes occur in the state of a thermodynamic system, i.e., the thermodynamic parameters of the system change over time, this is referred to as a thermodynamic process.Isothermal change is a change in the pressure ($P$) and volume ($V$) of gas without any change in its temperature. It is given as,$PV= \text{constant}$When no heat is allowed to enter or depart from a gas, it undergoes an adiabatic change in pressure ($P$) and volume ($V$). It is given as$PV^\gamma=\text{constant}$Melting ice and sitting in a crowded room are two real-life illustrations of thermodynamics. 2. Work done in Isothermal and Adiabatic Expansion Consider one gram mole of an ideal gas contained in a cylinder with flawlessly conducting walls and a completely frictionless and conducting piston. Let $P_1, V_1$ and $P_2, V_2$ be the initial and final pressure and volume of the gas and $T$ be its temperature then work done under isothermal expansion is given as,$W=2.303 RT \log_{10}\dfrac{V_2}{V_1}$Consider a gram mole of an ideal gas confined in a cylinder with flawlessly nonconductive walls and a perfectly frictionless, nonconducting piston. Let $P_1, V_1, T_1$ and $P_2, V_2, T_2$ be the initial and final pressure, volume and temperature of the gas then work done under adiabatic expansion is given as,$W=R\dfrac{(T_2-T_1)}{(1-\gamma)}$ 3. First Law of thermodynamics and its Limitations According to this law, the energy ($dQ$) supplied to a system increases partly the internal energy of the system ($dU$) and the rest is spent on doing work ($dW$) on the environment. It is written as,$dQ=dU+dW$There are three limitations of this law;This law doesn’t indicate the direction in which the change can occur.This law doesn’t give any idea about the extent of change.This law gives no information about the source of heat i.e, whether it is hot or cold. 4. Cyclic Process and Non-Cyclic Process A cyclic process consists of a series of changes which return the system back to its initial state while a non-cyclic process consists of a series of changes which don’t return the system back to its initial state. 5. Heat Engine A heat engine is a mechanical device that transforms heat energy into mechanical energy. It consists of a source of heat at a higher temperature, a working substance and a sink of heat at a lower temperature.The thermal efficiency ($\eta$) of a heat engine is defined as the ratio of work done per cycle by the engine to the total amount of heat absorbed per cycle by the working substance from the source. It is given as,$\eta=1-\dfrac{Q_1}{Q_2}$Here, $Q_1$ = the amount of heat absorbed by a working substance from the source and $Q_2$=amount of heat rejected to the sink. 6. Refrigerator or Heat Pump A heat pump is a device that is used for cooling things. It can be regarded as a heat engine working in the reverse direction.Thus in a refrigerator, the working substance would absorb a certain quantity of heat ($Q_2$) from the sink at a lower temperature ($T_2$) and reject a large amount of heat ($Q_1$) to the source at a higher temperature($T_1$).The coefficient of performance ($\beta$) of a refrigerator is defined as the ratio of the quantity of heat removed per cycle from the contents of the refrigerator ($Q_2$) to the energy spent per cycle ($W$) to remove this heat. It is given as,$\beta=\dfrac{Q_2}{W}=\dfrac{Q_2}{Q_1-Q_2}$ 7. Second Law of thermodynamics This law contains two important statements which are:Kelvin Planck Statement: It is possible to construct a heat engine which would absorb heat from a reservoir and convert 100% of the heat absorbed into work.Clausius Statement: It is impossible to design a self-acting machine unaided by any external agency, which would transfer heat from a body at a lower temperature to another body at a higher temperature. 8. Reversible and Irreversible processes A thermodynamic process taking a system from the initial state (i) to the final state (f) is reversible, if the process can be reversed, the system and its surroundings will revert to their previous conditions, with no additional changes occurring anywhere else in the universe.The following conditions have to be satisfied for a process to be reversible:The process should proceed at an extremely slow rate i.e, the process is quasi-static so that the system is in equilibrium with its surroundings at every state.The system should be free from dissipative forces like friction, inelasticity, viscosity etc.A process which doesn’t satisfy even one of the conditions for a reversible process is called an irreversible process. 9. Third Law of Thermodynamics According to this law, the entropy of a flawless crystal at zero Kelvin (absolute zero) is equal to zero. 10. Carnot Engine A Carnot engine is based upon the Carnot cycle which is explained below.Consider a cylinder containing one gram mole of an ideal gas. The initial volume, pressure, and temperature of the gas are $P_1,V_1,T_1$. The initial stage is represented by the point A on P-V diagram as shown in the diagram below:The Carnot cycle consists of the following four stages;Isothermal expansion: Let the cylinder be placed on the source and the gas be allowed to expand by slow outward motion of the piston. Since it absorbs the required amount of heat from the source through the conducting base of the cylinder hence temperature remains constant. This operation is called isothermal expansion and is represented by isothermal curve AB in the diagram. Let the amount of heat absorbed be $Q_1$ and $W_1$ be the corresponding work done which is written as,$Q_1=W_1=RT_1\log_{e}\dfrac{V_2}{V_1}=\text{area ABMKA}$Adiabatic expansion: The cylinder is now removed from the source and placed on a perfectly insulating pad. The gas is allowed to expand further from B($V_2,P_2$) to C($V_3,P_3$). The temperature of the gas falls to $T_2$, the expansion is adiabatic and is represented by curve BC. Let $W_2$ be the work done by the gas in expanding adiabatically from B($V_2,P_2$) to C($V_3,P_3$) is,$W_2=R\dfrac{(T_2-T_1)}{(1-\gamma)}=\text{area BCNMB}$Isothermal compression: The cylinder is now removed from the insulating pad and is placed on the sink at a temperature $T_2$. The piston is moved inwards slowly so that gas is compressed until its pressure is $P_4$ and volume is $V_4$. This process is isothermal and represented by CD. Let $Q_2$ be the amount of heat energy required to sink and $W_3$ be the amount of work done on the gas in compressing it isothermally from a state C($V_3,P_3$) to D($V_4,P_4$) is given as,$Q_2=W_3=-RT_2\log_{e}\dfrac{V_4}{V_3}=-\text{area CDLNC}$Adiabatic Compression: The cylinder is put on the insulating pad once more. The piston is further moved downwards so that the gas is compressed to its initial volume $V_1$ and pressure $P_1$. This is an adiabatic change and represented by DA. Let $W_4$ be the work done on the gas in compressing it adiabatically from D($V_4, P_4$) to the state A($V_1,P_1$) is given as,$W_4=-R\dfrac{(T_2-T_1)}{(1-\gamma)}=\text{area DAKLD}$Now the total work done($W$) is,$W=W_1-W_3= \text{area ABCDA}$ 11. Efficiency of Carnot Engine The efficiency ($\eta$) of a Carnot engine is defined as the ratio of net mechanical work done ($W$) per cycle by the gas to the amount of heat energy absorbed per cycle from the source($Q_1$). It is given as,$\eta=\dfrac{W}{Q_1} =1-\dfrac{T_2}{T_1}$

### List of Important Thermodynamics Formulas

 S.No. Name of the Concept Formula Thermodynamics equations for Isothermal and adiabatic change Isothermal change is given as,$PV=\text{ constant}$Adiabatic change is given as,$PV^\gamma=\text{constant}$ Work done in Isothermal and Adiabatic Expansion Work done under isothermal expansion is given as,$W=2.303 RT \log_{10}\dfrac{V_2}{V_1}$Work done under adiabatic expansion is given as,$W=R\dfrac{(T_2-T_1)}{(1-\gamma)}$ First Law of thermodynamics According to first law, we can write;$dQ=dU+dW$ Heat Engine The efficiency of the heat engine is given as,$\eta=1-\dfrac{Q_1}{Q_2}$ Refrigerator or Heat Pump The coefficient of performance of a refrigerator can be written as,$\beta=\dfrac{Q_2}{W}=\dfrac{Q_2}{Q_1-Q_2}$Relation between coefficient of performance of a refrigerator and efficiency of heat engine is given as,$\beta=\dfrac{1-\eta}{\eta}$ Carnot Engine Work done during isothermal expansion is,$W_1=RT_1\log_{e}\dfrac{V_2}{V_1}$Work done during adiabatic expansion is,$W_2=R\dfrac{(T_2-T_1)}{(1-\gamma)}$Work done during isothermal compression is,$W_3=-RT_2\log_{e}\dfrac{V_4}{V_3}$Work done during adiabatic compression is,$W_4=-R\dfrac{(T_2-T_1)}{(1-\gamma)}$Total work done by the engine per cycle is,$W=W_1-W_3=RT_1\log_{e}\dfrac{V_2}{V_1}-RT_2\log_{e}\dfrac{V_4}{V_3}$ Efficiency of Carnot Engine The efficiency ($\eta$) of a Carnot engine is given as,$\eta=\dfrac{W}{Q_1} =1-\dfrac{T_2}{T_1}$

### 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

### Conclusion

In this post, we have discussed various thermodynamics terms like heat, temperature and work. We have also discussed the three laws of thermodynamics which are very crucial from an exam point of view and also mention the principle of thermodynamics. We have also mentioned the application part of concepts which is discussed in this post, in the form of numerical.

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## FAQs on JEE Important Chapter - Thermodynamics

FAQ

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.