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Last updated date: 24th Mar 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.

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

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.

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.

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)

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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JEE Main 2023 January and April Session exam dates and revised schedule have been announced by the NTA. JEE Main 2023 January and April Session will now be conducted on 24-Jan-2023 to 31-Jan-2023 and 6-Apr-2023 to 12-Apr-2023, and the exam registration closes on 12-Jan-2023 and Apr-2023. You can check the complete schedule on our site. Furthermore, you can check JEE Main 2023 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.

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Last updated date: 24th Mar 2023

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NTA has announced the JEE Main 2023 January session application form release date on the official website https://jeemain.nta.nic.in/. JEE Main 2023 January and April session Application Form is available on the official website for online registration. Besides JEE Main 2023 January and April session application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details. April Session's details will be updated soon by NTA.

Last updated date: 24th Mar 2023

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It is crucial for the the engineering aspirants to know and download the JEE Main 2023 syllabus PDF for Maths, Physics and Chemistry. Check JEE Main 2023 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Main 2023 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.

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JEE Main 2023 Study Materials: Strengthen your fundamentals with exhaustive JEE Main Study Materials. It covers the entire JEE Main syllabus, DPP, PYP with ample objective and subjective solved problems. Free download of JEE Main study material for Physics, Chemistry and Maths are available on our website so that students can gear up their preparation for JEE Main exam 2023 with Vedantu right on time.

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Download JEE Main Question Papers & Answer Keys of 2022, 2021, 2020, 2019, 2018 and 2017 PDFs. JEE Main Question Paper are provided language-wise along with their answer keys. We also offer JEE Main Sample Question Papers with Answer Keys for Physics, Chemistry and Maths solved by our expert teachers on Vedantu. Downloading the JEE Main Sample Question Papers with solutions will help the engineering aspirants to score high marks in the JEE Main examinations.

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Last updated date: 24th Mar 2023

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In order to prepare for JEE Main 2023, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2023 exam so that they can grab the top rank in the all India entrance exam.

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JEE Main 2023 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2023 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.

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Last updated date: 24th Mar 2023

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NTA is responsible for the release of the JEE Main 2023 January and April Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2023 January and April Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.

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Last updated date: 24th Mar 2023

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NTA will release the JEE Main 2023 January and April sessions exam dates on the official website, i.e. {official-website}. Candidates can directly check the date sheet on the official website or https://jeemain.nta.nic.in/. JEE Main 2023 January and April sessions is expected to be held in February and May. Visit our website to keep updates of the respective important events of the national entrance exam.

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JEE Main 2023 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in IIT colleges. JEE Main 2023 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2023 state rank list on the official website or on our site.

Last updated date: 24th Mar 2023

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Want to know which Engineering colleges in India accept the JEE Main 2023 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.

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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.

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