JEE Exam > JEE Notes > Physics for JEE Main & Advanced > Solved Example: Thermodynamics
Solved Example: Thermodynamics | Physics for JEE Main & Advanced PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1 JEE Solved Example on Thermodynamics JEE Mains Q1: Find the approx. number of molecules contained in a vessel of volume 7 litres at ?? ° ?? at ?? . ?? × ???? ?? ???????????? (A) ?? . ?? × ???? ???? (B) ?? × ???? ???? (C) ?? × ???? ???? (D) ?? . ?? × ???? ???? Ans: (A) ?? = 7 L = 7 × 10 -3 m 3 ?? = 273 K P = 1.3 × 10 5 Pa ? m = PV RT = 1.3 × 10 5 × 7 × 10 -3 8.314 × 273 ? 0.4 moles ? No. of molecules = 0.4 × 6.02 × 10 23 = 2.4 × 10 23 molecules Q2: An ideal gas mixture filled inside a balloon expands according to the relation ???? ?? /?? = constant. The temperature inside the balloon is (A) Increasing (B) Decreasing (C) Constant (D) Can't be said Ans: (A) PV 2/3 = const. ? nRT V V 2/3 = const. ? TV -1/3 = const. ? T = conts. × V 1/3 ? On increasing volume, temperature increase Q3: A rigid tank contains ???? ???? of nitrogen at ?? ?????? . Sufficient quality of oxygen is supplied to increase the pressure to ?? ?????? , while the temperature remains constant. Amount of oxygen supplied to the tank is: (A) ?? ???? (B) ???? ???? (C) ???? ???? (D) ???? ???? Ans: (C) T = constant, V = constant, Page 2 JEE Solved Example on Thermodynamics JEE Mains Q1: Find the approx. number of molecules contained in a vessel of volume 7 litres at ?? ° ?? at ?? . ?? × ???? ?? ???????????? (A) ?? . ?? × ???? ???? (B) ?? × ???? ???? (C) ?? × ???? ???? (D) ?? . ?? × ???? ???? Ans: (A) ?? = 7 L = 7 × 10 -3 m 3 ?? = 273 K P = 1.3 × 10 5 Pa ? m = PV RT = 1.3 × 10 5 × 7 × 10 -3 8.314 × 273 ? 0.4 moles ? No. of molecules = 0.4 × 6.02 × 10 23 = 2.4 × 10 23 molecules Q2: An ideal gas mixture filled inside a balloon expands according to the relation ???? ?? /?? = constant. The temperature inside the balloon is (A) Increasing (B) Decreasing (C) Constant (D) Can't be said Ans: (A) PV 2/3 = const. ? nRT V V 2/3 = const. ? TV -1/3 = const. ? T = conts. × V 1/3 ? On increasing volume, temperature increase Q3: A rigid tank contains ???? ???? of nitrogen at ?? ?????? . Sufficient quality of oxygen is supplied to increase the pressure to ?? ?????? , while the temperature remains constant. Amount of oxygen supplied to the tank is: (A) ?? ???? (B) ???? ???? (C) ???? ???? (D) ???? ???? Ans: (C) T = constant, V = constant, ?? ?? = 35 28 = 1.25 Kmoles ? ?? ?? = ???? ?? = const. ? ?? ?? ?? ?? = ?? ?? ?? ?? ? ?? ?? = ?? ?? ?? ?? × ?? ?? = 9 6 × 35 28 ˜ 1.875 kmoles ? Moles of O 2 supplied = 1.875 - 1.25 = 0.625kmoles ? Amount of oxygen = 20 kg Q4: At temperature ???? , the pressure of ?? . ?? ?? argon in bulb is ?? . The bulb is put in a bath having temperature higher by ???? ?? than the first one. ?? . ?? ?? of argon gas had to be removed to maintained original pressure. The temperature ?? is equal to (A) ?????? ?? (B) ?????? ?? (C) ?????? ?? (D) ???? ?? Ans: (B) ???? = 4 ?? ???? ???? = ( 4 - 0.8) ?? ?? ( ?? + 50) = 3.2?? ?? ( ?? + 50) By (i) and (ii) we get, 4 ?? ???? = 3.2?? ?? ( ?? + 50) ? 4?? = 3.2?? + 3.2 × 50 ? ?? = 200 ?? Q5: When 2 gms of a gas are introduced into an evacuated flask kept at ???? ° ?? the pressure is found to be one atmosphere. If ???????? of another gas added to the same flask the pressure becomes 1.5 atmosphere. The ratio of the molecular weights of these gases will be (A) ?? : ?? (B) 3:1 (C) ?? : ?? (D) ?? : ?? Ans: (A) For P = 1 atm , m 1 = 2gms , T = 298 K ? 1 × ?? = 2?? ?? 1 × 298 For ?? = 1.5 atm, ?? 1 = 2gm , ?? 2 = 3gm , ?? = 298 K. Page 3 JEE Solved Example on Thermodynamics JEE Mains Q1: Find the approx. number of molecules contained in a vessel of volume 7 litres at ?? ° ?? at ?? . ?? × ???? ?? ???????????? (A) ?? . ?? × ???? ???? (B) ?? × ???? ???? (C) ?? × ???? ???? (D) ?? . ?? × ???? ???? Ans: (A) ?? = 7 L = 7 × 10 -3 m 3 ?? = 273 K P = 1.3 × 10 5 Pa ? m = PV RT = 1.3 × 10 5 × 7 × 10 -3 8.314 × 273 ? 0.4 moles ? No. of molecules = 0.4 × 6.02 × 10 23 = 2.4 × 10 23 molecules Q2: An ideal gas mixture filled inside a balloon expands according to the relation ???? ?? /?? = constant. The temperature inside the balloon is (A) Increasing (B) Decreasing (C) Constant (D) Can't be said Ans: (A) PV 2/3 = const. ? nRT V V 2/3 = const. ? TV -1/3 = const. ? T = conts. × V 1/3 ? On increasing volume, temperature increase Q3: A rigid tank contains ???? ???? of nitrogen at ?? ?????? . Sufficient quality of oxygen is supplied to increase the pressure to ?? ?????? , while the temperature remains constant. Amount of oxygen supplied to the tank is: (A) ?? ???? (B) ???? ???? (C) ???? ???? (D) ???? ???? Ans: (C) T = constant, V = constant, ?? ?? = 35 28 = 1.25 Kmoles ? ?? ?? = ???? ?? = const. ? ?? ?? ?? ?? = ?? ?? ?? ?? ? ?? ?? = ?? ?? ?? ?? × ?? ?? = 9 6 × 35 28 ˜ 1.875 kmoles ? Moles of O 2 supplied = 1.875 - 1.25 = 0.625kmoles ? Amount of oxygen = 20 kg Q4: At temperature ???? , the pressure of ?? . ?? ?? argon in bulb is ?? . The bulb is put in a bath having temperature higher by ???? ?? than the first one. ?? . ?? ?? of argon gas had to be removed to maintained original pressure. The temperature ?? is equal to (A) ?????? ?? (B) ?????? ?? (C) ?????? ?? (D) ???? ?? Ans: (B) ???? = 4 ?? ???? ???? = ( 4 - 0.8) ?? ?? ( ?? + 50) = 3.2?? ?? ( ?? + 50) By (i) and (ii) we get, 4 ?? ???? = 3.2?? ?? ( ?? + 50) ? 4?? = 3.2?? + 3.2 × 50 ? ?? = 200 ?? Q5: When 2 gms of a gas are introduced into an evacuated flask kept at ???? ° ?? the pressure is found to be one atmosphere. If ???????? of another gas added to the same flask the pressure becomes 1.5 atmosphere. The ratio of the molecular weights of these gases will be (A) ?? : ?? (B) 3:1 (C) ?? : ?? (D) ?? : ?? Ans: (A) For P = 1 atm , m 1 = 2gms , T = 298 K ? 1 × ?? = 2?? ?? 1 × 298 For ?? = 1.5 atm, ?? 1 = 2gm , ?? 2 = 3gm , ?? = 298 K. ? 1.5 × ?? = [ 2 ?? 1 + 3 ?? 2 ] ?? × 298 Using (ii)/(i) we get, 1.5 = [ 2 ?? 1 + 2 ?? 2 ] 2 ?? 1 ? 3 = [2 + 3?? 1 ?? 2 ] ? ?? 1 ?? 2 = 1 3 Q6: During an experiment, an ideal gas obeys an addition equation of state ?? ?? ?? = constant. The initial temperature and pressure of gas are ?? and ?? respectively. When it expands to volume ?? ?? , then its temperature will be? (A) T (B) v ?? ?? (C) ?? ?? (D) ?? v ?? ?? Ans: (B) ?? = ?????? ?? ? ( ?????? ?? ) 2 ?? = constant ? ?? 2 ?? = constant ? ?? ?? = ( ?? ?? ?? ?? ) 1 2 × ?? ?? = v 2?? Q7: A barometer tube, containing mercury, is lowered in a vessel containing mercury until only ???? ???? of the tube is above the level of mercury in the vessel. If the atmospheric pressure is ???? ???? of mercury, what is the pressure at the top of the tube? (A) ???? . ???????? (B) ???? . ???????? (C) ?? . ?????????? (D) ?? . ?????????? Ans: (A) ?? at top of tube + ?? due to 50 cm of Hg = ?? atmospheric ? ?? at top of tube = ( 75 - 50) cm of Hg = 25 cm of Hg = 25 × 10 5 75 Pa = 33.3kPa Page 4 JEE Solved Example on Thermodynamics JEE Mains Q1: Find the approx. number of molecules contained in a vessel of volume 7 litres at ?? ° ?? at ?? . ?? × ???? ?? ???????????? (A) ?? . ?? × ???? ???? (B) ?? × ???? ???? (C) ?? × ???? ???? (D) ?? . ?? × ???? ???? Ans: (A) ?? = 7 L = 7 × 10 -3 m 3 ?? = 273 K P = 1.3 × 10 5 Pa ? m = PV RT = 1.3 × 10 5 × 7 × 10 -3 8.314 × 273 ? 0.4 moles ? No. of molecules = 0.4 × 6.02 × 10 23 = 2.4 × 10 23 molecules Q2: An ideal gas mixture filled inside a balloon expands according to the relation ???? ?? /?? = constant. The temperature inside the balloon is (A) Increasing (B) Decreasing (C) Constant (D) Can't be said Ans: (A) PV 2/3 = const. ? nRT V V 2/3 = const. ? TV -1/3 = const. ? T = conts. × V 1/3 ? On increasing volume, temperature increase Q3: A rigid tank contains ???? ???? of nitrogen at ?? ?????? . Sufficient quality of oxygen is supplied to increase the pressure to ?? ?????? , while the temperature remains constant. Amount of oxygen supplied to the tank is: (A) ?? ???? (B) ???? ???? (C) ???? ???? (D) ???? ???? Ans: (C) T = constant, V = constant, ?? ?? = 35 28 = 1.25 Kmoles ? ?? ?? = ???? ?? = const. ? ?? ?? ?? ?? = ?? ?? ?? ?? ? ?? ?? = ?? ?? ?? ?? × ?? ?? = 9 6 × 35 28 ˜ 1.875 kmoles ? Moles of O 2 supplied = 1.875 - 1.25 = 0.625kmoles ? Amount of oxygen = 20 kg Q4: At temperature ???? , the pressure of ?? . ?? ?? argon in bulb is ?? . The bulb is put in a bath having temperature higher by ???? ?? than the first one. ?? . ?? ?? of argon gas had to be removed to maintained original pressure. The temperature ?? is equal to (A) ?????? ?? (B) ?????? ?? (C) ?????? ?? (D) ???? ?? Ans: (B) ???? = 4 ?? ???? ???? = ( 4 - 0.8) ?? ?? ( ?? + 50) = 3.2?? ?? ( ?? + 50) By (i) and (ii) we get, 4 ?? ???? = 3.2?? ?? ( ?? + 50) ? 4?? = 3.2?? + 3.2 × 50 ? ?? = 200 ?? Q5: When 2 gms of a gas are introduced into an evacuated flask kept at ???? ° ?? the pressure is found to be one atmosphere. If ???????? of another gas added to the same flask the pressure becomes 1.5 atmosphere. The ratio of the molecular weights of these gases will be (A) ?? : ?? (B) 3:1 (C) ?? : ?? (D) ?? : ?? Ans: (A) For P = 1 atm , m 1 = 2gms , T = 298 K ? 1 × ?? = 2?? ?? 1 × 298 For ?? = 1.5 atm, ?? 1 = 2gm , ?? 2 = 3gm , ?? = 298 K. ? 1.5 × ?? = [ 2 ?? 1 + 3 ?? 2 ] ?? × 298 Using (ii)/(i) we get, 1.5 = [ 2 ?? 1 + 2 ?? 2 ] 2 ?? 1 ? 3 = [2 + 3?? 1 ?? 2 ] ? ?? 1 ?? 2 = 1 3 Q6: During an experiment, an ideal gas obeys an addition equation of state ?? ?? ?? = constant. The initial temperature and pressure of gas are ?? and ?? respectively. When it expands to volume ?? ?? , then its temperature will be? (A) T (B) v ?? ?? (C) ?? ?? (D) ?? v ?? ?? Ans: (B) ?? = ?????? ?? ? ( ?????? ?? ) 2 ?? = constant ? ?? 2 ?? = constant ? ?? ?? = ( ?? ?? ?? ?? ) 1 2 × ?? ?? = v 2?? Q7: A barometer tube, containing mercury, is lowered in a vessel containing mercury until only ???? ???? of the tube is above the level of mercury in the vessel. If the atmospheric pressure is ???? ???? of mercury, what is the pressure at the top of the tube? (A) ???? . ???????? (B) ???? . ???????? (C) ?? . ?????????? (D) ?? . ?????????? Ans: (A) ?? at top of tube + ?? due to 50 cm of Hg = ?? atmospheric ? ?? at top of tube = ( 75 - 50) cm of Hg = 25 cm of Hg = 25 × 10 5 75 Pa = 33.3kPa Q. 8 A vessel contains 1 mole of ?? ?? gas (molar mass 32) at a temperature ?? . The pressure of the gas is ?? . An identical vessel containing one mole of He gas (molar mass 4) at a temperature ?? ?? has a pressure of (A) ?? /?? (B) ?? (C) ?? ?? (D) ???? Ans: (C) Q9: The ratio of average translational kinetic energy to rotational kinetic energy of a diatomic molecule temperature ?? is (A) 3 (B) ?? /?? (C) ?? /?? (D) ?? /?? Ans: (D) For diatomic molecule Translational degree of freedoms = 3 Rotational degree of freedoms = 2 Ratio = 3???? 2???? = 3 2 Page 5 JEE Solved Example on Thermodynamics JEE Mains Q1: Find the approx. number of molecules contained in a vessel of volume 7 litres at ?? ° ?? at ?? . ?? × ???? ?? ???????????? (A) ?? . ?? × ???? ???? (B) ?? × ???? ???? (C) ?? × ???? ???? (D) ?? . ?? × ???? ???? Ans: (A) ?? = 7 L = 7 × 10 -3 m 3 ?? = 273 K P = 1.3 × 10 5 Pa ? m = PV RT = 1.3 × 10 5 × 7 × 10 -3 8.314 × 273 ? 0.4 moles ? No. of molecules = 0.4 × 6.02 × 10 23 = 2.4 × 10 23 molecules Q2: An ideal gas mixture filled inside a balloon expands according to the relation ???? ?? /?? = constant. The temperature inside the balloon is (A) Increasing (B) Decreasing (C) Constant (D) Can't be said Ans: (A) PV 2/3 = const. ? nRT V V 2/3 = const. ? TV -1/3 = const. ? T = conts. × V 1/3 ? On increasing volume, temperature increase Q3: A rigid tank contains ???? ???? of nitrogen at ?? ?????? . Sufficient quality of oxygen is supplied to increase the pressure to ?? ?????? , while the temperature remains constant. Amount of oxygen supplied to the tank is: (A) ?? ???? (B) ???? ???? (C) ???? ???? (D) ???? ???? Ans: (C) T = constant, V = constant, ?? ?? = 35 28 = 1.25 Kmoles ? ?? ?? = ???? ?? = const. ? ?? ?? ?? ?? = ?? ?? ?? ?? ? ?? ?? = ?? ?? ?? ?? × ?? ?? = 9 6 × 35 28 ˜ 1.875 kmoles ? Moles of O 2 supplied = 1.875 - 1.25 = 0.625kmoles ? Amount of oxygen = 20 kg Q4: At temperature ???? , the pressure of ?? . ?? ?? argon in bulb is ?? . The bulb is put in a bath having temperature higher by ???? ?? than the first one. ?? . ?? ?? of argon gas had to be removed to maintained original pressure. The temperature ?? is equal to (A) ?????? ?? (B) ?????? ?? (C) ?????? ?? (D) ???? ?? Ans: (B) ???? = 4 ?? ???? ???? = ( 4 - 0.8) ?? ?? ( ?? + 50) = 3.2?? ?? ( ?? + 50) By (i) and (ii) we get, 4 ?? ???? = 3.2?? ?? ( ?? + 50) ? 4?? = 3.2?? + 3.2 × 50 ? ?? = 200 ?? Q5: When 2 gms of a gas are introduced into an evacuated flask kept at ???? ° ?? the pressure is found to be one atmosphere. If ???????? of another gas added to the same flask the pressure becomes 1.5 atmosphere. The ratio of the molecular weights of these gases will be (A) ?? : ?? (B) 3:1 (C) ?? : ?? (D) ?? : ?? Ans: (A) For P = 1 atm , m 1 = 2gms , T = 298 K ? 1 × ?? = 2?? ?? 1 × 298 For ?? = 1.5 atm, ?? 1 = 2gm , ?? 2 = 3gm , ?? = 298 K. ? 1.5 × ?? = [ 2 ?? 1 + 3 ?? 2 ] ?? × 298 Using (ii)/(i) we get, 1.5 = [ 2 ?? 1 + 2 ?? 2 ] 2 ?? 1 ? 3 = [2 + 3?? 1 ?? 2 ] ? ?? 1 ?? 2 = 1 3 Q6: During an experiment, an ideal gas obeys an addition equation of state ?? ?? ?? = constant. The initial temperature and pressure of gas are ?? and ?? respectively. When it expands to volume ?? ?? , then its temperature will be? (A) T (B) v ?? ?? (C) ?? ?? (D) ?? v ?? ?? Ans: (B) ?? = ?????? ?? ? ( ?????? ?? ) 2 ?? = constant ? ?? 2 ?? = constant ? ?? ?? = ( ?? ?? ?? ?? ) 1 2 × ?? ?? = v 2?? Q7: A barometer tube, containing mercury, is lowered in a vessel containing mercury until only ???? ???? of the tube is above the level of mercury in the vessel. If the atmospheric pressure is ???? ???? of mercury, what is the pressure at the top of the tube? (A) ???? . ???????? (B) ???? . ???????? (C) ?? . ?????????? (D) ?? . ?????????? Ans: (A) ?? at top of tube + ?? due to 50 cm of Hg = ?? atmospheric ? ?? at top of tube = ( 75 - 50) cm of Hg = 25 cm of Hg = 25 × 10 5 75 Pa = 33.3kPa Q. 8 A vessel contains 1 mole of ?? ?? gas (molar mass 32) at a temperature ?? . The pressure of the gas is ?? . An identical vessel containing one mole of He gas (molar mass 4) at a temperature ?? ?? has a pressure of (A) ?? /?? (B) ?? (C) ?? ?? (D) ???? Ans: (C) Q9: The ratio of average translational kinetic energy to rotational kinetic energy of a diatomic molecule temperature ?? is (A) 3 (B) ?? /?? (C) ?? /?? (D) ?? /?? Ans: (D) For diatomic molecule Translational degree of freedoms = 3 Rotational degree of freedoms = 2 Ratio = 3???? 2???? = 3 2 Q10 : One mole of an ideal gas at STP is heated in an insulated closed container until the average speed of its molecules is doubled. Its pressure would therefore increase by factor. (A) 1.5 (B) v ?? (C) 2 (D) 4 Ans: (D) We know ?? rms = 1.085v avg ? v 3?? ?? = 1.085v avg ; ?? is density In this case ?? is constant ? ?? ? ( ?? avg ) 2 ? ?? ?? ?? ?? = [ ( ?? avg ) ?? ( ?? avg ) ?? ] 2 = 4 Q11: One mole of an ideal gas is contained within a cylinder by a frictionless piston and is initially at temperature ?? . The pressure of the gas is kept constant while it is heated and its volume doubles. If ?? is molar gas constant, the work done by the gas in increasing its volume is? (A) ???? ???? ?? (B) ?? /?? RT (C) RT (D) ?? /?? RT Ans: (C) Frictionless piston means, no generation of heat due to the motion of piston. n = 1 T i = T P = const. ?? i = V ?? f = 2 V Initially: - ???? = ???? Work done = P?V = P( 2 V - V) = PV = RT Q12: A polyatomic gas with six degrees of freedom does ???? ?? work when it is expanded at constant pressure. The heat given to the gas is? (A) ?????? ?? (B) ?????? ?? (C) ?????? ?? (D) ?????? ?? Ans: (A) ?? = 6Read More
|
289 videos|635 docs|179 tests
|
FAQs on Solved Example: Thermodynamics - Physics for JEE Main & Advanced
| 1. What is the first law of thermodynamics? |  |
Ans. The first law of thermodynamics states that energy cannot be created or destroyed, only transformed from one form to another. It is also known as the law of conservation of energy.
| 2. Can you explain the difference between heat and temperature in thermodynamics? | |
Ans. Heat is the transfer of energy between two systems due to a temperature difference, while temperature is a measure of the average kinetic energy of the particles in a system.
| 3. What is entropy in thermodynamics? | |
Ans. Entropy is a measure of the disorder or randomness in a system. It is a thermodynamic quantity that increases in the direction of spontaneous change and is related to the second law of thermodynamics.
| 4. How does the second law of thermodynamics affect the efficiency of heat engines? | |
Ans. The second law of thermodynamics states that heat will naturally flow from a hot object to a cold object, and that no process is 100% efficient. This means that heat engines cannot convert all of the input heat into work, leading to limitations on their efficiency.
| 5. Can you provide an example of a thermodynamic process in everyday life? | |
Ans. An example of a thermodynamic process in everyday life is the operation of a refrigerator. The refrigerator works by removing heat from the inside to the outside, using a refrigeration cycle based on the principles of thermodynamics.
Related Exams
About this Document
|
4.97/5 Rating |
|
Nov 21, 2024 Last updated |
Document Description: Solved Example: Thermodynamics for JEE 2024 is part of Physics for JEE Main & Advanced preparation.
The notes and questions for Solved Example: Thermodynamics have been prepared according to the JEE exam syllabus. Information about Solved Example: Thermodynamics covers topics
like and Solved Example: Thermodynamics Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Solved Example: Thermodynamics.
Introduction of Solved Example: Thermodynamics in English is available as part of our Physics for JEE Main & Advanced
for JEE & Solved Example: Thermodynamics in Hindi for Physics for JEE Main & Advanced course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: Solved Example: Thermodynamics | Physics for JEE Main & Advanced
Description
Full syllabus notes, lecture & questions for Solved Example: Thermodynamics | Physics for JEE Main & Advanced - JEE | Plus excerises question with solution to help you revise complete syllabus for Physics for JEE Main & Advanced | Best notes, free PDF download
Information about Solved Example: Thermodynamics
In this doc you can find the meaning of Solved Example: Thermodynamics defined & explained in the simplest way possible. Besides explaining types of
Solved Example: Thermodynamics theory, EduRev gives you an ample number of questions to practice Solved Example: Thermodynamics tests, examples and also practice JEE
tests
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
MCQs
,study material
,mock tests for examination
,Extra Questions
,Exam
,video lectures
,practice quizzes
,shortcuts and tricks
,Semester Notes
,Objective type Questions
,ppt
,Important questions
,Solved Example: Thermodynamics | Physics for JEE Main & Advanced
,past year papers
,Previous Year Questions with Solutions
,Free
,Summary
,Viva Questions
,Sample Paper
,Solved Example: Thermodynamics | Physics for JEE Main & Advanced
,Solved Example: Thermodynamics | Physics for JEE Main & Advanced
;
Additional Information about Solved Example: Thermodynamics for JEE Preparation
Solved Example: Thermodynamics Free PDF Download
The Solved Example: Thermodynamics is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Solved Example: Thermodynamics now and kickstart your journey towards success in the JEE exam.
Importance of Solved Example: Thermodynamics
The importance of Solved Example: Thermodynamics cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Solved Example: Thermodynamics Notes
Solved Example: Thermodynamics Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Solved Example: Thermodynamics.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Solved Example: Thermodynamics Notes on EduRev are your ultimate resource for success.
Solved Example: Thermodynamics JEE Questions
The "Solved Example: Thermodynamics JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Solved Example: Thermodynamics on the App
Students of JEE can study Solved Example: Thermodynamics alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Solved Example: Thermodynamics,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Solved Example: Thermodynamics is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily