Answer: (b) 0.25п J Solution: It is a cyclic process so the net change in internal energy of the system will be zero. i.e., ΔU = 0 From first law of thermodynamics, ΔQ = ΔU + ΔW Therefore, ΔQ = ΔW ΔW is the area of the shaded region Q cycle = W cycle = π (25) (10) Kpa-cc = π (25) (10) × 10 3 × 10 -6 = 0.250 πJ MCQ (More than One Correct Answer) The correct option (s) about entropy (S) is (are) $ [\mathrm {R}=$ gas constant, $\mathrm {F}=$ Faraday constant, $\mathrm {T}=$ Temperature $]$ View Question
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View Question Which one of the following options correctly represents a thermodynamic process that is used as a correction in the determination of the speed of soun. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi = 105 Pa and volume Vi = 10-3 m3 ch. View Question Download PDF Study Material Exam Info Syllabus Courses Previous Year Question Paper Practice Materials Download Free PDF of JEE Advanced Important Questions of Heat and Thermodynamics on Vedantu Get here the most reliable study material with the important questions for IIT JEE Heat and Thermodynamics to help you make your studying easier. Thermodynamics and Thermochemistry is a crucial topic in JEE Advanced Chemistry, and mastering this topic can help students score well in the exam. One of the best ways to prepare for JEE Advanced Chemistry is by solving practice papers on Thermodynamics and Thermochemistry. Thermodynamics - JEE Advanced Previous Year Questions with Solutions JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.
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The succeeding operations that enable this transformation of states are (A) Heating, cooling, heating, cooling (B) cooling, heating, cooling, heating (C) Heating, cooling, cooling, heating (D) Cooling, heating, heating, cooling [JEE 2013] Ans. (C) Isochoric $\Rightarrow \mathrm {V}-$ constant Q. In thermodynamics, the system is defined as a quantity of matter or a region in space under investigation. Everything external to the system is the environment or surroundings. The imaginary or actual surface separating the system from its surroundings is known as the boundary. It may be either fixed or movable. Closed System (m = constant) Submit Your JEE MAIN 2nd Attempt Application Form👉https://bit.ly/JEEResults-YTCheck Bhishma JEE Advanced Batch 2023👉https://smart.link/g8nfubkjvvlzmBhishma. About this unit Equivalence of heat and work, First law of thermodynamics and its applications (only for ideal gases) I E Irodov Solutions for IIT JEE Advanced Physics - Thermodynamics General Physics by I E irodov can really test your understanding of fundamental of physics.
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The Thermodynamics Topic is one of the critical chapters for JEE aspirants to understand thoroughly to perform well in the Chemistry for JEE Advanced Section of the JEE Examination. Many aspirants find this section a little complicated and thus they can take help from EduRev notes for JEE, prepared by experts according to the latest JEE syllabus. Physics, Chemistry, Mathematics Previous Year Questions well organized subject wise, chapter wise and year wise with best solutions, provider ExamSIDE.Com
The Heat and Thermodynamics JEE Advanced Questions PDF includes different types of questions: single correct answer, matching column, true or false, assertion- reasoning, fill in the blanks and many more. JEE Advanced 2023 Revision Notes for Thermodynamics JEE Advanced Revision Notes Physics Thermodynamics Last updated date: 21st Dec 2023 • Total views: 31.8k • Views today: 0.31k Download PDF Study Material Exam Info Syllabus Courses Previous Year Question Paper Practice Materials
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With Thermodynamics and Thermochemistry JEE Advanced Questions with Solutions aspirants will be able to study productively and perform better in the exam. Additionally, the JEE Advanced Thermodynamics and Thermochemistry questions with solutions are available in the form of a PDF which can be downloaded instantly. Detailed Solution for JEE Advanced (Single Correct MCQs): Thermodynamics - Question 1. TIPS/Formulae : Heat capacity at constant volume (q v) = ΔE Heat capacity of constant pressure (q p) = ΔH. ΔH = ΔE +ΔnRT or ΔH - ΔE = ΔnRT. Δn = no. of moles of gaseous products - no. of moles of gaseous reactants. = 12 - 15 = -3.
