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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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Simulator Previous Years JEE Advanced Questions. Q. A real gas behaves like an ideal gas if its ? (A) pressure and temperature are both high (B) pressure and temperature are both low (C) pressure is high and temperature is low (D) pressure is low and temperature is high. Q. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of.
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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
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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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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
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Considering the air flow to be streamline, the steady mass flow rate of air exit JEE Advanced 2023 Paper 2 Online | Heat and Thermodynamics | Physics | JEE Advanced. ExamSIDE. Questions. Joint Entrance Examination.. Questions Asked from Heat and Thermodynamics (Numerical)
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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.
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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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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.