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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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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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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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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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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.
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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)
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