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SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 12. THERMODYNAMICS #QID# 41272 (1.) At 27°C a gas suddenly compressed such that its pressure becomes 1 8 th of original pressure. The temperature of the gas will be (γ = 5/3) (a.) −142°C (b.) 300K (c.) 327° (d.) 420 K ANSWER: a EXPLANATION: (a) In adiabatic process, the relation between temperature (T) and pressure(p) is T γ Pγ−1 = constant Where γ is ratio of specific heats. Given, T1 = 27°C = 27 + 273 = 300 K, p1 = p, p2 = p 8 , γ = 5 3 ∴ T1 T2 = ( p1 p2 ) γ−1 γ ⇒ T1 T2 = ( 8 1 ) 5 3 −1 5/3 = (8) 0.4 = 2.297
SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 ⇒ T2 = T1 2.297 = 300 2.297 = 130.6 K ≈ 131K ⇒ T2 = 131 − 273 = −142°C #QID# 41273 (2.) For a monoatomic gas, work done at constant pressure is W. The heat supplied at constant volume for the same rise in temperature of the gas is (a.) W/2 (b.) 3W/2 (c.) 5W/2 (d.) W ANSWER: b EXPLANATION: (b) For the process at constant pressure dQ = CpdT + dw dT = dQ − dW Cp For the process at constant volume, dQ = CvdT (∴ dW = 0) = Cv ( dQ − dW Cp ) = dQ − dW Cp/Cv = dQ − dW γ or (γ − 1)dQ = dW ( 5 3 − 1) dQ = W, dQ = 3 2 W #QID# 41274
SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 (3.) Which statement is incorrect (a.) All reversible cycles have same efficiency (b.) Reversible cycle has more efficiency than an irreversible one (c.) Carnot cycle is a reversible one (d.) Carnot cycle has the maximum efficiency in all cycles ANSWER: a EXPLANATION: (a) Efficiency of all reversible cycles depends upon temperature of source and sink which will be different. #QID# 41275 (4.) For an isothermal expansion of a perfect gas, the value of ∆P P is equal (a.) −γ 1/2 ∆V V (b.) − ∆V V (c.) −γ ∆V V (d.) −γ 2 ∆V V ANSWER: b
SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 EXPLANATION: (b) Differentiate PV = constant w. r.t. V ⇒ P∆V + V∆P = 0 ⇒ ∆P P = − ∆V V #QID# 41276 (5.) A thermodynamical system is taken from state A to state B along ACB and is brought back to A along BDA as shown in figure. Net work done during one complete cycle is given by area. (a.) ACBDA (b.) ACB p2p1A (c.) A V1 V2 BDA (d.) BD Ap1p2 B ANSWER: a EXPLANATION: (a) In a cycle process, work done is equal to area of the loop ACBDA, representing the cycle of changes #QID# 41277 (6.) A mass of dry air at NTP. is compressed to 1 20 th of its original volume suddenly. If γ = 1.4, the final pressure would be

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