Question

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9. Consider a heat engine uses a 0.50 moles of a monoatomic ideal gas. The initial temperature is where V1= 1.000L V2= 3.375 L, V3=8.210 L and V4=2.433 L while the highest temperature was obser during adiabatic process is 450 K 12 (10) a) For reversible adiabatic expansion calculate P3 w.. AD, AS

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ANSWER I Griven information and data mole of monoatomie gasis (n) = 0-50 moles Ti = 2ook V, 1. OVOLV. - 3375 L V2 = 8.2104 ; VP = 2.433 L during adiabatic expansion process Temperahre = 450k = For reversible adiabatic expansion calculate P3, W, q, BU, AS = P = ? is for reversible adiabatic expansion, so T-450k because reversible adiabatic expansion is highest work, so more highest temperature increases, so 73 =450k V3 = 8.210L (abready given) . n=0.50 hole. (already given) Ideal gas given fo P₂ V3 = nRT3 13 = nRTS = (0.50 mote) * (8.314 IK"mal)x( 450K) V3 8.220L ": [1L = 0.001 m3 d 11: 1kg m?s-2) Rulting P = 0.50 × 8.314 Kg 452x450 0.001m? P3 = 1870650 Kg mt 52 ; I Pa = The Ryman I 1 P = 1870650 Pa

13: 18 70650 Pa 1 Pa = 9.869 X 10 Calm ) 13 = 1970650 X 9-869 xlö atm. 13 = 18.46 atm Cm 12 = OIl because adiabate proces, do no change. > A8 = qvev = 01503K") AS = 07*" → AU En cvett . --.--Ci heat capacity at constand volume AT= 450K - 200k = 150k 07 = 150K] ; ma 0.50 mal abready given gas. { C u 2 = 3 2 1 1 2 3 4 | Cv z 3Rl for ideal monoatomic gas ni dua det for det er Now putting value .. of n, co, At in equation (i, AU- 0.50 poté x 3488148 JK"78074 x 150k IBU = 935.3253) & W = 9 we know that first law of thermodynamics AU= qtw here qão So DU=w Theis lw= 935.325 sty