Question

A 1.00 mole sample of an ideal monatomic gas, originally a pressure of 1.00 atm, undergoing a three-step process: • Expands adiabatically from T1 = 588 K to T2 = 300 K • It is compressed at constant pressure until its temperature reaches T3; • Then it returns to its original pressure and temperature using a constant volume process. Calculate cycle efficiency Select one: (Quickly, please :()

Calculate cycle efficiency Select one: to. 30.4% b. None of the above options c. 35.7% d. 13.6% and. 39.1% F. 36.1% g. 14.5% h. 21.9% i. 81.21% j. 33.61%

Answer 1

P Ti = 300k 8= 5 3 2 3 MRTI V V 18:31:4 x 588 1.0 X 10 105 دی :: P2 = 10x10 0.13219) T = 588P, = 1.01x10 - N/m² (mono-atomie) = 1.67 V, using Universal gas law, P, 0.0484 Ny using adiabatie for midla, 0,67 T T2 (1096)0167 0.0484 Com 0.0486 y- V2 TO 588 300 0.0484 V2 3 DY, m V 0.1321 1.67 7 For, 5 eye Pi V2 4 1.888 X10 N/m² (1

from charles law, V3. V2. Tq oy, Tz = N3 x T2 Tz V2 V2 0.0484 * 300 0.1321 109,92 K Heat Energy absorbed = Ep dT In 31 5 *8-314 *(588 -109.92) Process 2 3 j 9.936 9 x 10 work done in 1 +2, no Wis (t. T2) 1x8.314 0.67 x(588 - 300) 3 3. § 7 38 x 10 J work done in 2 + 3 Wz 4 1.888X10 P₂. (V3 -ve) (0.0484 - 0. 1321) xios -1.5802 x 10 j 3 . Net work done W, +W₂ = 1.9935 x 10 J 3 1.9935 x 10 Cycle Efficiency :. Work done Heat absorbed 9.9369 x103 0.2006 20.06 7. So, option .
P= Pressure, V=Volume, T=Temperature, n= no. of moles, R=universal gas constant=8.314 J/mol/K

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