Question

(3). A sample of 1.00 mol ideal gas molecules with Cp, m = 7/2 R is initially at p = 1.00 bar and V = 22.44 L and
then put thought the following cycle in reversible processes: (a) constant-pressure expansion to twice its
initial volume, (b) constant-volume cooling to its initial temperature, (c) isothermal-compression back to
1.00 bar. Calculate q, w, ΔU, ΔH, ΔS for each process and for the whole cycle.

Answer 1

som Geiven that - a) Process 1- A B ~ const. pressure Expansion. Pialben, v= 22.442, cim- z R halmol ...V2 = 2V, = 44.88L for consti pressene. It - 7 2 2 3 3 = 7 2 y =T2 = 27, Ta bv 1ben x 22:44L I max 8.314x lo-2 L ban. - Komel 3270k . T2 = suok Qp = AH = CAMAT = 1 x Zx8.314 Jom X 2.70K 12-0H = 7.857 KJ) AV = n CM,ĐT= n(2+ ) AT - malx q *8.314 Trel x 270 AU = 100 102KJ) AU=q+W = W =DU-q = 10.102-70857 W2 20 25 RJ AS = nomen ( 1 ) = 1x Zx 8.314 X dm 2 AS = 20.14 TLR)

(6) Constant volume cooling to its initial temp. (B-C) AV-0, Tg = 7, W=0]=Pdro AU=q = nCumAT = 1x 9 R X (T3-T2) = 1xqx8.314 x (270 - 540) Au= q =-10.102KJ Akan Commste 1x Zx 8i314x :(270-540) D --7857 kJ) As an cuiment = 1x9x8.314 Initi) TAS =-25.94 J/ k7 Isothermal compression back to I bam, le-A) . AT=0 , Py = P, = 1 ban Plq LAU - AH=07 q=-Wa Vyav, W==nRT ln V --1X8.314 X 270 4 44.88 W = 1.56 kJ 19=-1.56 KJ] AS = nRAN () = 2X 8314 In () = -50775/ .: 052 -50771/5)