Question

The figure shows a cycle undergone by 1.41 mol of an ideal monatomic gas. For 1 to 2, what are (a) heat Q, (b) the change in internal energy ΔEint, and (c) the work done W? For 2 to 3, what are (d)Q, (e) ΔEint, and (f)W? For 3 to 1, what are (g)Q, (h) ΔEint, and (i)W? For the full cycle, what are (j)Q, (k) ΔEint, and (l)W? The initial pressure at point 1 is 1.09 atm (where 1 atm = 1.013 × 105 Pa). What are the (m) volume and (n) pressure at point 2 and the (o) volume and (p) pressure at point 3?

Tg = 600 K Pressure Adiabatic T1 = 300 K Tg = 455 K Volume

Please only answer parts j,k,l,m,n,o,p

Answer 1

ginen malul mol Ti = Book T2 = Gook T3z455k 2 is a constant volume process O11->2 = MCVAT 21:41XBR (T2-40) 21:41*2*3.304 Immal t) (600-300) = 5275-233 Similarly change in internal energy for the Process is Dein = CUMAT = 5275.23J G workdone in the process is 6,7220-9 Eint Wazo) d) The process is from 2-33 is adiabatic So A2-33 -0. I change in internal energy for the process 2 3 is semt 2-3 = MCVAT = 1.41X2x8.314x(455-609 2-2.54963

no 3 is an adiabatic process W273 = Eust 2-33 5-6-2-549kJ) 22.949kJ 9. For the process, 3 1, we have B13-= mcrot enER (11-13) =1-4145 x8:314 (300-455) 233) = -4.542 kg) ho change in untermal energy is given by 371 A fint en crat : :=1.413 *8.314% (300-165k) 5 = -2.5 kg 725 I work done. W-8-4E = -4.54 -2-2.54 ks) 2-1.895%) u For the full cycle, we have · OrzO11-12 + O2337013-21 25.275k7+ 0 + (-4.54) 20.735 kg

- change in internal enegy Aegint = AE intant Ajut as t Eintra 25.275 + -2.549) + (-2.325 0 I work done for the full cycle is W-O-Eust. =0.735-0 We 0.735kJ) ME ginen julial Pressure P121.03 x16 pa VIP MRT 2 1:41%8:314x30 : = 0.0347/m3 => B = PA TI = 1.013X105XLDO Ti 300 P2 = 2.03X105 Pa c va 2 VITE = 0.03431x455% =0,06265 -0.03471 x 455k e Pi 300k 0,05265 m3 Dessure at point 3= plessure at 121.013X105ra