Question

Please answer all the questions

Consider the ideal Stirling cycle shown above, working between a maximum temperature of 770 K and a minimum temperature of 420 K and a minimum volume of 0.3 L and a maximum volume of 1.2 L. The working gas of the cycle is 0.10 mol of an ideal gas with c′_v = 20.8 J/mol-K.

Note, +Q = heat absorbed by the gas and +W = work done by the gas.

isothermal Pressure W isothermal V2 Volume Consider the ideal Stirling cycle shown above, working between a maximum temperature of 770 K and a minimum temperature of 420 K and a minimum volume of 0.3 L and a maximum volume of 1.2 L. The working gas of the cycle is 0.10 mol of an ideal gas with c'y = 20.8 J/mol-K. Note, +Q = heat absorbed by the gas and +W = work done by the gas.

Answer 1

Cv = 20.8 n=0.10 mo) Lotha d T-Th=770K 3.Me IL a To=T,= 4200 1,-0.3 W 12 c) Qcd = nR Tn in (*) = 0.10×8.314x710 in (12) .887.47 J D) Qda = 0 A) Qab = nRT, In ( ) = 0.10x8314x 42010 (92) E-484.07J B) Q bc = 0. PV=nRT | E) Pa Va Forta Pa=nRTA = 0.10*8.314x420 1.2 = 290.99 TE PGVG EAR TO P6= 0.10+8.314x420 0.3 1163.96 G) Pc = 0.10*8.314 x 770 = 2133.92 0.3 H) Wab = -AR Te In ( ) = -0.10 «8.914x 4201002) - 484.07 J I Wbc = ARE NECT-T.) = 0.16x20.8( 770-420) =7285

5) Wed = -ART, In (42) = -887.47 K) Wda = nC. (Tion-Tn) = 0.10*20.8 (420-770) = - 728 2) in lefficiency) = (1 - In)x100 = - 420)<c00 = 46% PLEASE RATE IT UP IT THAT IT UP 11 THANKSI!