A cylindrical vessel with rigid adiabatic walls is separated intotwo parts by a

Question

A cylindrical vessel with rigid adiabatic walls is separated intotwo parts by a frictionless adiabatic piston. Each part contains50.0 L of an ideal monatomic gas with CV, m =3R/2 (R= gas constant ~ 8.314 J/molK). Initially,Ti= 325 K and Pi= 2.50 x105 Pa in each part. Heat is slowly introducedinto the left part using an electrical heater until the piston hasmoved sufficiently to the right to result in a final pressurePf= 7.50 barr in the right part. Consider thecompression of the gas in the right part to be a reversibleprocess.

(a) Calculate the work done on the right part in this process andthe final temperature in the right part.
(b) Calculate the final temperature in the left part and the amountof heat that flowed into this part.

Please try to show your steps.

Explanation / Answer

Right side is adiabatic compression P*V = constant     =Cp/Cv = (Cv + R)/Cv = (1.5 R+ R)/(1.5 R) = (2.5/1.5) = 5/3 constant = Pi*Vi = (2.50 e5Pa)*(50 L)5/3 = 169651101 Pa*L5/3 Pf = 7.50 bar = 7.50 bar *(10e5 Pa/1 bar) = 7.5e5 Pa Vf = [constant/P]1/ = (169651101Pa*L5/3/7.5e5 Pa)3/5 = 25.8640929 Liters =25.86 L W = -integral [P*dV] = -constant *integral [ dV/V]= -constant/(-1)*(-1/V-1)   evaluated from V final to Vinitial -1 = 5/3 - 1 = 2/3 W = [constant/(2/3)]*[1/(25.86 L)2/3 - 1/(50L)2/3] W = (3/2)*169651101 Pa*L5/3[1/(25.86 L)2/3 -1/(50 L)2/3] W = 10347104.5 Pa*L W = 10347104.5 Pa*L *(1 m3/1000L) *( 1 J/ [1 Pa*m3]) *(1 kJ/1000J)= 10.347 kJ ~ 10.3 kJ B) Pf on the left = Pf on the right (mechanical equilibrium) = 7.5 bar= 7.5 e5 Pa Vf on the left + Vf on the right = Vi on left +Vi on right         (totalvolume does not change) Vf on the left = Vi on left + Vi on right - Vf on the right = 50L +50 L - 25.8640929 L Vf on the left = 74.1359071 Liters Final Temperature Pi*Vi/Ti = Pf*Vf/Tf Tf = (Pf/Pi)*(Vf/Vi)*Ti Tf = (7.5e5 Pa / 2.5e5 Pa)*(74.135907 L / 50 L)*325 K Tf = 1445.65019 Kelvin ~ 1446 K no. moles = PV/(RT) = (2.5e5 Pa*50 e-3 m3)/(8.314 J/mol/K *325K) =4.62611721 moles U = n*Cv*T = 4.62611721 moles*(3*8.314 J/mol/K /2)*(1445.65019 K - 325 K) U = 64652.8956 Joules = 64.65 kJ W_left = -W_right = - 10.3kJ       (mechanicalreversibility) U = q + W_right q = U - W_right = 64.65 kJ - (- 10.3 kJ) ~ 75.0 kJ I hope this has helped you. PM me if you are still confused. Alsoplease rate my answer for the time it took me to complete thequestion.

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