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For 1 rightarrow 2, what are the energy transferred as heat Q, J the change in i

ID: 2245772 • Letter: F

Question

For 1 rightarrow 2, what are the energy transferred as heat Q, J the change in interval energy delta Eint, J and the work done W? J For 2 rightarrow 3, what are the energy transferred as heat Q, J the change in internal energy delta Eint, J and the work done W? J For 3 rightarrow 1, what are the energy transferred as heat Q, J anthe change in internal energy delta Eint, J and the work done W? J For the full cycle, what are the net energy transferred as heat Q, J the change in internal energy delta Eint, J and the net work done W? J At point 2 what are the (m) volume and m3 (n) pressure? atm. At point 3 what are the volume and m3 pressure? atm.

Explanation / Answer

1) for 1-----> 2



the process is a isochoric process .


As the volume is constant , work done is zero .


So work done w = 0



So Q= delta E internal = nCv dT


Q= 1.60 x 3R/2 x (600-300)


Q= 1.60 x 3R/2 x 300


Q = 5986.08 J


Q = delta E = 5.98 kJ




2) for 2-----> 3


it is a adiabatic process .


heat in a adiabatic system is zero


so Q=0


dE = nCv dT


= 1.60 x 3R/2 x -145


dE = 2893.72


dE = - 2.89 kJ


work done W = 2.89 kJ




3) for 3-----> 1



it is a isobaric process ,



heat q = nCp dT


= 1.60 x 5R/2 x -155


Q= -5154.68 J


Q= - 5.15 kJ


dE = nCv dT


=1.60 x 3R/2 x -155


= -3092.80


dE = -3092.80 J


dE = -3.09 kJ



work done W = -2061.87 J


work done = -2.06 kJ



net internal enrgy = 5.98 -2.89 - 3.09 = 0


net heat = 5.98 +0 -5.15


net Heat = 0.8314 kJ


net Heat = 831. 4 J


net work done = 828.50 J


at point 1


PV =nRT


V= nRT/P


V= 1.60 x 8.314 x 300/ 1.3 x 1.013 x 105


V = 0.03 m3



So volume at point is also 0.03 m3



we know that at constant volume


P1/T1= P2/T2


P2 = P1T2/T1


= 1.30 x 600/300


P2 = 2.60



pressure at point 2 is 2.60 atm .



pressure at point 3 = pressure at point 1



so pressure at point 3 = 1.30 atm



we know that at constant pressure


V3/t3 = V1/T1


V3 = V1T3/T1


V3 = 0.03 x 455/ 300


V3= 0.0455 m3



volume at point 3 is 0.0455 m3




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