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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