A heat engine does work by using a gas at an initial pressure of 1000 Pa and vol
ID: 1303409 • Letter: A
Question
A heat engine does work by using a gas at an initial pressure of 1000 Pa and volume .1m 3. Step-by-step, it then increases the pressure to 10,000 Pa (at constant volume), increases the volume to .15m3 (at constant pressure), decreases the pressure back to 1,000 Pa (at constant volume) and returns the volume back to .1m3 (at constant pressure).
How much work is done by the gas during the first step?
W = 0 J
W = 450 J
W = 500 J
W = 750 J
W = 1250 J
W = 1350 J
How much work is done by the gas during the second step?
W = 0 J
W = 450 J
W = 500 J
W = 750 J
W = 1250 J
W = 1350 J
How much work is done by this heat engine in one complete cycle?
W = 0 J
W = 450 J
W = 500 J
W = 750 J
W = 1250 J
W = 1350 J
What is the change in internal energy during the first step?
DU = 0 J
DU = 450 J
DU = 500 J
DU = 750 J
DU = 1250 J
DU = 1350 J
What is the change in internal energy during the second step?
DU = 0 J
DU = 450 J
DU = 500 J
DU = 750 J
DU = 1250 J
DU = 1350 J
What is the change in internal energy over one complete cycle?
DU = 0 J
DU = 450 J
DU = 500 J
DU = 750 J
DU = 1250 J
DU = 1350 J
How much heat is added to the gas in the first step?
Q = 0 J
Q = 450 J
Q = 500 J
Q = 750 J
Q = 1250 J
Q = 1350 J
How much heat is added to the gas in the second step?
Q = 0 J
Q = 450 J
Q = 500 J
Q = 750 J
Q = 1250 J
Q = 1350 J
How much heat is added to the gas in one complete cycle?
Q = 0 J
Q = 450 J
Q = 500 J
Q = 750 J
Q = 1250 J
Q = 1350 J
Explanation / Answer
a) at A, PV = n RT >> n=1 gram mole R = 8.314 put the value of P & V calculate T(A)
-----------------
during AB constant Pressure, >>
PV(A) = R T(A) and PV(B) = R T(B)
divide >> T(B) = [V(B)/V(A)]*T(A) = ?? T(A) calculate as T(A) known
------------------------
during CA constant volume, >>
P(c) V = R T(c) and P(a) V = R T(a)
divide >> T(c) = [P(c)/P(a)]*T(a) calculate as T(A) known
===============
what you find relation in T(c) & T(b)
if equal then B-C process is Isothermal compression >> dU = 0
---------------------
net work done = area of ABCA
W = 0.5*AB*AC
----------------------
change in internal energy = 1 cycle = 0 because system started fro T(A) temp and reached back at T(A) U = f (T) and depends on end points not the path taken
internal energy = 1 cycle = same
dU = >> u = constant
---------
head added/removed >> dQ = dU + dW = 0 +work done in cylce
for this >> dW1 = P dV (constant pressure A-B process) its expansion, work done by gas >>> i take that +ve
C-A process> dV=0 constant volume dw2 =0
B-C process >> if isothermal (Tb=Tc)
w(bc) = RT(b) log [vc/vb] >>> will be negative
add all works in proper sign
dw = + or - ve will decide
dQ = + or -ve
-------------------------
before cycle >> Ta = ?
heat content Q(initial) >>> S(initial) = entropy = Q(ini)/Ta
after cycle Ta = ? same
heat content Q2 = Q(initial) + or - dQ from above
S(final) = Q2/Ta
result & calculations will fix things precisly.
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