Please explain (a) How much energy must be added to the system by heat as it goe
ID: 1440220 • Letter: P
Question
Please explain
(a) How much energy must be added to the system by heat as it goes from A through B and on to C?
1_________ J
(b) If the pressure at point A is five times that of point C, what is the work done by the system in going from C to D?
2________ J
(c) What is the energy exchanged with the surroundings by heat as the cycle goes from C to A along the direct path?
___________ J
(d) If the change in internal energy in going from point D to point A is +590 J, how much energy must be added to the system by heat as it goes from point C to point D?
4____________ J
Explanation / Answer
(a)
Change of internal energy equals heat transferred to the gas plus work done on the gas:
U_AC = Q_ABC + W_ABC
Hence:
Q_ABC = U_AC - W_ABC
= 810J - (+520J)
= 290J
Note:
Internal energy is a state function. So the change of internal energy associated with a process going from state A to state C is always same irrespective of the path you choose. That's why i wrote
U_AC and not U_ABC.
(b)
You can find this work from the given work along the path W_ABC
Work done on the gas is given by the integral:
W = - p dV
for constant pressure processes like on path AB and CD this simplifies to
W = - p dV = - p V
So
W_AB = - p_A V_AB
W_CD = - p_C V_CD
=>
W_CD / W_AB = (p_C/p_A) (V_CD / V_AB)
The pressure ratio is given in the problem. From the diagram you can read, that the volume change has the same magnitude but opposite sign:
W_CD / W_AB = (p_C/(5p_C) (V_CD / -V_CD) = -1/5
Recall the work integral and you find that on the path BC no work is done because the volume does not change. So the work done on the path ABC is equal to the work done on the part from A to B
W_AB = W_ABC = 520J
Therefore:
W_CD = (-1/5)W_AB = (-1/5) (520J) = -104J
(c)
Well I can't see a green path on your picture. I guess you think of the path C-D-A
As pointed out in part (a) the change of internal energy between A nd C does not depend on the path. So going back from C to A changes the internal energy by the same magnitude but opposite sign:
U_CA = -U_AC = -810J
Since no work is done in the section D to A (no volume change = no work). the work done on the whole path equals word done on section CD calculated in part(b)
W_CDA = W_CD + W_DA = -104J + 0 = -104J
So the heat transferred to the gas on this path is:
Q_CDA = U_CA - W_CDA = -810J - (-104)J = -706J
(d)
From given internal energy change on path DA and the value for the whole path CDA from part (c) you can calculate the change between C and D:
U_CA = U_CD + U_DA
=>
U_CD = U_CA - U_CD
= -810J - 590J
= -1400J
Work done we calculated in part (b).
Q_CD = U_CD - W_CD
= -1400J -(-) 104J
= -1296J