In the figure below, the change in internal energy of a gas that is taken from A
ID: 1561842 • Letter: I
Question
In the figure below, the change in internal energy of a gas that is taken from A to C along the blue path is +755 J. The work done on the gas along the red path ABC is -445 J.
(a) How much energy must be added to the system by heat as it goes from A through B to C?
J
(b) If the pressure at point A is five times that of point C, what is the work done on the system in going from C to D?
J
(c) What is the energy exchanged with the surroundings by heat as the gas goes from C to A along the green path?
J
(d) If the change in internal energy in going from point D to point A is +520 J, how much energy must be added to the system by heat as it goes from point C to point D?
J
Explanation / Answer
(a) Change of internal energy equals heat transferred to the gas plus work done on the gas:
UAC = QABC + WABC
=> QABC = UAC - WABC
=> QABC = 755 - (-445J)
=> QABC = 1200J
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(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
WAB = - PA * VAB
WCD = - Pc * VCD
=> WCD/ WAB = (Pc/PA) (VCD / VAB)
=> WCD/ WAB = (Pc/5PC) (VCD / VCD)
=> WCD/ WAB = -(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
WAB = WABC = - 400J
Therefore:
WCD = (-1/5)WAB = (-1/5) (-445J)
=> WCD = 89J
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(c)
Change of internal energy between A and 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:
UCA = -UAC = -755 J
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)
WCDA = WCD + WDA = 89J + 0 = 89J
So the heat transferred to the gas on this path is:
QCDA = UCA - WCDA = -755 J - 89 J
=> QCDA = -844 J
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(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:
UCA = UCD + UDA
=> UCD = UCA - UCD
=> UCD = -755 J - 520J
=> UCD = -1275J
Work done we calculated in part (b).
QCD = UCD - WCD
=> QCD = -1275J - 89J
=> QCD = -1364 J