Calculate the total work done in the cyclical process ABCDA shown in Figure 7(b) by the following two methods to verify that work equals the area inside the closed loop on the PVPV size 12{ ital "PV"} {} diagram. (Take the data in the figure to be precise to three significant figures.) (a) Calculate the work done along each segment of the path and add these values to get the total work. (b) Calculate the area inside the rectangle ABCDA.

*Strategy*

To find the work along any path on a PVPV size 12{ ital "PV"} {} diagram, you use the fact that work is pressure times change in volume, or W=PΔVW=PΔV size 12{W=PΔV} {}. So in part (a), this value is calculated for each leg of the path around the closed loop.

*Solution for (a)*

The work along path AB is

W
AB
=
P
AB
ΔV
AB
=
(
1
.
50
×
10
6
N/m
2
)
(
5
.
00
×
10
–4
m
3
)
=
750
J.
W
AB
=
P
AB
ΔV
AB
=
(
1
.
50
×
10
6
N/m
2
)
(
5
.
00
×
10
–4
m
3
)
=
750
J.
alignl { stack {
size 12{W rSub { size 8{"AB"} } =P rSub { size 8{"AB"} } DV rSub { size 8{"AB"} } } {} #
= \( 1 "." "50"´"10" rSup { size 8{6} } " N/m" rSup { size 8{2} } \) \( 5 "." "00"´"10" rSup { size 8{4} } " m" rSup { size 8{3} } \) ="750"" J" "." {}
} } {}

(5)Since the path BC is isochoric, ΔVBC=0ΔVBC=0 size 12{DV rSub { size 8{"BC"} } =0} {}, and so WBC=0WBC=0 size 12{W rSub { size 8{"BC"} } =0} {}. The work along path CD is negative, since ΔVCDΔVCD size 12{DV rSub { size 8{"CD"} } } {} is negative (the volume decreases). The work is

W
CD
=
P
CD
ΔV
CD
=
(
2
.
00
×
10
5
N/m
2
)
(
–5
.
00
×
10
–4
m
3
)
=
–
100
J
.
W
CD
=
P
CD
ΔV
CD
=
(
2
.
00
×
10
5
N/m
2
)
(
–5
.
00
×
10
–4
m
3
)
=
–
100
J
.
alignl { stack {
size 12{W rSub { size 8{"CD"} } =P rSub { size 8{"CD"} } DV rSub { size 8{"CD"} } } {} #
= \( 2 "." "00"´"10" rSup { size 8{5} } " N/m" rSup { size 8{2} } \) \( 5 "." "00"´"10" rSup { size 8{4} } " m" rSup { size 8{3} } \) "=-""100"" J" "." {}
} } {}

(6)Again, since the path DA is isochoric, ΔVDA=0ΔVDA=0 size 12{DV rSub { size 8{"DA"} } =0} {}, and so WDA=0WDA=0 size 12{W rSub { size 8{"DA"} } =0} {}. Now the total work is

W
=
W
AB
+
W
BC
+
W
CD
+
W
DA
=
750 J
+
0
+
(
−
100
J
)
+
0
=
650 J.
W
=
W
AB
+
W
BC
+
W
CD
+
W
DA
=
750 J
+
0
+
(
−
100
J
)
+
0
=
650 J.

(7)*Solution for (b)*

The area inside the rectangle is its height times its width, or

area
=
(
P
AB
−
P
CD
)
ΔV
=
(
1.50
×
10
6
N/m
2
)
−
(
2
.
00
×
10
5
N/m
2
)
(
5
.
00
×
10
−
4
m
3
)
=
650 J.
area
=
(
P
AB
−
P
CD
)
ΔV
=
(
1.50
×
10
6
N/m
2
)
−
(
2
.
00
×
10
5
N/m
2
)
(
5
.
00
×
10
−
4
m
3
)
=
650 J.
alignl { stack {
size 12{"area"= \( P rSub { size 8{"AB"} } -P rSub { size 8{"CD"} } \) DV} {} #
= left [ \( 1 "." "50"´"10" rSup { size 8{6} } " N/m" rSup { size 8{2} } \) - \( 2 "." "00"´"10" rSup { size 8{5} } " N/m" rSup { size 8{2} } \) right ]´ \( 5 "." "00"´"10" rSup { size 8{-4} } " m" rSup { size 8{3} } \) {} #
="750"" J" "." {}
} } {}

(8)Thus,

area=650 J=W.area=650 J=W. size 12{"area"="650"" J"=W} {}

(9)*Discussion*

The result, as anticipated, is that the area inside the closed loop equals the work done. The area is often easier to calculate than is the work done along each path. It is also convenient to visualize the area inside different curves on PVPV size 12{ ital "PV"} {} diagrams in order to see which processes might produce the most work. Recall that work can be done to the system, or by the system, depending on the sign of WW size 12{W} {}. A positive WW size 12{W} {} is work that is done by the system on the outside environment; a negative WW size 12{W} {} represents work done by the environment on the system.

Figure 8(a) shows two other important processes on a PVPV size 12{ ital "PV"} {} diagram. For comparison, both are shown starting from the same point A. The upper curve ending at point B is an isothermal process—that is, one in which temperature is kept constant. If the gas behaves like an ideal gas, as is often the case, and if no phase change occurs, then PV=nRTPV=nRT size 12{ ital "PV"= ital "nRT"} {}. Since TT size 12{T} {} is constant, PVPV size 12{ ital "PV"} {} is a constant for an isothermal process. We ordinarily expect the temperature of a gas to decrease as it expands, and so we correctly suspect that heat transfer must occur from the surroundings to the gas to keep the temperature constant during an isothermal expansion. To show this more rigorously for the special case of a monatomic ideal gas, we note that the average kinetic energy of an atom in such a gas is given by

1
2
m
v
¯
2
=
3
2
kT
.
1
2
m
v
¯
2
=
3
2
kT
.
size 12{ { {1} over {2} } m { bar {v}} rSup { size 8{2} } = { {3} over {2} } ital "kT" "." } {}

(10)The kinetic energy of the atoms in a monatomic ideal gas is its only form of internal energy, and so its total internal energy UU size 12{U} {} is

U=N12mv¯2=32NkT, (monatomic ideal gas),U=N12mv¯2=32NkT, (monatomic ideal gas), size 12{U=N cdot { {1} over {2} } m { bar {v}} rSup { size 8{2} } = { {3} over {2} } ital "NkT",} {}

(11)where NN size 12{N} {} is the number of atoms in the gas. This relationship means that the internal energy of an ideal monatomic gas is constant during an isothermal process—that is, ΔU=0ΔU=0 size 12{ΔU=0} {}. If the internal energy does not change, then the net heat transfer into the gas must equal the net work done by the gas. That is, because ΔU=Q−W=0ΔU=Q−W=0 size 12{ΔU=Q - W=0} {} here, Q=WQ=W size 12{Q=W} {}. We must have just enough heat transfer to replace the work done. An isothermal process is inherently slow, because heat transfer occurs continuously to keep the gas temperature constant at all times and must be allowed to spread through the gas so that there are no hot or cold regions.

Also shown in Figure 8(a) is a curve AC for an adiabatic process, defined to be one in which there is no heat transfer—that is, Q=0Q=0 size 12{Q=0} {}. Processes that are nearly adiabatic can be achieved either by using very effective insulation or by performing the process so fast that there is little time for heat transfer. Temperature must decrease during an adiabatic process, since work is done at the expense of internal energy:

U=32NkT.U=32NkT. size 12{U= { {3} over {2} } ital "NkT"} {}

(12)(You might have noted that a gas released into atmospheric pressure from a pressurized cylinder is substantially colder than the gas in the cylinder.) In fact, because Q=0, ΔU=–WQ=0, ΔU=–W size 12{Q=0, DU"=-"W} {} for an adiabatic process. Lower temperature results in lower pressure along the way, so that curve AC is lower than curve AB, and less work is done. If the path ABCA could be followed by cooling the gas from B to C at constant volume (isochorically), Figure 8(b), there would be a net work output.

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