H.2 Euler-Maclaur in Sum Formula
2.4 Work Due to Expansion of an Ideal Gas
We calculate the work due to expansion of one mole of an ideal gas that obeys the equation of state Eq. (2.12). For simplicity, we will further assume that the gas has a constant heat capacityCV at constant volume. According to Eq. (2.9), this results in
dU=CVdT; U=CVT+constant. (2.17)
2.4.1 Reversible Isothermal Process
For a reversible isothermal process, the path in theV,pplane is an equilateral hyperbola, pV =constant, where the value of the constant depends onT. We assume that this path joins two states that satisfyp1V1=p2V2, so the quasistatic work is
W=
T=constantpdV=RT V2
V1
dV
V =RTln(V2/V1), one mole. (2.18) ForV2>V1the gas expands and does positive work, as shown inFigure 2–2. For the reverse transformation fromV2toV1, the gas contracts and does negative work; in this case, the environment of the gas does positive work on the gas. SinceUdepends only onT, we have U1=U2soU=0. Therefore, by the first law,Q=Wfor this process.
2.4.2 Reversible Isobaric Expansion Followed by Isochoric Transformation
We assume the path to be a reversible expansion from V1 to V2 at constant pressure p1 (isobaric expansion) followed by lowering the pressure to p2 at constant volume V2 (isochorictransformation). This is illustrated by the dashed line in Figure 2–3. The quasistatic work is
W=p1
V2 V1
dV+ V2 V2
pdV=p1(V2−V1) (2.19) because the second integral is zero. The temperature will change throughout this process.
In general, the end points will have different temperatures,T1=p1V1/RandT2=p2V2/R, and the change in internal energy will beU = CV(T2−T1). If the end points happen to satisfyp1V1 = p2V2, thenT1 = T2, butduring the processT will not be constant. In general,Q=U+W, but ifT1 =T2, thenU =0 andQ=W. Then the work given by Eq. (2.19) can also be written asRT1(V2−V1)/V1 = RT2(V2−V1)/V1and is greater than that given by Eq. (2.18) withT = T1 = T2. The reader is invited to prove this statement mathematically.
2.4.3 Isochoric Transformation Followed by Reversible Isobaric Expansion
We assume the path to consist of lowering the pressure to p2 at constant volume V1
followed by reversible expansion fromV1toV2at constant pressurep2. This is illustrated by the dot-dashed line inFigure 2–3. The quasistatic work is
V p
V1 V2
V1, p1
V2, p2
FIGURE 2–2 Illustration of quasistatic work for isother- mal expansion of one mole of an ideal gas. The system makes a quasistatic transition from a state atV1,p1 to a stateV2,p2 such that pV = RT. The work done by the gas is equal to the area under the curve.
V p
V1 V2
V1, p1
V2, p2
FIGURE 2–3 Illustration of quasistatic work for one mole of an ideal gas. The dashed line represents an isobaric expansion at pressurep1followed by an isochoric transformation atV2. The dot-dashed line represents an isochoric transformation atV1followed by an isobaric transformation at pressure p2. The full line represents an isothermal transformation from V1,p1toV2,p2, which is only possible ifV1p1=V2p2.
W= V1 V1
pdV+p2
V2 V1
dV =p2(V2−V1), (2.20) which is clearly smaller than that given by Eq. (2.19). If the end points happen to be at the same temperature, the work given by Eq. (2.20) can be writtenRT1(V2 −V1)/V2 = RT2(V2−V1)/V2and is less than that given by Eq. (2.18) withT =T1=T2.
2.4.4 Reversible Adiabatic Expansion
We assume that the gas is perfectly insulated from its surroundings so thatδQ = 0 at each stage of the process. Such processes are calledadiabaticprocesses.11 We allow the gas to expand quasistatically, and therefore reversibly, from a stateV2,p2to a stateV3,p3. Applying the first law to each stage of this process gives
CVdT= −pdV, (2.21)
which by Eq. (2.12) may be rewritten in the form CVdT
T +RdV
V =0, one mole of ideal gas. (2.22)
11Some authors use the word adiabatic to mean thatδQ= 0andthat the process is reversible, but we use adiabatic to mean onlyδQ=0. An irreversible adiabatic process is illustrated inSection 2.4.5. See Eq. (3.13) for the entropy change of an adiabatic process.
26 THERMAL PHYSICS
Taking the logarithmic derivative of Eq. (2.12) gives dT/T = dp/p+dV/V which allows Eq. (2.22) to be recast in the form
CV
dp
p +(CV+R)dV
V =0. (2.23)
Eq. (2.23) is a differential equation for the path in theV,pplane. It can be integrated to give
lnp+γlnV=constant, (2.24)
whereγ :=(CV +R)/CV =Cp/CV >1. Exponentiating Eq. (2.24) gives a more usual form pVγ =p2V2γ =p3V3γ =K=constant. (2.25) The path of the system is represented by the solid line inFigure 2–4. The quasistatic work is therefore
W=K V3 V2
V−γdV =K V1−γ 1−γ
V3
V2
=
p3V3−p2V2
1−γ =CV(T2−T3). (2.26) We have labored to produce this result which, however, could have been derived simply by applying the first law withQ =0 to giveW = −U = −CV(T3−T2). Nevertheless, we see clearly how the quasistatic work integral depends on path.
Just as Eq. (2.23) is a differential equation for the path in the V,pplane, Eq. (2.22) is the differential equation of the path in the T,V plane. It could be integrated di- rectly, but the same result can be obtained by substitution of Eq. (2.12) into Eq. (2.25) to obtain
TVγ−1=constant. (2.27)
V p
V3 V3∗
V2, p2
V3, p3
V3∗, p3
FIGURE 2–4 The solid line represents a reversible adiabatic expansion for one mole of an ideal gas forγ =5/3.
The dotted line represents, for the sake of comparison, an isothermal expansion from the same initial state. The point atV3∗,p3is the final state for an irreversible adiabatic expansion at constantexternal pressurep3. In this irreversible case, the system starts in the stateV2,p2, “leaves the page” as it progresses through non-equilibrium states, and “reenters the page,” ultimately coming to equilibrium at the stateV3∗,p3, which is represented by a circled point.
Note thatγ −1=R/CV, consistent with Eq. (2.22). Similarly, T
p(γ−1)/γ =constant. (2.28)
2.4.5 Irreversible Adiabatic Expansion
Here again we assume that the gas is perfectly insulated from its surroundings so that δQ = 0 at each stage of the process. We start out at the same state V2,p2 as for the reversible adiabatic process treated above, but we allow the gas to expand suddenly against a constant reducedexternalpressurep3that is chosen to have the same value as p3for the final state of the reversible adiabatic expansion considered above. During this expansion, the pressure of the gas is not well-defined, so we cannot represent this process by a path inFigure 2–4. Because this process is irreversible, it will come to equilibrium in a state having temperatureT3∗ and volumeV3∗ different from those for the reversible case. The work done will beW = p3(V3∗−V2)and the change in internal energy will be U=CV(T3∗−T2). SinceQ=0 we will haveW= −U, which becomes
p3(V3∗−V2)=CV(T2−T3∗). (2.29) By using Eq. (2.12), we can writep3V3∗=RT3∗andp3V2=RT2p3/p2, in which case Eq. (2.29) can be written in the form
T3∗
T2 =CV+R p3/p2
CV+R =1−q+qr, (2.30)
wherer := p3/p2 andq :=(γ −1)/γ. In this same notation, Eq. (2.28) for the reversible adiabatic expansion leads to
T3
T2 =rq. (2.31)
We shall see thatT3∗>T3. This is illustrated inFigure 2–5.
0.5 1 1.5 2
0.6 0.8 1.2 1.4
T3∗/T2
r
T3/T2
FIGURE 2–5 Graphs ofT3/T2for a reversible adiabatic process, Eq. (2.30), andT3∗/T2for an irreversible adiabatic process, Eq. (2.31), versusr=p3/p2forq=2/5, which corresponds toγ =5/3. The straight line corresponds to T3∗/T2and shows thatT3∗>T3forr=1.
28 THERMAL PHYSICS
We first note forr = 1 thatT3∗ = T3 = T2as expected. Then we take derivatives with respect torto obtain
d dr
T3∗ T2
=q; d dr
T3 T2
=qrq−1. (2.32)
These derivatives are also equal forr=1, so the curve represented by Eq. (2.31) is tangent to the line represented by Eq. (2.30) atr =1. Sinceq−1= −1/γ is negative, we see that the slope of a graph ofT3∗versusris less than that ofT3versusrfor anyr <1. Moreover, the slope of a graph ofT3∗versusris greater than that ofT3versusrfor anyr >1. Hence, T3∗ > T3for anyr = 1, which means that the irreversible adiabatic expansion results in a final state with greater temperature than the reversible adiabatic expansion. The same would be true for contraction, in which caseV3<V2andr>1. For the end points of the two processes, Eq. (2.12) can be writtenp3V3 =RT3andp3V3∗ =RT3∗. Taking the ratio of these equations gives
V3∗ V3 =T3∗
T3
. (2.33)
From this result, we see thatV3∗ > V3. In summary, irreversible adiabatic expansion or contraction against a constant external pressurep3results in a different final state (larger temperature and volume) than a reversible adiabatic expansion to a final state12 having pressurep3.