Volumetric Properties of Pure Fluids
3.3 IDEAL GAS AND IDEAL-GAS STATE
infinity. Thus, it is a law only at limiting conditions. As these limits are approached, the mol- ecules making up a gas become more and more widely separated, and the volume of the mol- ecules themselves becomes a smaller and smaller fraction of the total volume occupied by the gas. Furthermore, the forces of attraction between molecules become ever smaller because of the increasing distances between them. In the zero-pressure limit, molecules are separated by infinite distances. Their volumes become negligible compared with the total volume of the gas, and the intermolecular forces approach zero. The ideal gas concept extrapolates this behavior to all conditions of temperature and pressure.
The internal energy of a real gas depends on both pressure and temperature. Pressure dependence results from intermolecular forces. If such forces did not exist, no energy would be required to alter intermolecular distances, and no energy would be required to bring about pressure and volume changes in a gas at constant temperature. Thus, in the absence of inter- molecular forces, internal energy would depend on temperature only.
These observations are the basis for the concept of a hypothetical state of matter designated the ideal-gas state. It is the state of a gas comprised of real molecules that have negligible molecular volume and no intermolecular forces at all temperatures and pressures.
Although related to the ideal gas, it presents a different perspective. It is not the gas that is ideal, but the state, and this has practical advantages. Two equations are fundamental to this state, namely the “ideal-gas law” and an expression showing that internal energy depends on temperature alone:
∙ The equation of state:
PV ig = RT (3.7)
∙ Internal energy:
U ig = U( T) (3.8) The superscript ig denotes properties for the ideal-gas state.
The property relations for this state are very simple, and at appropriate conditions of T and P they may serve as suitable approximations for direct application to the real-gas state.
However, they have far greater importance as part of a general three-step procedure for calcu- lating property changes for real gases that includes a major step in the ideal-gas state. The three steps are as follows:
1. Evaluate property changes for the mathematical transformation of an initial real-gas state into the ideal-gas state at the same T and P.
2. Calculate property changes in the ideal-gas state for the T and P changes of the process.
3. Evaluate property changes for the mathematical transformation of the ideal-gas state back to the real-gas state at the final T and P.
This procedure calculates the primary property-value changes resulting from T and P changes by simple, but exact, equations for the ideal-gas state. The property-value changes for transitions between real and ideal-gas states are usually relatively minor corrections. These transition calculations are treated in Chapter 6. Here, we develop property-value calculations for the ideal-gas state alone.
3.3. Ideal Gas and Ideal-Gas State 79
Property Relations for the Ideal-Gas State
The definition of heat capacity at constant volume, Eq. (2.15), leads for the ideal-gas state to the conclusion that CVig is a function of temperature only:
CVig ≡ ( ∂ ____U ig
∂ T )
V
= d______U ig ( T)
dT = CVig ( T) (3.9) The defining equation for enthalpy, Eq. (2.10), applied to the ideal-gas state, leads to the conclusion that H ig is also a function only of temperature:
H ig ≡ U ig + PV ig = U ig ( T) + RT = H ig ( T) (3.10) The heat capacity at constant pressure CPig , defined by Eq. (2.19), like CVig , is a function of temperature only:
CPig ≡ ( ____∂ H ig
∂ T )
P
= d___________H ig ( T)
dT = CPig ( T) (3.11) A useful relation between CPig and CVig for the ideal-gas state comes from differentiation of Eq. (3.10):
CPig ≡ _____dH ig dT = ____dU ig
dT + R = CVig + R (3.12) This equation does not mean that CPig and CVig are themselves constant for the ideal-gas state, but only that they vary with temperature in such a way that their difference is equal to R. For any change in the ideal-gas state, Eqs. (3.9) and (3.11) lead to:
dU ig = CVig dT (3.13a) Δ U ig = ∫ C Vig dT (3.13b)
dH ig = CPig dT (3.14a) Δ H ig = ∫ CPig dT (3.14b)
Because both U ig and CVig for the ideal-gas state are functions of temperature only, Δ U ig for the ideal-gas state is always given by Eq. (3.13b), regardless of the kind of process causing the change. This is illustrated in Fig. 3.5, which shows a graph of internal energy as a function of V ig at two different temperatures.
The dashed line connecting points a and b represents a constant-volume process for which the temperature increases from T1 to T2 and the internal energy changes by Δ U ig = U 2ig − U 1ig . This change in internal energy is given by Eq. (3.13b) as Δ U ig = ∫ CVig dT . The dashed lines connecting points a and c and points a and d represent other processes not occurring at constant volume but which also lead from an initial temperature T1 to a final temperature T2. The graph shows that the change in U ig for these processes is the same as for the constant-volume process, and it is therefore given by the same equation, namely, Δ U ig = ∫ CVig dT . However, Δ U ig is not equal to Q for these processes, because Q depends not only on T1 and T2 but also on the path of the process. An entirely analogous discussion applies to the enthalpy H ig in the ideal-gas state.
Process Calculations for the Ideal-Gas State
Process calculations provide work and heat quantities. The work of a mechanically reversible closed-system process is given by Eq. (1.3), here written:
dW = −P dV ig (1.3)
For the ideal-gas state in any closed-system process, the first law as given by Eq. (2.6) written for a unit mass or a mole may be combined with Eq. (3.13a) to give:
dQ + dW = CVig dT
Substitution for dW by Eq. (1.3) and solution for dQ yields an equation valid for the ideal-gas state in any mechanically reversible closed-system process:
dQ = CVig dT + PdV ig (3.15)
This equation contains the variables P, Vig, and T, only two of which are independent.
Working equations for dQ and dW depend on which pair of these variables is selected as independent; i.e., upon which variable is eliminated by Eq. (3.7). We consider two cases, eliminating first P, and second, Vig. To eliminate P, we substitute P = RT / V ig into Eqs. (3.15) and (1.3) to obtain:
Figure 3.5: Internal energy changes for the ideal-gas state. Because U ig is independent of V ig , the plot of U ig vs. V ig at constant temperature is a horizontal line. For different temperatures, U ig has different values, with a separate line for each temperature. Two such lines are shown, one for temperature T 1 and one for a higher temperature T 2 .
U2 T2
U1 U
Vig
T1
b
a
d c
dQ = CVig dT + RT d____V ig
V ig (3.16) dW = −RT ____dV ig
V ig (3.17) To eliminate dV ig we take a differential of V ig = RT / P, obtaining dVig = R
__ P (dT − T ___dP P ).
Substituting for dV ig and for CVig = CPig − R transforms Eqs. (3.15) and (1.3) into:
dQ = CPig dT − RT dP___
P (3.18) dW = −RdT + RT ___dP
P (3.19)
These equations apply to the ideal-gas state for various process calcula- tions. The assumptions implicit in their derivation are that the system is closed and the process being considered is mechanically reversible.
3.3. Ideal Gas and Ideal-Gas State 81
Isothermal Process
By Eqs. (3.13b) and (3.14b),
Δ U ig = Δ H ig = 0 ( const T) By Eqs. (3.16) and (3.18), Q = RT ln V___2ig
V1ig = RT ln ___P 1 P 2 By Eqs. (3.17) and (3.19), W = RT ln ___V1ig
V2ig = RT ln P___ 2 P 1
Because Q = −W , a result that also follows from Eq. (2.3), we can write in summary:
Q = −W = RT ln V_2ig
V1ig = RT ln _P 1
P 2 ( const T) (3.20)
Isobaric Process
By Eqs. (3.13b) and (3.19) with dP = 0,
Δ U ig = ∫ CVig dT and W = −R ( T 2 − T 1 ) By Eqs. (3.14b) and (3.18),
Q = Δ H ig = ∫ CPig dT ( const P) (3.21)
Isochoric (Constant-V) Process
With dV ig = 0, W = 0 , and by Eqs. (3.13b) and (3.16),
Q = Δ U ig = ∫ CVig dT (const V ig ) (3.22)
Adiabatic Process; Constant Heat Capacities
An adiabatic process is one for which there is no heat transfer between the system and its sur- roundings; i.e., dQ = 0. Each of Eqs. (3.16) and (3.18) may therefore be set equal to zero.
Integration with CVig and CPig constant then yields simple relations among the variables T, P, and V ig , valid for mechanically reversible adiabatic compression or expansion in the ideal-gas state with constant heat capacities. For example, Eq. (3.16) becomes:
dT
___T = − ___R CVig ____dV ig
V ig
Integration with CVig constant gives:
T 2
___T 1 = ( ___V1ig V2ig )
R/ C Vig
Similarly, Eq. (3.18) leads to:
T 2
___T 1 = ( ___P 2 P 1 ) R/ C Pig These equations may also be expressed as:
T ( V ig ) γ − 1 = const (3.23a) TP ( 1 − γ ) /γ = const (3.23b) P ( V ig ) γ = const (3.23c) where Eq. (3.23c) results by combining Eqs. (3.23a) and (3.23b) and where by definition,4
γ ≡ CP ig_
CVig (3.24)
Equations (3.23) apply for the ideal-gas state with constant heat capaci- ties and are restricted to mechanically reversible adiabatic expansion or compression.
The first law for an adiabatic process in a closed system combined with Eq. (3.13a) yields:
dW = dU = CVig dT
For constant CVig ,
W = Δ U ig = CVig ΔT (3.25)
Alternative forms of Eq. (3.25) result if CVig is eliminated in favor of the heat-capacity ratio γ:
γ ≡ ___CPig
CVig = _____CVig + R
CVig = 1 + ___R
CVig or CVig = ___R γ − 1 and
W = CVig ΔT = ____RΔT γ − 1
4If C pig and C vig are constant, γ is necessarily constant. The assumption of constant γ is equivalent to the assumption that the heat capacities themselves are constant. This is the only way that the ratio C pig / C vig and the difference C pig – C vig = R can both be constant. Except for the monatomic gases, both C pig and C vig actually increase with temperature, but the ratio γ is less sensitive to temperature than the heat capacities themselves.
3.3. Ideal Gas and Ideal-Gas State 83 Because RT 1 = P 1 V1ig and RT 2 = P 2 V2ig , this expression may be written:
W = R_______T 2 − RT 1 γ − 1 =
P 2 V2ig − P 1 V1ig
___________
γ − 1 (3.26)
Equations (3.25) and (3.26) are general for adiabatic compression and expansion processes in a closed system, whether reversible or not, because P, Vig, and T are state functions, independent of path. However, T2 and V2ig are usually unknown. Elimination of V 2ig from Eq. (3.26) by Eq. (3.23c), valid only for mechanically reversible processes, leads to the expression:
W = _____P 1 V1ig γ − 1 [ ( ___P 2
P 1 )
(γ − 1)/γ
− 1] = ___γR − 1 T 1 [ ( P___ 2 P 1 )
(γ − 1)/γ
− 1] (3.27)
The same result is obtained when the relation between P and Vig given by Eq. (3.23c) is used for the integration, W = −∫ Pd V ig .
Equation (3.27) is valid only for the ideal-gas state, for constant heat capacities, and for adiabatic, mechanically reversible, closed-system processes.
When applied to real gases, Eqs. (3.23) through (3.27) often yield satisfactory approxi- mations, provided the deviations from ideality are relatively small. For monatomic gases, γ = 1.67; approximate values of γ are 1.4 for diatomic gases and 1.3 for simple polyatomic gases such as CO2, SO2, NH3, and CH4.
Irreversible Processes
All equations developed in this section have been derived for mechanically reversible, closed-sys- tem processes for the ideal-gas state. However, the equations for property changes— dUig, dHig, ΔUig, and ΔHig—are valid for the ideal-gas state regardless of the process. They apply equally to reversible and irreversible processes in both closed and open systems, because changes in prop- erties depend only on initial and final states of the system. On the other hand, an equation for Q or W, unless it is equal to a property change, is subject to the restrictions of its derivation.
The work of an irreversible process is usually calculated by a two-step procedure.
First, W is determined for a mechanically reversible process that accomplishes the same change of state as the actual irreversible process. Second, this result is multiplied or divided by an efficiency to give the actual work. If the process produces work, the absolute value for the reversible process is larger than the value for the actual irreversible process and must be multiplied by an efficiency. If the process requires work, the value for the reversible process is smaller than the value for the actual irreversible process and must be divided by an effi- ciency. In this approach, efficiencies are always less than 1, approaching 1 as a process approaches reversibility.
Applications of the concepts and equations of this section are illustrated in the examples that follow. In particular, the work of irreversible processes is treated in Ex. 3.5.
Example 3.3
Air is compressed from an initial state of 1 bar and 298.15 K to a final state of 3 bar and 298.15 K by three different mechanically reversible processes in a closed system:
(a) Heating at constant volume followed by cooling at constant pressure.
(b) Isothermal compression.
(c) Adiabatic compression followed by cooling at constant volume.
These processes are shown in the figure. We assume air to be in its ideal-gas state, and assume constant heat capacities, C Vig = 20.785 and C Pig = 29.100 J·mol−1·K−1. Calculate the work required, heat transferred, and the changes in internal energy and enthalpy of the air for each process.
0 5 10 15 20 25
2 4 6
c
a b
Vig 103 m3
P bar 2
1
Solution 3.3
Choose the system as 1 mol of air. The initial and final states of the air are identi- cal with those of Ex. 2.7. The molar volumes given there are
V1ig = 0.02479 m 3 V2ig = 0.008263 m 3
Because T is the same at the beginning and end of the process, in all cases, Δ U ig = Δ H ig = 0
(a) The process here is exactly that of Ex. 2.7(b), for which:
Q = −4958 J and W = 4958 J
(b) Equation (3.20) for isothermal compression applies. The appropriate value of R here (from Table A.2 of App. A) is R = 8.314 J· mol −1 · K −1 .
Q = −W = RT ln ___P 1
P 2 = ( 8.314 ) ( 298.15 ) ln __1
3 = −2723 J
(c) The initial step of adiabatic compression takes the air to its final volume of 0.008263 m3. By Eq. (3.23a), the temperature at this point is:
3.3. Ideal Gas and Ideal-Gas State 85
T′ = T 1 ( ___V1ig V2ig )
γ − 1
= ( 298.15 ) ( ________0.02479
0.008263 ) 0.4 = 462.69 K For this step, Q = 0, and by Eq. (3.25), the work of compression is:
W = CVig ΔT = CVig ( T′ − T 1 ) = ( 20.785 ) ( 462.69 − 298.15 ) = 3420 J For the constant-volume step, no work is done; the heat transfer is:
Q = Δ U ig = CVig ( T 2 − T′ ) = 20.785 ( 298.15 − 462.69 ) = −3420 J Thus for process (c),
W = 3420 J and Q = −3420 J
Although the property changes ΔUig and ΔHig are zero for each process, Q and W are path-dependent, and here Q = −W. The figure shows each process on a PVig diagram. Because the work for each of these mechanically reversible processes is given by W = −∫ Pd V ig , the work for each process is proportional to the total area below the paths on the PVig diagram from 1 to 2. The relative sizes of these areas correspond to the numerical values of W.
Example 3.4
A gas in its ideal-gas state undergoes the following sequence of mechanically reversible processes in a closed system:
(a) From an initial state of 70°C and 1 bar, it is compressed adiabatically to 150°C.
(b) It is then cooled from 150 to 70°C at constant pressure.
(c) Finally, it expands isothermally to its original state.
Calculate W, Q, ΔUig, and ΔHig for each of the three processes and for the entire cycle.
Take C Vig = 12.471 and C Pig = 20.785 J· mol −1 ·K −1 .
Solution 3.4
Take as a basis 1 mol of gas.
(a) For adiabatic compression, Q = 0, and
Δ U ig = W = CV ig ΔT = ( 12.471 ) ( 150 − 70 ) = 998 J Δ H ig = CPig ΔT = ( 20.785 ) ( 150 − 70 ) = 1663 J Pressure P2 is found from Eq. (3.23b):
P 2 = P 1 ( ___T 2 T 1 )
γ/ ( γ − 1 )
= ( 1 ) ( 150 + 273.15___________
70 + 273.15 ) 2.5 = 1.689 bar
(b) For this constant-pressure process,
Q = Δ H ig = CPig ΔT = ( 20.785 ) ( 70 − 150 ) = −1663 J ΔU = CVig ΔT = ( 12.471 ) ( 70 − 150 ) = −998 J
W = Δ U ig − Q = −998 − ( −1663 ) = 665 J
(c) For this isothermal process, ΔUig and ΔHig are zero; Eq. (3.20) yields:
Q = −W = RT ln ___P 3
P 1 = RT ln ___P 2
P 1 = ( 8.314 ) ( 343.15 ) ln _____1.689
1 = 1495 J For the entire cycle,
Q
=
0 − 1663 + 1495 = −168 J W
= 998 + 665 − 1495 = 168 J Δ U ig = 998 − 998 + 0 = 0 Δ H ig
=
1663 − 1663 + 0 = 0
The property changes ΔUig and ΔHig both are zero for the entire cycle because the initial and final states are identical. Note also that Q = −W for the cycle. This follows from the first law with ΔUig = 0.
Example 3.5
If the processes of Ex. 3.4 are carried out irreversibly but so as to accomplish exactly the same changes of state—the same changes in P, T, Uig, and Hig—then different values of Q and W result. Calculate Q and W if each step is carried out with a work efficiency of 80%.
Solution 3.5
If the same changes of state as in Ex. 3.4 are carried out by irreversible processes, the property changes for the steps are identical with those of Ex. 3.4. However, the values of Q and W change.
P
Vig
70°C 150°C
70°C
a b
c
3 2
1
3.3. Ideal Gas and Ideal-Gas State 87 (a) For mechanically reversible, adiabatic compression, the work is Wrev = 998 J.
If the process is 80% efficient compared with this, the actual work is larger, and W = 998/0.80 = 1248 J. This step cannot here be adiabatic. By the first law,
Q = Δ U ig − W = 998 − 1248 = −250 J
(b) The work required for the mechanically reversible cooling process is 665 J. For the irreversible process, W = 665/0.80 = 831 J. From Ex. 3.4(b), ΔUig = −998 J, and Q = Δ U ig − W = −998 − 831 = −1829 J
(c) As work is done by the system in this step, the irreversible work in absolute value is less than the reversible work of −1495 J, and the actual work done is:
W = ( 0.80 ) ( −1495 ) = −1196 J Q = Δ U ig − W = 0 + 1196 = 1196 J For the entire cycle, ΔUig and ΔHig are zero, with
Q = −250 − 1829 + 1196 = −883 J W = 1248 + 831 − 1196 = 883 J
A summary of these results and those for Ex. 3.4 is given in the following table;
values are in joules.
Mechanically reversible, Ex. 3.4 Irreversible, Ex. 3.5
ΔUig ΔHig Q W ΔUig ΔHig Q W
(a) 998 1663 0 998 998 1663 −250 1248
(b) −998 −1663 −1663 665 −998 −1663 −1829 831
(c) 0 0 1495 −1495 0 0 1196 −1196
Cycle 0 0 −168 168 0 0 −883 883
The cycle is one which requires work and produces an equal amount of heat.
The striking feature of the comparison shown in the table is that the total work required when the cycle consists of three irreversible steps is more than five times the total work required when the steps are mechanically reversible, even though each irreversible step is assumed to be 80% efficient.
Example 3.6
Air flows at a steady rate through a horizontal pipe to a partly closed valve. The pipe leaving the valve is enough larger than the entrance pipe that the kinetic-energy change of the air as it flows through the valve is negligible. The valve and connecting pipes are well insulated. The conditions of the air upstream from the valve are 20°C and 6 bar, and the downstream pressure is 3 bar. If the air is in its ideal-gas state, what is the temperature of the air some distance downstream from the valve?
Solution 3.6
Flow through a partly closed valve is known as a throttling process. The system is insulated, making Q negligible; moreover, the potential-energy and kinetic- energy changes are negligible. No shaft work is accomplished, and Ws = 0. Hence, Eq. (2.31) reduces to:
ΔH ig = H2ig − H1ig = 0. Because Hig is a function of temperature only, this requires that T2 = T1. The result that ΔHig = 0 is general for a throttling process, because the assumptions of negligible heat transfer and potential- and kinetic- energy changes are usually valid. For a fluid in its ideal-gas state, no temperature change occurs. The throttling process is inherently irreversible, but this is immaterial to the calculation because Eq. (3.14b) is valid for the ideal-gas state whatever the process.5
Example 3.7
If in Ex. 3.6 the flow rate of air is 1 mol·s–1 and if both upstream and downstream pipes have an inner diameter of 5 cm, what is the kinetic-energy change of the air and what is its temperature change? For air, C Pig = 29.100 J· mol −1 and the molar mass is ℳ = 29 g·mol−1.
Solution 3.7
By Eq. (2.23b),
u = ___n ∙
Aρ = ____n ∙V ig A where
A = π__
4 D 2 = ( _π
4 ) ( 5 × 10 −2 ) 2 = 1.964 × 10 −3 m 2
The appropriate value of the gas constant for calculating the upstream molar volume is R = 83.14 × 10−6 bar·m3·mol−1·K−1. Then
V1ig = ___RT 1 P 1 =
( 83.14 × 10 −6 ) ( 293.15 K )
___________________ 6 bar = 4.062 × 10 −3 m 3 ·mol −1 Then,
u 1 = ( 1 mol· s −1 ) ( 4.062 × 10 −3 m 3 ·mol −1 )
___________________________1.964 × 10 −3 m 2 = 2.069 m· s −1
If the downstream temperature is little changed from the upstream temperature, then to a good approximation:
V2ig = 2 V1ig and u 2 = 2 u 1 = 4.138 m· s −1
5The throttling of real gases may result in a relatively small temperature increase or decrease, known as the Joule/
Thomson effect. A more detailed discussion is found in Chapter 7.