The First Law and Other Basic Concepts
2.8 HEAT CAPACITY
2.8. Heat Capacity 43 independent of path, and can therefore be calculated by equations for a truly constant- volume process regardless of the actual process. Equation (2.17) therefore has general validity, because U, CV, T, and V are all state functions. On the other hand, Q and W depend on path. Thus, Eq. (2.18) is a valid expression for Q, and W is in general zero, only for a constant- volume process. This is the reason for emphasizing the distinction between state functions and path- dependent quantities such as Q and W. The principle that state functions are path- and process-independent is an essential concept in thermodynamics.
For the calculation of property changes, but not for Q and W, an actual process can be replaced by any other process that accomplishes the same change in state. The choice is made based on convenience, with simplicity a great advantage.
Heat Capacity at Constant Pressure
The constant-pressure heat capacity is defined as:
C P ≡ ( ∂ ___H
∂ T )
P (2.19)
Again, the definition accommodates both molar and specific heat capacities, depending on whether H is the molar or specific enthalpy. This heat capacity relates in an especially simple way to a constant-pressure, closed-system process, for which Eq. (2.19) is equally well written:
dH = C P dT (const P) (2.20)
which is integrated to yield:
ΔH = ∫T1T 2 C P dT (const P) (2.21) For a mechanically reversible, constant-P process, this result can be combined with Eq. (2.12):
Q = ΔH = ∫T1T 2 C P dT (const P) (2.22) Because H, CP, T, and P are state functions, Eq. (2.21) applies to any process for which P2 = P1 whether or not it is actually carried out at constant pressure. However, only for the mechan- ically reversible, constant-pressure process can the amount of heat transferred be calculated by Eq. (2.22) and work by Eq. (1.3), here written for 1 mole, W = −P ΔV.
Example 2.7
Air at 1 bar and 298.15 K is compressed to 3 bar and 298.15 K by two different closed-system mechanically reversible processes:
(a) Cooling at constant pressure followed by heating at constant volume.
(b) Heating at constant volume followed by cooling at constant pressure.
Calculate the heat and work requirements and ΔU and ΔH of the air for each path. The following heat capacities for air may be assumed independent of temperature:
C V = 20.785 and C P = 29.100 J·mol −1 ·K −1
Assume also that air remains a gas for which PV/T is a constant, regardless of the changes it undergoes. At 298.15 K and 1 bar the molar volume of air is 0.02479 m3· mol−1.
Solution 2.7
In each case take the system as 1 mol of air contained in an imaginary piston/
cylinder arrangement. Because the processes are mechanically reversible, the piston is imagined to move in the cylinder without friction. The final volume is:
V 2 = V 1 ___ P 1
P 2 = 0.02479 ( _1
3 ) = 0.008263 m 3
The two paths are shown on the V vs. P diagram of Fig. 2.3(I) and on the T vs. P diagram of Fig. 2.3(II).
Figure 2.3: V vs. P and T vs.
P diagrams for Ex. 2.7.
(I) (II)
0 1 2 3
P bar
0 300 600 900
T K
Path (a)
Path (b)
Initial
State Final
State
0 1 2 3
V m3·mol–1
P bar
0 0.01 0.02
Path (a)
Path (b) Initial State
Final State
(a) During the first step of this path, air is cooled at a constant pressure of 1 bar until the final volume of 0.008263 m3 is reached. The temperature of the air at the end of this cooling step is:
T′ = T 1 V___ 2
V 1 = 298.15 ( _0.008263
0.02479 ) = 99.38 K Thus, for the first step,
Q
= ΔH = C P ΔT = (29.100) (99.38 − 298.15) = −5784 J W = − P ΔV = − 1 × 10 5 Pa × (0.008263 − 0.02479) m 3 = 1653 J ΔU
= ΔH − Δ(PV) = ΔH − P ΔV = − 5784 + 1653 = − 4131 J
2.8. Heat Capacity 45 The second step is at constant V2 with heating to the final state. Work W = 0, and for this step:
ΔU
= Q = C V ΔT = (20.785) (298.15 − 99.38) = 4131 J V ΔP = 0.008263 m 3 × (2 × 10 5 ) Pa = 1653 J
ΔH
= ΔU + Δ(PV) = ΔU + V ΔP = 4131 + 1653 = 5784 J For the overall process:
Q
= −5784 + 4131 = −1653 J W = 1653 + 0 = 1653 J ΔU = − 4131 + 4131 = 0 ΔH
= − 5784 + 5784 = 0
Notice that the first law, ΔU = Q + W, applied to the overall process is satisfied.
(b) Two different steps of this path produce the same final state of the air. In the first step air is heated at a constant volume equal to V1 until the final pressure of 3 bar is reached. The air temperature at the end of this step is:
T′ = T 1 ___P 2 P 1 = 298.15 ( 3_
1 ) = 894.45 K For this first constant-volume step, W = 0, and
Q
= ΔU = C V ΔT = ( 20.785 ) ( 894.45 − 298.15 ) = 12,394 J V ΔP = ( 0.02479 ) ( 2 × 10 5 ) = 4958 J
ΔH
= ΔU + V ΔP = 12,394 + 4958 = 17,352 J
In the second step air is cooled at P = 3 bar to its final state:
Q
= ΔH = C P ΔT = ( 29.10 ) ( 298.15 − 894.45 ) = − 17,352 J W = −P ΔV = − ( 3 × 10 5 ) ( 0.008263 − 0.02479 ) = 4958 J ΔU
= ΔH − Δ ( PV) = ΔH − P ΔV = − 17,352 + 4958 = − 12,394 J
For the two steps combined,
Q
= 12,394 − 17,352 = − 4958 J W
= 0 + 4958 = 4958 J ΔU = 12,394 − 12,394 = 0 ΔH
= 17,352 − 17,352 = 0
This example illustrates that changes in state functions (ΔU and ΔH) are indepen- dent of path for given initial and final states. On the other hand, Q and W depend on the path. Note also that the total changes in ΔU and ΔH are zero. This is because the input information provided makes U and H functions of temperature only, and T1 = T2. While the processes of this example are not of practical interest, state-function changes (ΔU and ΔH) for actual flow processes are calculated as illustrated in this example for processes that are of practical interest. This is possi- ble because the state-function changes are the same for a reversible process, like the ones used here, as for a real process that connects the same states.
Example 2.8
Calculate the internal energy and enthalpy changes resulting when air is taken from an initial state of 5°C and 10 bar, where its molar volume is 2.312 × 10−3 m3·mol−1, to a final state of 60°C and 1 bar. Assume also that air remains a gas for which PV/T is con- stant and that CV= 20.785 and CP= 29.100 J·mol−1·K−1.
Solution 2.8
Because property changes are independent of process, calculations may be based on any process that accomplishes the change. Here, we choose a two-step, mechan- ically reversible process wherein 1 mol of air is (a) cooled at constant volume to the final pressure, and (b) heated at constant pressure to the final temperature. Of course, other paths could be chosen, and would yield the same result.
T 1 = 5 + 273.15 = 278.15 K T 2 = 60 + 273.15 = 333.15 K With PV = kT, the ratio T/P is constant for step (a). The intermediate temperature between the two steps is therefore:
T′ = (278.15) (1 / 10) = 27.82 K
and the temperature changes for the two steps are:
Δ Δ TTa b = 27.82 − 278.15 = −250.33 K = 333.15 − 27.82 = 305.33 K For step (a), by Eqs. (2.17) and (2.14),
Δ U a
= C V Δ T a = ( 20.785 ) ( −250.33 ) = −5203.1 J Δ H a = Δ U a + V Δ P a
= −5203.1 J + 2.312 × 10 −3 m 3 × ( −9 × 10 5 ) Pa = − 7283.9 J For step (b), the final volume of the air is:
V 2 = V 1 ____P 1 T 2
P 2 T 1 = 2.312 × 10 −3 ( _10 × 333.15
1 × 278.15 ) = 2.769 × 10 −2 m 3 By Eqs. (2.21) and (2.14),
Δ H b = C P Δ T b = (29.100) (305.33) = 8885.1 J Δ U b = Δ H b − P Δ V b
= 8885.1 − (1 × 10 5 ) (0.02769 − 0.00231) = 6347.1 J For the two steps together,
ΔU
= − 5203.1 + 6347.1 = 1144.0 J ΔH = − 7283.9 + 8885.1 = 1601.2 J
These values would be the same for any process that results in the same change of state.9
9You might be concerned that the path selected here goes through an intermediate state where, in reality, air would not be a gas, but would condense. Paths for thermodynamic calculations often proceed through such hypothetical states that cannot be physically realized but are nonetheless useful and appropriate for the calculation. Additional such states will be encountered repeatedly in later chapters.