Internal Energy (du) Relation

2. Internal Energy (du) Relation

Combining the First and Second laws du = T ds – P dv Using Eq. (32) to eliminate ds, we obtain the expression du = c v dT + (T( ∂P/∂T) v – P) dv,

(38) which implies that the change in internal energy equals the energy stored in the form of trans-

lational, vibrational, and rotational energies plus the intermolecular potential energy.

a. Remarks In the case of an ideal gas P = RT/v, so that

T( ∂P/∂T) v – P = (RT/v) - P = 0 Therefore, the potential energy term in Eq. (38) equals zero for ideal gases. Recall

from the discussion regarding the Lennard–Jones potential energy curve in Chapter 1 that the intermolecular potential energy term equals zero at larger intermolecular dis- tances at which the intermolecular attraction force is zero. Consequently, energy is not stored in the form of potential energy as the volume is changed at large volumes, and a change in the gas volume (or the intermolecular distance) does not affect the intermolecular potential energy. Equation (38) is called the “Calorific Equation of State” and indicates that the internal energy of a simple compressible substance (the solid, liquid, or gas phases) is a func- tion of temperature and volume (cf. Figure 7 ). In it, the first term, c v dT, represents the thermal portion of the energy (which varies due to changes in the translational, vi- brational, and rotational energies, with each degree of freedom contributing an

amount equal to 2k B T per molecule, as discussed in Chapter 1). Applying the First law to a closed system undergoing a reversible process, in the context of Eq. (38)

δQ rev = dU + PdV = mc v dT + (T ∂P/∂T – P)dV + PdV (38’) Eq.(38’) states that energy transfer δQ rev is used to raise the thermal energy by the in-

cremental amount mc v dT (e.g., te, re, ve), to overcome the intermolecular potential energy barrier (T ∂P/∂T – P)dV (i.e., ipe required to move the molecules against at-

tractive forces) and to perform boundary PdV work. For an ideal gas, the intermo- lecular potential energy equals zero so that heat transfer is used for PdV work if the process is isothermal.

In an adiabatic system the internal energy change c v dT + (T ∂P/∂T – P)dv equals the work performed on the system. Note that if V and T are held fixed, δQ rev =0. How- ever, this does not imply that Q = Q(V,T) since the functional form for the relation

changes, depending upon the process path between the initial and final states (i.e., Q is not a property). Consider Example 6 in Chapter 2. As we place a large weight on the piston, the gas is adiabatically compressed and we perform Pdv work. The potential energy decrease of the weight is converted into internal energy increase of the gas. The intermolecular potential energy decreases during compression (T ∂P/∂T – P)dv < 0, since the second

term in Eq. (38) decreases as the intermolecular spacing becomes closer. Eq. (38) im- plies that the term c v dT must increase. If gas is ideal, then change in ipe =0; this im- plies that temperature changes to a greater extent in a real gas than in an ideal gas). For incompressible substances, dv = 0 and, consequently from Eq. (38), du = c v dT. Attractive forces are very strong in such substances so that the intermolecular poten- tial energy remains virtually constant. In the case of solids and liquids it is useful to express Eq. (38) in terms of the isobaric expansivity and isothermal compressibility. Since ( ∂P/∂v) T ( ∂v/∂T) P ( ∂T/∂P) v = –1, and

( ∂P/∂T) v = β P / β T , Eq. (38) can be written in the form du = c v dT + (T β P / β T – P) dv. The values of β P and β T are approximately constant for most solid and liquid sub-

stances. The volume of a two-phase mixture can be changed at a specified temperature and pressure (e.g., that of water at 100ºC and 1 bar pressure) by altering its quality. There- fore, while changes in the mixture volume are possible during phase transition at given T, those in the pressure are not. Consequently, ( ∂v/∂P) T (= a finite value ÷ 0)

→∞, i.e., β T → ∞. Similarly, (∂v/∂T) P → ∞ and hence β p → ∞ for a two-phase mix- ture. Thus as we approach the critical point, β T → ∞ and β P →∞. Since the relation for du in Eq. (38) is an exact differential, applying the appropriate

criterion,

2 2 2 ∂c 2

v / ∂v = (∂/∂T)(T(∂P/∂T) v – P) = T ∂ P/ ∂T + ∂P/∂T – ∂P/∂T = T(∂ P/ ∂T ) v .

u(T,v 2 )

Figure 7: Illustration of the variation of the internal en- ergy with respect to temperature.

Applying the state relations for liquids and solids (cf. Chapter 6) it can be shown that T( ∂ 2 P/ ∂T 2 ) v ≈ 0, i.e.,

c v =c v (T) alone. Integrating Eq. (39) between the limits v → ∞ (i.e., as P → 0) and v = v, we obtain an expression for the deviation of the constant volume specific heat

from its ideal gas value, i.e., cv cvo = v −

∞ T ( ∂ ∫ 2 P / ∂ T 2 ) v dv

If the P–v–T behavior of a gas is known from either measurements or theory, and data for c vo is available, the above relation can be integrated to evaluate values of c v . At T = 0 K, it will be later shown that ∂P/∂T = 0. At that condition, du → (–Pdv), i.e.,

∂u/∂v → –P. l. Example 12

The state of a copper bar is initially at a pressure of 1 bar and temperature of 250 K. It is compressed so that the exerted pressure is 1000 bar. Assume that the compression is adiabatic and reversible (i.e., the material reverts to its original state once the load is removed), and that

×10 bar , v = 1.11 ×10 m kg ,

c –1 p = 0.372 kJ kg K , and c v = 0.364 kJ kg K . Determine the change in the inter- nal energy and the intermolecular potential energy of the solid.

Integrating the relation

β T =−

v  ∂ PT 

ln ( v 2 / v 1 ) =− β T ( P 2 − P 1 ) or v 2 − v 1 ≈− β T vP 1 ( 2 − P 1 ) (A)

2 = 1.1092 ×10 m kg and which are similar to the results obtained in Example 7.

Therefore, v 2 -v 1 = -0.845x10 -7

Integrating Eq. (37),

( ln( T 2 / T 1 ) =− { β P / β T cv v } ( 2 − v 1 ) or ( T 2 − T 1 ) ≈− ( β P / β T cv T v ) 12 ( − v 1 ) (B)

Using the results from Eq. (A),

T 2 − T 1 = ( T 1 β P / cv v P ) 12 ( − P 1 ) = . 0 365 K

Consequently, the temperature following compression is 250.365 K. For an adiabatic reversible process, the First law yields,

∆u = – ∫P dv = –∫P (∂v/∂P) dP = ∫β 2

T P v dP = ( P 2 2 − P 1 ) β/ T v 2

= 0.00423 kJ kg –1 or 4.23 J kg –1 (similar to example 7) (C) We may also use following equation: du = c v dT + (T β P / β T – P) dv.

(D) The thermal portion of change in “u” is given as,

(E) The intermolecular potential energy of the solid

c dT = 0.364

× 0.365 = 0.1329J kg .

∆(ipe) = (Tβ P / β T – P) dv where the second term is same as on the RHS of Eq. (C). Therefore

∆(ipe) = (250 K × 48×10 –7 ÷ 7.62x10 bar –1 )

–6 K –1

×100 kPa bar -1 + 0.00423, or ∆ (ipe) = -0.1331 + 0.00423 = -0.1288 kJkg -1

× (-8.45x10 -8 m 3 kg -1 )

(F) The net change in the internal energy of the solid is

du = 0.1329 - 0.1288 = 0.0041 kJ kg –1 . Which is approximately the same as the answer in Eq. ( C). In this example, the tem-

perature increases by 0.365 K, increasing the thermal portion of the internal energy by 0.1329 kJ, but the IPE decreases by 0.1288 kJ (Eq.(F)) . At the minimum intermo- lecular potential energy (Chapter 1) , compression should cause the "ipe" to increase. Since the ipe decreases with compression, this indicates that the solid is not at that minimum value.

m. Example 13

A substance undergoes an adiabatic and reversible process. Obtain an expression for ( ∂T/∂v) s in terms of c v , β P , β T and T. What is the value of ( ∂T/∂v) s for copper, given

that 3 β P =5 ×10 K , β T = 8.7 ×10 bar ,c=c v = 0.386 kJ kg K , v = 1.36 ×10 m

kg –7 , and the temperature is 25ºC? What is the temperature rise if dv = –8.106 ×10 m 3 kg –1 ?

Solution We will use the relations

ds = c v dT/T + ( ∂P/∂T) v dv, and (A) ( ∂P/∂T) v ( ∂T/∂v) P ( ∂v/∂P) T = – 1.

(B) Therefore,

( ∂P/∂T) v =–( ∂v/∂T) P /( ∂v/∂P) T = β P / β T , i.e., (C) ds = c v dT/T + ( β P / β T ) dv

(D) For an adiabatic and reversible process, ds = 0, and

(E) For copper,

( ∂T/∂v) s =.–T β P /( β T c v ).

∂T/∂v) –1 s =–5 ×10 K –1

× 298K ÷ (8.7 × 10 –1 bar ÷ 100 kPa bar )

6 ÷ 0.386 kN m kg –3 K = –4.4370 ×10 K kg m .

On compression dT

s =( ∂T/∂v) dv = –4.437 ×10 × (–8.106×10 s –7 ) = 3.6 K. Remarks

The compression of a solid in the elastic regime is isentropic. The First law indicates that the work done (–Pdv) during adiabatic compression and du >0. Equation (D) provides the answer for the change in T when the volume changes for any simple compressible substance. Typically β T > 0 for any substance. Those substances that

expand upon heating (e.g., gases, steel, water above 4ºC at P = 1 bar) β P > 0. From Eq. (E), (dT/dv) s < 0. Thus, upon compression (dv < 0), and dT > 0. Those sub- expand upon heating (e.g., gases, steel, water above 4ºC at P = 1 bar) β P > 0. From Eq. (E), (dT/dv) s < 0. Thus, upon compression (dv < 0), and dT > 0. Those sub-

(F) which is the same as Eq.(B) in Example 12.

ln (T 2 /T 1 ) = (v 2 –v 1 ) β P /( β T c v ).