First Law for a Closed System

2. First Law for a Closed System

We will present the First law of thermodynamics for a closed system, and illustrate applications pertaining to both reversible and irreversible processes.

a. Mass Conservation For closed systems the mass conservation equation is simply that the mass

m = Constant, (1) In the field of atomic physics, mass and energy E are considered convertible into each another

and, taken together, are conserved through the well–known Einstein relation E = mc 2 , where c denotes the light speed. However, in the field of thermodynamics it is customary to assume that the conversion of mass and energy into each other is inconsequential and, therefore, either is separately conserved.

b. Energy Conservation An informal statement regarding energy conservation is as follows: “Although energy assumes various forms, the total quantity of energy is constant, with the consequence that when energy disappears in one form, it appears simultaneously in others”.

i. Elemental Process For a closed system undergoing an infinitesimally slow process, ( Figure 1a ) during which the only allowed interactions with its environment are those involving heat and work, the first law can be expressed quantitatively as follows

δQ – δW = dE, (2) where δQ denotes the elemental (heat) energy transfer across the system boundaries due to

temperature differences ( Figure 1a ), δW the elemental (work) energy in transit across the boundaries (e.g., the piston weight lifted due to the expansion of the system), and dE the en-

ergy change in the system. The "E" includes internal energy U (=TE+VE+RE etc.) which re- sides in the matter, kinetic energy KE and potential energy PE. Note that Q and W are transi- tory forms of energy and their differentials are written in the inexact forms δQ and δW (see

Chapter 1) while differential of resident energy E is written as an exact differential. Dividing Eq. (2) by m,

δq – δw = de, (3a) where q denotes the heat transfer per unit mass Q/m, w the analogous work transfer W/m, and,

likewise, e = E/m. It is customary to choose a sign convention for the work and heat transfer that follows common sense. In the absence of work transfer, i.e., δW = 0, addition of heat causes an in-

crease in energy. Therefore, it is usual to accord a positive sign for heat transfer into a system. For an adiabatic system ( δQ = 0), if the work done by the system is finite and conferred with a positive sign (W > 0), then, from Eq. (2), dE < 0. This is intuitively appropriate, since in order

to perform work, the system must expend energy. On the other hand, if the system of Fig. 2 is adiabatically compressed, work is done on the system (so that W < 0), and the stored energy in the system increases (dE > 0).

The system energy consists of the internal, potential, and kinetic energies. Equation (2) may be rewritten for a static system in the form

δQ – δW = dU. (3b)

ii. Internal Energy At a microscopic level the internal energy is due to the molecular energy which is the sum of the (1) molecular translational, vibrational and rotational energies (also called the ther- mal portion of the energy), (2) the molecular bond energy (also called the chemical energy), and the (3) intermolecular potential energy, ipe (cf. Chapter 1). At a given temperature the energy depends upon the nature of a substance and, hence, is known as an intrinsic form of energy.

iii. Potential Energy The potential energy of a system is due to the work done on a system to adiabatically move its center of gravity through a force field. The potential energy of a system whose center of gravity is slowly raised vertically (so as not to impart a velocity to it) in the earth’s gravity field through a distance of dz increases by a value equal to mgdz. The first law

δQ - δW = dU + d(PE) + d(KE), where PE and KE denote the potential and kinetic energies, can be applied after noting that for

this case δQ = dU = d(KE) = 0, so that 0– δW = 0 + d(PE) + 0.

Now, δW = –F dz. The negative sign arises since work is done on the system by a force F that lifts it through a distance dz. In raising the mass, the direction of the force is vertically upward.

In the absence of any acceleration of the mass, this force is also called a body force. Using the relation F = mg for the force with g denoting the local gravitational acceleration, the work done W = –mg dz, and using Eq. (4) d(PE) = – δW = mg dz. Integrating this expression across

a vertical displacement that extends from z 1 to z 2 , the potential energy change is given as

∆PE = mg(z 2 –z 1 ).

The potential energy per unit mass due to the gravitational acceleration at a location z above a stipulated datum is also called the gravitational potential pe, i.e.,

pe = gz. (5)

In SI units, pe can be expressed in J kg –1 or in units of m 2 s –2 , where

pe (in units of kJ kg –1 ) = g(in units of m s –2 ) z(in units of m)/1000. In English units, pe can be expressed as BTU lb –1 = g(ft s –2 ) z(ft) ÷(g c J), where g c = 32.174 (lb

ft s –2 lbf –1 ) is the gravitational constant, and J denotes the work equivalent of heat of value 778.1(ft lbf BTU –1 ).

iv. Kinetic Energy In order to move a mass along a level frictionless surface, a boundary or surface force must be exerted on it. Applying the first law, namely, δQ - δW = dU + d(PE) + d(KE), the

adiabatic work due to these forces can be expressed as 0– δW = 0 + 0 + d(KE).

(6) The work performed in moving the center of gravity of a system through a distance dx is (– F

dx), where the force F = m dV/dt, the velocity V = dx/dt, and t denotes time. In order to be con- sistent with the standard sign convention, the work done on the system is considered negative. Therefore, d(KE) = m (dV/dt) ×(V dt) = mVdV. Upon integration, the kinetic energy change of

the system as it changes its velocity from a value V 1 to V 2 is ∆KE = (1/2)m( V 2 2 2 − V 1 ).

The kinetic energy per unit mass ke is ke =1/2 V 2 .

(7) In SI units, ke is expressed in J kg –1 , namely, ke(J kg –1 ) = (1/2)V 2 (m 2 s –2 ). Often, it is prefer-

able to express ke as

ke (kJ kg –1 ) = (1/2000)V 2 .

In English units, ke(BTU lb –1 )=V 2 (ft 2 s –2 ) ÷(g c J). The kinetic and potential energies are inde- pendent of the nature of the matter within a system, and are known as extrinsic forms of en-

ergy. v. Integrated Form

Integrating Eq. (2) between any two thermodynamic states (1) and (2) we have

(8) The heat and work transfers are energy forms in transit and, hence, do not belong to the matter

Q 12 –W 12 =E 2 –E 1 = ∆E.

within the system with the implication that neither Q nor W is a property of matter. Therefore, while it is customary to write the energy change for a process E 12 =E 2 –E 1 , we cannot write Q 12

= (Q 2 –Q 1 ) or W 12 = (W 2 –W 1 ). Since for a cycle the initial and final states are identical,

∫ δ Q = ∫ W=0.

Writing Eq. (8) on a unit mass basis q 12 –w 12 =e 2 –e 1 = ∆e. (9)

The application of the first law to systems require these to be classified as either cou- pled systems in which the transit energy modes, namely, Q and/or W, affect particular storage forms of energy, or as uncoupled systems if the heat and/or work transfer affect more than one mode of energy as illustrated below.

vi. Uncoupled Systems Consider an automobile that is being towed uphill on a frictionless road during a sunny summer afternoon from initial conditions Z 1 = V 1 =U 1 = 0 to an elevation Z 2 , velocity

V 2 and energy U 2 . Taking the automobile as a system, the heat transfer Q 12 from the ambient to the car is determined by applying Eq. (8), i.e.,

Q 12 –W 12 =E 2 –E 1 = ∆U + ∆PE + ∆KE,

so that Q 12 = ∆U. Therefore, the heat transfer across the boundary increases the system internal energy by ∆U which changes the static state of the system. The work performed to tow the automobile is

–W 12 = ∆ PE + ∆ KE, which influences the dynamic state of the system.

a. Example 1

A car of mass 2000 kg is simultaneously accelerated from a velocity V = 0 to 55 mph (24.6 m s –1 ) and elevated to a height of 100 m. Determine the work required. Treat the problem as being uncoupled.

Solution

Q 12 –W 12 ==U 2 –U 1 + KE 2 – KE 1 + PE 2 – PE 1 .

0–W 12 = (0+(2000 ÷2)(24.6 m s –1 2 ) –0+2000 ×(9.81×100)–0)÷1000 = 2568 kJ. Remark

All of the work can be recovered if the car is made to slide down on a frictionless road to ground level (i.e., to zero potential energy) so that the potential energy is completely converted into kinetic energy. Upon impact against a spring the vehicle kinetic energy is fur- ther transformed into the spring potential energy, thereby recovering the work. Hence, the process is uncoupled.

vii. Coupled Systems In coupled systems two or more interactions across the system boundary (e.g., heat and work) influence the same energy mode. For example, if a tow truck pulls a car on a rough high-friction road, the work performed is higher than that for an uncoupled system, since addi- tional work is required in order to overcome the external friction. Frictional heating can cause the internal energy of the car tires to increase, and if the tires do not serve as good insulators, heat transfer to the road can occur. Therefore, the work is coupled with both internal energy and heat transfer. This is illustrated in the following example.

b. Example 2

A car of mass 2000 kg that is simultaneously accelerated from a velocity V = 0 to 55 mph (24.6 m s –1 ) and elevated to a height of 100 m requires a work input of 3000 kJ. If the car is well insulated, what is the change in the internal energy of the car?

Solution

Q 12 –W 12 ==U 2 –U 1 + KE 2 – KE 1 + PE 2 – PE 1 . 0+3000=U 2 –U 1 +((2000 ÷2)(24.6ms –1 2 ) –0+2000 ×(9.81×100)–0)÷1000=2568 kJ,

i.e., U 2 –U 1 = 3000 – 2568 = 432 kJ.

Remarks The work input is more than ∆KE and ∆PE. Thus additional work is used to over-

come friction. Frictional work results in heating. If the tires (which are part of the car) are well insulated, their internal energy increases by 432 kJ. In this case the work is coupled to changes in the internal, kinetic, and potential energies of the system. Dividing the work into intrinsic and extrinsic contributions

W 12 =W 12,int +W 12 , ext ,

we find that W 12,int = 432 kJ, which results in the change in U, and that W 12 , ext = 3000 – 432 = 2568 kJ, which results in a change in the kinetic and potential energies, (diathermic) and the tire remains at fixed temperature. Then there is no change in the internal energy. Hence heat must be lost from the tires, i.e.,

Q 12 –W 12 = ∆U + ∆PE + ∆KE = 0 + ∆PE + ∆KE, where Q 12 = –432 kJ, and W 12 = –3000 kJ. In this case the work is coupled with the

heat transfer. The heat transfer affects the intrinsic energy by changing U. When the car moves at a high velocity, frictional drag due to the atmosphere can cause its body to heat, thereby increasing the internal energy. The work done on the car also increases its potential and kinetic energies, and the process becomes coupled. The work cannot be recovered, since the car will contain a higher internal energy even after impacting it against the spring, as illustrated in the previous example.

The heating of matter offers another example involving a coupled system. Consider constant pressure heating that causes a system of gases to expand and lift a weight of 100 kg through a distance of 2 m, if Q 12 = 10 kJ, W 12 = 1.96 kJ. (Here we neglect any change in the center of gravity of the matter contained in the system.) If there is no change in the system kinetic energy, from Eq. (8)

(10) In this case Q 12 ≠ ∆U. As a result of the work and heat interaction, ∆U = 10 – 1.96 = 8.04 kJ. If

Q 12 –W 12 =U 2 –U 1 = ∆U.

the system is confined to include only the moving boundary and the lifted weight, and these are considered adiabatic, then (–W 12 )= ∆(PE), so that the work performed alters the system potential energy.

The illustrations of coupled and uncoupled systems demonstrate that it is necessary to understand the nature of a problem prior to applying the mathematical equations.

c. Systems with Internal Motion Consider a mass of warm water contained in a vessel. If it is stirred, the entire effort imparts kinetic energy to that mass in the absence of frictional forces, and the center of gravity of each elemental mass of water moves with a specific kinetic energy ke. If the kinetic energy distribution is uniform throughout the system, its total value equals m x ke. Such a situation exists in an automobile engine when fresh mixture is admitted or when the exhaust valve opens. Oftentimes, the kinetic energy is destroyed due to internal frictional forces between the system walls and moving matter, which converts the kinetic energy into internal energy, as in coupled systems.

dZ system

(a)

combustor

W net =1000kJ

Pump

Turbine

Generator

condenser

(b)

atmosphere

300kJ

Figure 1: Illustration of the first law for a cyclical process.

viii. Adiabatic Work and Caratheodary Axiom I The work performed during all adiabatic processes (Q12 = 0) between two given

states is the same. Applying Eq. (8)

W 12 =E 2 –E 1 = ∆E.

This statement is called the Caratheodary Axiom I (see Postulate III of Chapter 5). For exam- ple, the electrical work (= voltage × charge current) used to heat a fluid adiabatically between two temperatures is identical to the mechanical work (e.g., performed using a pulley–paddle assembly to stir the water) required for a similar adiabatic heating process.

d. Cyclical Work and Poincare Theorem ix. Cyclical Work

For a closed system undergoing a thermodynamic cyclical process,

(11) Hence, the first law (Eq. (2)) yields,

∫ dE = 0.

and Eq. (12) implies that if ∫ δQ ≠0, then ∫ δW ≠0. On a unit mass basis ∫ δ q = ∫ δ w .

Figure 1b illustrates the cyclical process in a steam power plant for which the heat transfer during the various processes is indicated. Applying Eq. (12) for all processes, i.e., 1–2, 2–3, 3–4,..., and 8–1,

∫ δ Q = Q 12 + Q 23 +

L + Q 81 = ∫ δ W = W 12 + W 23 + L + W 81 ,

so that

∫ δW = 0 –300 – 2000 + 0 + 0 –200 + 0 + 4000 – 500 = 1000 kJ.

Therefore, by considering the net heat transfer for this cyclical process, the net work output of the plant can be determined.

x. Poincare Theorem Consider an adiabatic system containing water and a mechanical stirrer. Work transfer through the stirrer is used to raise the water temperature from a quiescent state 1 to an- other motionless state 2. Since

Q 12 = 0, the change in internal energy can be obtained by ap-

plying the first law (Eq. (10)),

and U 2 – U 1 = ∆U = |W 12 |.

Next, if the insulation is re-

moved and the water allowed to cool to its initial state, Eq.(10) can again be used to

determine the heat flow Q 21 . In this case Q 21 = |U 2 – U 1 | =

|W 21 | as a consequence of the Poincare theorem of thermo- dynamics, which states that during a cyclical process the net heat interactions equal the

Figure 2: P–v diagram for quasiequilibrium and nonquasie- net work interactions. While

quilibrium processes.

the Caratheodary axiom states the First Law in context of a single adiabatic process, the Poin- care theorem expresses it for a cyclical process.

xi. Rate Form Equation (2) can be used to express the change in state over a short time period δt (i.e., δQ = ˙Q δt, and δW = W ˙ δt to obtain the First Law in rate form, namely,

Q – ˙ W ˙ = dE/dt. (13)

P 2R

V 2R

Figure 3: a. Quasiequilibrium process; b. Nonquasiequilibrium process. The rate of work W ˙ is the energy flux crossing the boundary in the form of macro-

scopic work (e.g., due to the system boundary motion through a distance dz as illustrated in Figure 2 ). The heat flux Q is a consequence of a temperature differential, and does not itself ˙ move the boundary, but alters the amplitude of molecular motion that manifests itself in the

form of temperature. We see that energy conservation can be expressed in various forms (e.g., Eqs. (2), (3), (8), (9), (12) and (13)). The laws of thermodynamics are constitutive equation independent. It

is possible to determine dE/dt accurately if W ˙ and ˙ Q are measured. Calculations of ˙ Q and/or W ˙ may require constitutive relations. In the context of the relation ˙ Q = - λ∇T, a constitutive equation for heat transfer is employed with W ˙ = 0. Therefore, the value of dE/dt depends upon

the accuracy of the Fourier law and can differ from actual experimental data.

e. Quasiequilibrium Work Consider an adiabatic frictionless piston–cylinder assembly on which infinitesimal weights are placed as illustrated in Figure 3a and Figure 3b . If the small weights are slowly removed, the system properties remain almost uniform throughout the removal process. There- fore, at any instant following the removal of an infinitesimal weight, if the system is isolated, it is in an equilibrium state (i.e., its properties are invariant with respect to time). Since the inten- sive state can be determined during any part of the process involving the successive removal of weights, the path along which the process proceeds can be described (e.g., as illustrated in Figure 2 for a quasiequilibrium process that moves the system from state 1 to 2R along the path ABC). Due to their nature, quasiequilibrium processes are also termed quasistatic.

However, not all quasistatic processes are at quasiequilibrium. Consider the example of a gasoline–air mixture (system) contained in a piston–cylinder assembly. At the end of a compression process, spark is initiated, hot region develops around the spark plug, while the remainder of the mixture is much colder. Even though the piston moves slowly (i.e., it is qua- sistatic) during this process, the spark initiation results in a non–equilibrium state, since the temperature distribution is nonuniform, and it is not possible to assign a single system tem- perature.

The consequences of the quasiequilibrium processes illustrated in Figure 3 are as fol- lows: If the infinitesimal weights are slowly removed, then at any time the force of the weights F ≈ PA, where A denotes the piston surface area. A force of P×A is exerted by the system.

Therefore, the infinitesimal work W performed by the system as the individual weights are removed, and the piston moves infinitesimally through a displacement dx, is

δW = F dx = PAdx = PdV. (14) Consequently, the work done during the process 1–2 is

(15) This is an illustration of reversible work.

W 12 = ∫ 1 PdV .

The work performed by the system results in a p\otential energy gain for the remaining weights that are placed in the environment outside the system. Since energy is transferred to the environment, according to the first law the system loses internal energy. The process can be reversed by slowly placing the weights back on the piston. This action will push the piston inward into the system, reduce the potential energy of the weights placed in the system environment, and restore the system to its initial state.

A quasiequilibrium process is entirely reversible, since the initial states of both the system and environment can be completely restored without any additional work input or heat in- teraction. We will see later that a totally reversible process is always a quasiequilibrium process. The work done on or by the system PdV is due to the matter contained within it. The sign convention follows, since it is positive for expansion (when work is done by the system), and negative during compres-

sion (when work is performed on the

system).

It can be mathematically shown that W is an inexact differential. Equation (14) may be written in the form δW = PdV +

0 ×dP. Using the criteria for exact differ-

entials (discussed in Chapter 1) with M = P, and N = 0, it is readily seen that ∂M/∂P = 1, and ∂N/∂V = 0. Therefore,

∂M/∂P ≠ ∂N/∂V.

c. Example 3 Figure 4: P–v diagram with P expressed in units of bar and v in m 3 Air is isobarically expanded from –1 kg .

3 state 1 (P –1

1 = 1 bar, v 1 =1m kg ),

to state 2 (P

2 = 1 bar, v 2 = 3 m

kg –1 ), and then compressed isometrically to state 3 (P 3 = 3 bar, v =3m 3 kg 3 –1 ). De- termine the final temperature and the net work. Air is isometrically compressed from state 1 (P = 1 bar, v =1m 3 1 kg 1 –1 ), to state 4 (P 4 = 3 bar, v =1m 3 kg 4 –1 ), and then expanded isobarically to state 3 (P 3 = 3 bar, v 3 =3

m 3 kg –1 ). Determine the final temperature and the net work. Solution

The P–v diagram for this example is illustrated in Figure 4 . The final temperature T 3 is independent of the work path, and

T 3 =P 3 v 3 /R = 300 ×3÷0.287 = 3136 K. The work along the two paths

w 123 =P 1 (v 2 –v 1 )=1 × 100 × (3 – 1) = 200 kJ kg –1 , and

(A)

w 143 =P 4 (v 3 –v

2 )=3 × 100 × (3 – 1) = 600 kJ kg .

(B)

Remarks The net work in the second case, i.e., w 143 , is larger compared to W 123 . The tempera- ture represents the state of the system, and its functional form, e.g., T 3 =P 3 v 3 /R, is in- dependent of the path selected to reach that state. However, the work expressions w 123 and w 143 (Eqs. A and B) depend upon the path selected to reach the same final state, even though the expressions for work (contain variables that only represent proper- ties. Therefore, the final temperature is path independent, but the net work is not. The inexact differential W integrated between two identical states along dissimilar paths 1–2–3 and 1–4–3 yields different results. An inexact differential can only be integrated if its path is known.

f. Nonquasiequilibrium Work In the context of Figure 3 , the initial pressure in the system is such that P 1 1 A = F , where F 1 denotes the combined weight of the piston and the aggregate weights placed upon it. If all of the weights are abruptly removed, rather than slowly as discussed previously, the force exerted on the system near the piston will be much smaller than F 1 . The difference between these two forces results in an acceleration of the piston due to Newton’s law, and the piston mass acquires kinetic energy. Thereupon, the system pressure in the vicinity of the piston rap- idly decreases. The translational energy of these molecules decreases with the pressure reduc- tion.However, molecules further removed from the piston still possess their initial velocities (i.e. higher T, higher P), and the system is in an internally nonequilibrium state. The matter adjacent to the piston also acquires kinetic energy (e.g., Section A in Figure 5 ) while that re- moved from it does not (e.g., Section B). Hence at any instant the system properties are non- uniform and, consequently, the process is not at quasiequilibrium.

The time taken for the system to equilibrate, also called its relaxation time t relax , is of the order of the distance divided by average molecular velocity (that approximately equals the -1 sound speed V s ). Typically, the sound speed in air at room temperature is 350 m s . It follows

that if L = 10 cm, t relax = 3 Η10 -3 s. Consequently, a disturbance near the piston, such as a de- creased pressure or decreased molecular velocity is communicated through random molecular

motion to molecules located 10 cm away after roughly 0.3 ms. If the piston is displaced by 10 P 2

Figure 5: Illustration of a nonquasiequilibrium process.

cm every 0.3 ms, the disturbance perpetuates, and a non-equilibrium condition continually prevails.

This behavior is similar to that of a disturbance due to stones dropped into a placid pond. The disturbance is always present unless the time interval between two sequentially dropped stones is long. If the rate of stones being dropped is fast enough, the disturbance strengthens. In the piston–cylinder example, if the piston moves with a velocity of 1 m s –1 (which is much lower than the sound speed), the typical time scale involving motion through a

10 cm displacement (=L/V P ) is 100 ms, which is much larger than the relaxation time. In this case, the pressure rapidly conforms to a uniform value within the whole system. Therefore, a process may be assumed to be in quasiequilibrium as long as its relaxation timescale (=L/Vs) is considerably smaller than the process timescale (=L/V P ) responsible for the property gradients that are the source of nonequilibrium system conditions. If V P = 350 m s –1 , quasiequilibrium cannot be assumed, and the system properties (i.e., its state) along the process path cannot be described. Therefore, an uncertain path is used to illustrate such a process in Figure 2 .

During the quasiequilibrium process 1–2R described in Figure 2 , the system performs more work than any corresponding non-equilibrium process 1–2 (path D-E-F-2), since part of the non-equilibrium work imparted to the piston in the form of kinetic energy is converted into thermal energy. As a consequence, even if expanded to the same final volume, the temperature

at the end of a non-equilibrium process is higher, and applying the ideal gas law P 2 >P 2R (state 2). This may also be understood by envisioning the frictional effects that dissipate and raise the system internal energy (therefore, temperature) during non-equilibrium processes. These proc- esses are irreversible, since the original system state cannot be reverted to its original state by simply reversing the work transfer. An additional amount of work is required to overcome the effects due to friction.

Placing the system boundary immediately around the piston and the external weights (that are respectively, of mass m p and m w ), the force experienced by the system is

m dV P /dt = P b A–P R A,

(16a) where dV P /dt denotes the piston acceleration, P b the pressure at the system boundary, P R =P o +

mg/A is the sum of the ambient pressure and the pressure due to the piston weight, and m = m p +m w . The work performed to move the accelerating mass m through a displacement dz is the difference between the work performed by the system and that performed to overcome the resistance to its motion. Multiplying Eq. (16a) by dZ

m (dV P /dt) dz = δW – (P o dV + mg dz),

where the boundary work δW = P b dV. Therefore,

δW = m (dV P /dt) dz + P o dV + mg dz. (16b) Using the relation dz = V P dt in Eq. (16a), and integrating appropriately,

W=m ∆ke + P o ∆V + m∆pe, (17) where P

2 o ke = V P /2 and ∆pe = gz. If the system pressure is uniform (e.g., t relax « L/Vp), then P b = P and the work per-

formed by system δW ≈ PdV,

which requires a functional relation between P and V for the matter contained in the system. The following example illustrates a nonquasiequilibrium process.

d. Example 4:

A mass of air is contained in a cylinder at P = 10 bar, and T = 600 K. A mass of 81.5 kg is placed on the piston of area 10 cm 2 and the piston is constrained with a pin. If A mass of air is contained in a cylinder at P = 10 bar, and T = 600 K. A mass of 81.5 kg is placed on the piston of area 10 cm 2 and the piston is constrained with a pin. If

81.5 kg pressure to be negligi- ble:

Determine the piston acceleration just after

81.5 kg

the pin is released. Write an expression

Piston area for the work per-

A=10cm 3

10 bar

formed on the sur-

600 K

roundings. Write an expression

for the work done by the system matter if it exists at a uniform state. Why is there a differ-

Figure 6: A nonquasiequilibrium process due to the release ence between the an-

of a mass accelerated by a pressurized system. swers to the questions above?

What are the effects of a frictional force of 0.199 kN? (See also Figure 6 .) Solution The force due to mass of 81.5 kg placed on the piston equals 81.5 ×9.81÷1000 = 0.8

kN. The pressure due to a weight of 0.8 kN equals 800 kPa (or 8 bar). Since the sys- tem pressure is 10 bar, there is a force imbalance equivalent to 2 bars, and the mass is accelerated. The force F = m ×a = mdV/dt, i.e.,

F = (10–8)bar –1 ×100 kN m –2 –1 –4 bar ×10 cm ×10 2 m cm ×1000 N kN = 200 N. Hence, the initial acceleration dV/dt = F/m = 2.45 m s –2 .

The work δW = 800×dV, which changes the potential energy of the mass. If the process is internally reversible, the matter is internally in a quasiequilibrium

state, and δW = PdV. The difference between the work performed by the system and that transmitted to the

weight in the form of potential energy increases the kinetic energy of the weight. If the imparted kinetic energy is zero (or dV/dt = 0), the work done by the system equals that done on its surroundings, i.e., there are no losses. In the case of a frictional force of 0.199 kN, the resistance force F = 0.8 + 0.199 =

0.999 kN so that the resistance pressure P = 0.999 kN /10 –3 m 2 = 999 kPa, which is virtually identical to the system pressure. Therefore, the force imbalance is negligible, and m dV/dt ≈ 0. If the process is internally reversible, the work done by the system

δW system = P dV, and that done on the weight δW W ≈ 800×dV. Hence, the frictional work

W F = P dV – 800 ×dV = (P – 800) ×dV.

e. Example 5

A mass of 50 kg is placed on a 10 cm 2 area weightless piston (cf. Figure 7 ). The am- bient is a vacuum, i.e., the pressure is zero in it. The initial gas pressure is 100 bar, and the initial volume is 10 cm 3 . The cylinder height is 10 cm. A pin, constraining the piston in place is suddenly released.

Consider the gases in the piston–cylinder assembly to constitute a system A. If the process in system A is internally reversible and isothermal, determine the work output

of the gas. Let system B be such that it includes the piston, weight, and ambient, but excludes the gases. What is the velocity of the piston when its position is at the cylinder rim? As- sume system B to be adiabatic.

Solution System A delivers work to system B during the process 1–2.

V = 10 cm 3 ,V = 10 cm × 10 cm 2 = 100 cm 1 3 2 .

The work done by system A is: W A = ∫PdV = ∫(mRT/V)dV = mRT ln(V 2 /V 1 )=P 1 V 1 ln(V 2 /V 1 )

(A)

∴W = 100 bar × 100 kN m –2 bar –1 × 10 cm A 3 × 10 –6 m 3 cm –3 ln(100/10)

= 0.230 kJ.

The work input from system A into system B results in an increase of the kinetic and potential energies of the weight. The initial and final heights of the piston in the cyl- inder are:

(B) Applying Eq. (8) to system B, i.e.,

1 =V 1 /A = 10 cm ÷ 10 cm 2 = 1 cm, Z 2 = 10 cm.

Q 12 –W 12 =E 2 –E 1 = ∆U + ∆PE + ∆KE,

(C) where ∆PE = 50 kg×9.81 m s –2 ×(10–1)cm×0.01 m cm –1 ÷ (1000 J kJ –1 ) = 0.044 kJ,

and ∆U = 0.

Using this result and Eq. (A) in Eq. (C),

0 – (–0.230) = 0 + ∆KE + 0.044, i.e., ∆KE = (1/2)m(V 2 2 2 –V 1 ) = 0.230 – 0.044 = 0.186 kJ. Since the initial velocity V 1 is zero, (1/2)mV 2 2 /1000 = 0.186 kJ, and substituting m=50 kg,

2 = 2.73 m s . Remarks Instead of the 50 kg weight, a projectile of

V –1

very small mass can be

similarly used. If the

projectile were fired

Z 2 (2)

from the

chamber

using, say,

gunpowder, the gases would ex- pand, al- though the

Figure 7: An analysis of a nonequilibrium process.

high temperature would remain unchanged over the period of interest due to the com- busting powder. In that case, the projectile velocity can be determined using the above example. Since the velocity in the example is of the order of 2.73 m s –1 , which is much slower than the room temperature molecular velocity of 350 m s –1 , one can assume rapid equilibration within the system. However, at lower temperatures, the quasiequilibrium assumption is invalid, since the molecular velocity can approach the process velocity. If the ambient pressure P o is 1 bar, the work transmitted to the matter, which is also called useful work, is given by the relation

W u = ∫PdV – ∫P o dV = ∫(P – P o )dV = ∫PdV – P o (V 2 –V 1 ), i.e., W

u = 0.230 – 1 × 100 × 90 × 10 = 0.221 kJ. Therefore, the kinetic energy change is

∆KE = (0.221 – 0.044) = 0.177 kJ, and

V 2 = (2 × 1000 × 0.177 ÷ 50) 1/2

= 2.66 m s –1 .

g. First Law in Enthalpy Form If the kinetic and potential energies are neglected, Eq. (2) transforms into

δQ – δW = dU. The enthalpy can replace the internal energy in this equation. The enthalpy of any substance is

defined as

H = U + PV, or (18)

h = u + Pv. For ideal gases PV = mRT and, hence, H = U + mRT. Substituting Eq. (18) in Eq. (3’) δQ – δW = d(H – PV).

For a quasiequilibrium process δW = PdV + δW other . Therefore, δQ – PdV – δW other = d(H – PV). Simplifying, this expression

δQ + VdP – δW other = dH. (19) If δW other =0 δQ + VdP = dH. The First Law can be written in the form

δQ– δW' = dH, where for a reversible process

δW' = – VdP. For a quasiequilibrium process at constant pressure

δQ P = dH. (20)

If an electric resistor is used to heat a gas contained in an adiabatic piston–cylinder–weight assembly, as shown in Figure 8b , the constant pressure electrical work

– δW elec,P = dH. (21) The constant volume work (cf. Figure 8a ) is

– δW elec = dU. (22) Note that the First Law is valid whether a process is reversible or not. However, once the

equality δW = P dV is accepted, a quasiequilibrium process is also assumed. xii. Internal Energy and Enthalpy

Experiments can be performed to measure the internal energy and enthalpy using Eqs. (21) and (22). For instance, electrical work can be supplied to a fixed volume adiabatic piston cylinder assembly (cf. Figure 8a ), and Eq. (22) used to determine the internal energy change dU or du. Alternately, using a constant pressure adiabatic assembly (cf. Figure 8b ), the electric work input equals the enthalpy change, and Eq. (21) can be utilized to calculate dH or dh.

The internal energy is the aggregate energy contained in the various molecular energy modes (translational, rotational, vibrational) which depend upon both the temperature and the intermolecular potential energy which is a function of intermolecular spacing or volume (see Chapter 1). Therefore, u = u(T,v) or u = u(T,P), since the specific volume is a function of pres- sure. While differences in internal energy can be determined, its absolute values cannot be obtained employing classical thermodynamics. However, we are generally interested in differ-

ences. For tabulation purposes a reference condition is desired. If the initial condition u 1 =u ref is the reference condition and u 2 = u during a process 1–2, the difference

∆u = u(T,P) – u ref (T ref ,P ref ). We normally set u ref = 0 at the reference temperature and pressure T ref , and P ref , which

characterize the reference state. For example, for tabulation of steam properties, the triple point (T tp = 0.01ºC, P tp = 0.006 bar) is used as the reference state. Once u is calculated with respect to the reference condition u ref = 0, Eq. (18) can be used to determine h. From the relation

h ref =u ref +P ref v ref =0+P ref v ref .

Adiabatic work addition - Adiabatic work addition -

constant pressure constant volume

Figure 8: (a) Constant volume, b) Constant pressure processes.

we note that h ref ≠ 0 even though u ref = 0. However, a separate reference condition can be used for the enthalpy so that

∆h = h(T,P) – h ref (T ref ,P ref ). The internal energy can be separately calculated at this reference state. Property tables for

many substances set h ref = 0 at (T ref ,P ref ) (Steam tables usually use T ref = 0.01 C and P ref = 0.0061 bar for liquid water).

f. Example 6 One kilogram of water at a temperature T = T ref =T tp = 0.01ºC is contained in an adiabatic piston cylinder assembly. The assembly resides in an evacuated chamber

and a weight is placed on top of the piston such that P = P ref 3 = 0.61 kPa. At these ref- erence conditions, the specific volume v(T ,P ) = 0.001 m kg –1 ref ref is assumed to be in-

dependent of temperature. During an isobaric process, a current of 0.26 A provided at

a potential of 110 V over a duration of 60.96 min raises the water temperature to 25ºC. Determine the enthalpy of water at that state if h ref = 0. Solution We will use the energy conservation equation δQ – δW = dU

and select the water mass as the system. In general, the work term will include a volumetric change component in addition to the electrical work so that δQ – PdV – δW elec = dU.

At constant pressure, δQ P – δW elec,P = dU + PdV = dH, and on a unit mass basis δq P – δw elec,P = du + Pdv = dh . Recalling that the system is adiabatic (q P = 0), and integrating the latter expression

–w elec,P =h –h ref . Now, W elec = 0.26 × 110 × 60.96 × 60 = 104.6 kJ. Therefore, – (–104.6) = h – 0, and

h (25ºC, 0.61 kPa) = 104.6 kJ kg –1 . Furthermore, u = h – Pv = 104.6 – 0.61 × 0.001 ≈ 104.6 kJ kg –1 .

Remarks The experiments may be repeated at different pressures for the same temperature range, and the enthalpy tabulated as a function of pressure. If the specific volume is known, applying the relation u = h – Pv, the internal energy can also be tabulated, as is done in the Steam tables. Through experiments performed on ideal gases, it is found that h = h(T) which is in- dependent of the pressure, e.g., the enthalpy of air at 25ºC and 1 bar is identical to that at 25ºC and 10 bar ( ≈ 300 kJ kg –1 ).

Denoting the enthalpy of an ideal gase by h(T), u = h(T) – Pv = h(T) – RT = u(T).

(23) (Later in this text, ideal gas properties will be denoted as u 0 ,h 0 , etc.). In general, for

any substance u = u(T,v). However, when an ideal gas is isothermally heated in a piston–cylinder assembly, the molecular translational, rotational, and vibrational en- ergies remain constant, while the gas expands, thereby increasing the intermolecular spacing. Under these conditions, the intermolecular potential energy for ideal gases is also unchanged, since intermolecular attractive forces are absent. Therefore, the inter- nal energy of an ideal gas is a function of temperature alone. A more detailed discus- sion of this is contained in Chapters 6 and 7.

xiii. Specific Heats at Constant Pressure and Volume As the matter contained within a system is heated, the temperature and internal energy change. Applying the First Law to a constant volume closed system δq v = du v . The specific

heat at constant volume c v is defined as

c v =( ∂u/∂T) v = δq v /dT v . (24) If instead of heating, electrical work is supplied to an adiabatic system (as in Figure 8 )

c v =( ∂u/∂T) v = (| δw elec v |/dT) v . If the matter contained in a piston–cylinder–weight assembly that ensures isobaric processes is

likewise heated (as illustrated in Figure 8b ), the constant pressure specific heat c p is defined as

c p =( ∂h/∂T) P = (| δw elec v |/dT) p . (25) For any substance, the values of the properties c p and c v can be experimentally meas-

ured. In general, incompressible liquids and solids are characterized by a single specific heat c which is a function of the temperature alone, i.e., c p ≈c v = c(T). A more detailed discussion is

contained in Chapters 3 and 7. The enthalpy at a given pressure can be determined as a func- tion of temperature by integrating Eq. (25), namely,

dh p =c p dT. (26) The ratio of the two specific heats k = c p /c v is an important thermodynamic parameter. Typi-

cally the value of k is 1.6 for monatomic gases (such as Ar, He, and Ne), 1.4 for diatomic gases (such as CO, H 2 ,N 2 ,O 2 ) and 1.3 for triatomic gases (CO 2 , SO 2 ,H 2 O).

g. Example 7

6 Consider an electron gas, the enthalpy of which is h = 3CT 2 /P . Obtain an expression for c p .

Solution

c p =( ∂h/∂T)

P = 18CT /P = f(T,P).

Remarks Although the differentiation is carried out at constant pressure, c p is a function of both pressure and temperature. If water is isobarically heated at 100 kPa from 25 to 60ºC its specific heat at constant pressure averaged over that temperature range is measured to be 4.184 kJ kg –1 K –1 . If the water is isobarically heated at 2 bars (e.g., in a pressure cooker) over the same temperature range, the average value of c p is 4.17 kJ kg –1 K –1 , illustrating that the spe- cific heat varies with pressure within the same temperature range.

For ideal gases, since u and h are functions of temperature alone, so are the two spe- cific heats, rendering the subscripts somewhat meaningless, i.e.,

c vo = du/dT = c vo (T), and c po = (dh/dT) = c po (T). (27) For ideal gases the subscript v is to be interpreted as differentiation of u with respect to T,

while the subscript P may be interpreted as differentiation of h with respect to T. Substituting Eq. (23) in Eq. (27)

c po =c vo + R. (28) Table A-6F presents relations for c po (T) for many ideal gases while Table A-6C provides c po

values at specific temperatures. The internal energy and enthalpy of an ideal gas can be calcu- lated using Eq. (27), i.e., values at specific temperatures. The internal energy and enthalpy of an ideal gas can be calcu- lated using Eq. (27), i.e.,

∫ T c (T) dT , ref v,o

ref p,o

where h ref =u ref = 0. Once either the enthalpy or internal energy is known, the other property can be calculated from the ideal gas relation u = h – RT. For instance, if c po (T) is specified ( Tables A-6F ), one can generate h and u tables for ideal gases ( Tables A-7 for air and A-8 to A-19 for many other ideal gases).

h. Example 8 In order to determine c p for an unknown ideal gas, 0.1 kg of its mass is deposited into an adiabatic piston–cylinder–weight assembly and electrically heated (cf. Figure 8b )

by a current of 0.26 A at 110V for a duration of 30 seconds. The resultant temperature rise is measured to be 10ºC. Calculate c p , assuming it to be constant. The experiment is repeated by removing the weight, but constraining the assembly with a pin so that the volume is kept constant (cf. Figure 8a ). For the same tempera- ture rise of 10ºC to occur, the current must now be applied for 23 seconds. Determine

c v . Determine the molecular weight of the unknown gas from the measured specific heats assuming the gas to be ideal.

Solution δW elec – P dV = dU, or – δW elec = d(H – PV) + P dV

Since the pressure is held constant, – δW elec = dH = m c p dT.

Assuming c p ≈ constant in the narrow temperature range, and integrating

–W elec = mc p (T 2 –T 1 ). Substituting for (T 2 – T 1 ) = 10ºC, and using a negative sign for the electrical work transfer to the system

c –1 p = (30 ×0.26×110÷1000)÷(0.1×10) = 0.85 kJ kg K . Since V = constant, – δW elec = dU = m c v dT, and

c v = (23 ×0.26×110÷1000)÷(0.1×10) = 0.65 kJ kg –1 K –1 . With the ideal gas assumption c p =c po , and c v =c vo , using the relation,

c po –c vo =R= R /M, R = 0.85 – 0.65 = 0.2 kJ kg –1 K –1 , and M = 8.314 ÷0.2 = 42 kg kmole –1 .

Remarks An alternative method to determine the molecular weight of an unknown gas is by charging a known mass of that gas into a bulb of known volume, measuring the tem- perature and pressure, and employing the relation M = m R T ÷(PV).

If the gas molecular weight is known, c vo can be determined if c po is known, and vice versa, since c vo =c po – R. At higher pressures, close to critical pressure, the intermolecular spacing becomes small, and the effects of intermolecular potential energy on u and h, and, therefore, c p and c v , become significant for gases. This is discussed in Chapter 7.

The temperature remains constant for a liquid being vaporized at a fixed pressure. Since, ac- cording to the First Law, the heat transfer per unit mass of liquid equals its latent heat of va-

porization, namely, q p = h fg , the enthalpy change is finite while dT = 0. Therefore, c p = ( ∂h/∂T) p → ∞ during vaporization. (Although c p for both the liquid and vapor phases has a

finite value, that value is infinite during phase change. Therefore, c p is discontinuous during phase change.) finite value, that value is infinite during phase change. Therefore, c p is discontinuous during phase change.)

adiabatic reversible process involving ideal gases

0 - P dv = c v0 dT Using ideal gas law P = RT/v and simplifying with the relations R = c p0 -c v0 and k = c p0 /c v0 -(k -1)dv/v = dT/T Assuming constant specific heats and integrating,

-(k-1) ln v = ln T + B´, i.e., ln T + (k-1) ln v = C´, or ln T v k-1 = C´. Therefore,

Tv k-1 = C˝, (29a) where C˝ = exp (C´). Using the relation T = Pv/R, we find that (Pv/R) v k-1 = C, or

Pv k = C. (29b) For air c p0 = 1, c v0 = 0.714, i.e., k = 1 ÷0.714 = 1.4.

Note that if a gas is compressed adiabatically and reversibly from state 1 to 2 and then expanded back adiabatically and reversibly from state 2 to 1, the net cyclic work is zero. For the cyclic work to be finite, one must add heat at the end of the adiabatic compression process; since the expansion line is parallel. In this case, the cycle cannot be closed unless heat is re- jected after the reversible expansion, which is manifest through the Second Law (cf. Chapter 3).

We now discuss why the temperature increases during adiabatic compression. Con- sider a 1 kg mass that is compressed for which δq -δw = du, where δw = Pdv. If the system is

adiabatic, δq =0. The deformation or boundary work (which is an organized form of energy with motion in a specified direction) is used to raise the internal energy of the 1 kg mass,

thereby raising the internal energy (manifest through the random energy of molecules that equals te+ve+re) and, hence, temperature. For an adiabatic process, if δw =0 then the internal

energy is unchanged, i.e., u = u (T) (as for an incompressible substance). The temperature does not change during the adiabatic compression of an incompressible substance.

xv. Polytropic Process In practical situations, processes may not be adiabatic. It is possible to determine the relation between P and v and find for most substances that Pv n =C where n may not necessarily equal k. Note that n = 1 for an isothermal process involving an ideal gas and n = 0 for an isobaric process.

i. Example 9 Air is contained in an adiabatic piston cylinder assembly at P 1 = 100 kPa, V 1 = 0.1 m 3 ,

1 = 300 K. The piston is constrained with a pin, and its area A is 0.01 m . Vac- uum surrounds the assembly. A weight Wt of 2 kN is rolled on to the piston, and the pin is released. Assuming that k o (=c p /c v ) = 1.4, and c vo = 0.7 kJ kg –1 K –1 ,

and T 2

Is the process 1–2 reversible or irreversible? What are the final pressure, volume, and temperature?

Solution We will select our system to include both the air and the weight rather than the air alone because the sudden process by which it changes state cannot be completely characterized. The process is clearly irreversible, since the system cannot be restored to its initial state unless the weight is lifted back to its original position, which re- quires extra work.

P 2 A = Wt, or P 2 = Wt /A.

(A)

With Wt = 2 kN, P 2 =2 ÷ 0.01 = 200 kPa.

Applying the First Law to the system,

Q 12 –W 12 =0=E 2 –E 1 , or E 2 =E 1 .

Neglecting the kinetic energy,

E 2 =U 2 + Wt Z 2 , and E 1 =U 1 + Wt Z 1 .

(B)

Substituting Eq. (A) in (B), since E 2 =E 1 ,

U 2 –U 1 = Wt (Z 2 –Z 1 ) = Wt (V 2 –V 1 ) ÷A=P 2 (V 1 –V 2 ), or

(C) Treating the air as an ideal gas, Eq. (C) may be written in the form

m (u 2 –u 1 ) = Wt (V 1 –V 2 ) ÷ A.

mc vo (T 2 –T 1 ) = Wt (V 1 –V 2 ) ÷ A.

(D) The two unknowns in Eq. (D) are T 2 ,V 2 , so that an additional equation is required to

solve the problem. Invoking the ideal gas law for the fixed mass

P 1 V 1 /RT 1 =P 2 V 2 /RT 2 ,

(E) Equations (D) and (E) provide the solution for V 2 and T 2 . Substituting for V 2 from Eq.

(E) in (D), we obtain a solution for T 2 /T 1 , namely,

(F) Using R = R /M = 8.314 ÷ 28.97 = 0.287 kJ kg –1 K –1 ,

T 2 /T 1 = (P 2 /P 1 +c vo /R)/(1 + c vo /R).

T 2 /T 1 = (200 ÷100 + 0.7÷0.287) ÷ (1 + 0.7 ÷ 0.287) = 1.29, or T 2 = 387 K. Substituting this result in Eq. (E),

(G) ∴V 2 /V 1 = (1+ 0.7 × 100 ÷ (0.287 × 200)) ÷ (1+ 0.7 ÷ 0.287) = 0.65, and

V 2 /V 1 = (1 + (c vo /R)(P 1 /P 2 ))/(1 + c vo /R).

V 2 3 = 0.65 × 0.1 = 0.065 m . Remarks

The potential energy of the weight is converted into thermal energy in air. Once P 2 and T 2 are known, it is possible to determine k o (= c po /c vo ) for an ideal gas using Eq. (F). Furthermore, employing the identity R = c po , − c vo , it is possible to cal-

culate the molar specific heats. The gas molecular weight is required in order to as- certain the mass–based specific heats. If the ambient pressure is finite, then Eq.

(F) and (G) remain unaffected, but P 2 =

W/A + P o .

A machine that violates the first law of thermodynamics is termed a perpetual motion ma- chine of the first kind (PMM1) (e.g., the “magician” David Copperfield lifting a man and, thus, changing potential energy without performing any work). Such a machine cannot exist.

We have thus far presented the First Law in the context of closed systems containing fixed masses. This analysis is applicable, for example, to expansion and compression processes within auto- mobile engines, and the heating of matter in en- closed cooking pots. Most of the practical systems

Figure 9: Nonuniform property within involve open systems such as compressors, turbines,

a control volume.

heat exchangers, biological species, etc. In the next section we will examine the derivation of the first law for an open system.