U = Q W, (4.1) W = F dr (4.2)

Transcription

1 ... there is in the physical world one agent only, and this is called Kraft [energy or work]. It may appear, according to circumstances, as motion, chemical affinity, cohesion, electricity, light and magnetism; and from any one of these forms it can be transformed into any of the others. Karl Friedrich Mohr - Zeitschrift für Physik (1837). Chapter 4 The First Law 4.1 Work, Heat and The First Law Taking to heart Professor Murray Gell-man s observation that opened Chapter 3, a respite from quantum mechanics is in order and the discussion begun in Chapter 1 is continued. Restating the First Law of Thermodynamics U = Q W, (4.1) where U is the change in U, the internal energy. U is a thermodynamic state function with quantum foundations. In this chapter the First Law s classical terms W (work) and Q (heat) are discussed, followed by a few elementary applications. 1. W (Work): W is energy transferred between a system and its surroundings by mechanical or electrical changes at its boundaries. Work is a notion imported from Newton s mechanics with the definition W = F dr (4.2) where F is an applied force and dr is an infinitesimal displacement. integral is carried out over the path of the displacement. Work is neither stored within a system nor does it describe any state of the system. Therefore W =? W final W initial has absolutely no thermodynamic meaning! 1 The

2 2 CHAPTER 4. THE FIRST LAW Incremental work, on the other hand, can be defined as d W, with finite work being the integral W = d W (4.3) The bar on the d is a warning that d W is not a true mathematical differential. 1 Note: Macroscopic parameters are further classified as: A. extensive proportional to the size or quantity of matter (e.g. energy, length, area, volume, number of particles, polarization, magnetization) B. intensive independent of the quantity of matter (e.g. force, temperature, pressure, density, tension, electric field, magnetic field.) Incremental work has the generalized meaning of energy expended by/on the system due to an equivalent of force (an intensive parameter) acting with an infinitesimal equivalent of displacement (an extensive parameter) which, in accord with Eq.4.2, is written as the conjugate pair, d W = j F j intensive Examples of conjugate work pairs are 1. pressure (p) volume ( dv ), 2. tension (τ) elongation ( dχ), 3. stress (σ) strain ( dɛ), 4. surface tension (s) area ( da), 5. electric displacement field (D) polarization ( dp), 6. magnetic field (H ) magnetization ( dm) chemical potential (µ) particle number ( dn ) extensive dxj. (4.4) 1 Incremental work is not a function of independent variables. It is not an exact differential. 2 The proper fields used in representing electric and magnetic work will be further examined in Chapter 12.

3 4.1. WORK, HEAT AND THE FIRST LAW 3 It is a helpful rule in thermodynamics to think of work as the equivalent of raising and lowering weights external to the system. Finite work, as expressed in Eq.4.3, depends on the the integration path the details of the process and not just on the process end points. The sign convention adopted here 3 is d W > 0 system performs work (d W gas by = pdv ) d W < 0 work done on the system (d Won gas = pdv ) 2. Q Heat : Q is energy transferred between a system and any surroundings with which it is in diathermic contact. Q is not stored within a system and is not a state parameter of a macroscopic system. Therefore Q? = = Q final Q initial has no thermodynamic meaning! On the other hand, incremental heat is defined as d Q, with its integral being finite heat Q Q = d Q The bar on the d is a warning that d Q is not a true mathematical differential. Therefore derivatives like d Q have no meaning.4 dt Finite heat depends on the details of the process, i.e. the integration path, and not just its end points. 3. Temperature T : Temperature is a parameter intuitively associated with a subjective sense of hotness or coldness. Empirical ideas about temperature have objective meaning through thermometers the height of a column of mercury, the pressure of an ideal gas in a container with fixed volume, the electrical resistance of a length of wire all vary in a way that allows a temperature scale to be defined. Since there is no quantum hermitian temperature operator a formal, consistent definition of temperature is elusive, but is achievable in practice through thermodynamic results of the least bias (maximum uncertainty) postulate 5 which shows, with remarkable consistency, how and where temperature appears and that it measures the same property as a common thermometer. 3 With system pressure and volume as the conjugate pair. 4 Q is not a function of T nor of any other independent variables. 5 This crucial subject will be discussed in Chapter 6.

4 4 CHAPTER 4. THE FIRST LAW 4. Internal Energy U: As discussed in Chapter 1, internal energy U is a state of the system. Therefore U does have meaning and du is an ordinary mathematical differential 6 with U = U(B) U(A) = A B du. (4.5) where A and B represent different macroscopic equilibrium states. Clearly, U depends only on the end points of the integration and not on any particular thermodynamic process (path) connecting them Exact Differentials All thermodynamic state variables are true (exact) differentials with a change in state defined as in Eq.4.5. Moreover, state variables are not independent and can be functionally expressed in terms of other state variables. Usually only a few are needed to completely specify any state of a system. The precise number required is the content of an important theorem which is stated here without proof. The number of independent parameters needed to specify a thermodynamic state is equal to the number of distinct conjugate pairs of quasistatic work [see Eqs.below (4.4)] needed to produce a differential change in the internal energy plus ONE 7 For example, if a gas undergoes mechanical expansion or compression, the only quasistatic work is mechanical work, d W QS = p dv. Therefore only two variables are required to specify a state. We could then write the internal energy of a system as a function of any two other state parameters. For example we can write U = U (T, V ) or U = U (P, V ). To completely specify the volume of a gas we could write V = V (P, T ) or even V = V (P, U). The choice of functional dependences is determined by the physical situation and computational objectives. If physical circumstances also allow particle number to change, say by evaporation of liquid into a gas phase or condensation of gas into the liquid phase or chemical reactions where molecular species are lost or gained, a thermodynamic description requires chemical work d W QS = µ dn where µ is the chemical potential and dn 6 It is an exact differential. 7 The one of the plus one originates from a conjugate pair T ds, where S is Entropy, about which much will be said throughout the remainder of this book.

5 4.1. WORK, HEAT AND THE FIRST LAW 5 is an infinitesimal change in particle number. Chemical potential µ is an intensive variable associated with energy per particle while particle number N is obviously an extensive quantity. In cases like this quasi-static work 8 d W QS = p dv µ dn (4.6) has two kinds of contributions and therefore three variables are required, e.g. U = U (T, V, N) or U = U (P, V, µ). Incorporating the definitions above, an incremental form of the First Law of Thermodynamics is du = d Q d W (4.7) Equations of State Equations of state are functional relationships among a system s equilibrium state parameters usually those that are most easily measured and controlled in the laboratory. Determining equations of state, by experiment or theory, is often a primary objective in practical research for then thermodynamics can fulfill its role in understanding and predicting state parameter changes for macroscopic systems. 1. Equations of state for gases (often called gas laws) take the functional form f(p, V, T ) = 0 where p is the equilibrium gas pressure, V is the gas volume, T is the gas temperature and f defines the functional relationship. For example a gas at low pressure and low particle density, n = N/V, where N is the number of gas particles (atoms, molecules, etc.), has the ideal gas equation of state or, more familiarly, where k B is Boltzmann s constant. 9 p (N/V ) k B T = 0 (4.8) pv = Nk B T (4.9) 2. An approximate equation of state for two and three dimensional metallic materials is usually expressed as g(σ, ɛ, T ) = 0 where σ = τ/a is the stress, 8 With a gas as an example. 9 For gases under realistic conditions Eq.4.9 may be inadequate and more complicated empirical or theoretical equations of state are used. Nevertheless, because of its simplicity and zeroth order validity, the ideal gas law is used in a variety of models and exercises.

6 6 CHAPTER 4. THE FIRST LAW ɛ = (l l 0 )/ l 0 is the strain 10, T is the temperature, A is the cross-sectional area and κ is a characteristic constant. An approximate equation of state for a metal rod is the Hookean behavior ɛ = κ σ. (4.10) 3. An equation of state for a uniform one-dimensional elastomeric material has the form Λ(τ, χ, T ) = 0 where τ is the equilibrium elastic tension, χ is the sample elongation, T is the temperature and Λ is the functional relationship. For example, a rubber-like polymer that has not been stretched beyond its elastic limit can have the simple equation of state or, alternatively σ K T ɛ = 0 (4.11) τ K T χ = 0, (4.12) where K and K are material specific elastic constants. 11,12 4. Non-polar dielectrics in weak polarizing fields have the equation of state P i = α ij (T ) E j (4.13) j where P i is an electric polarization vector component, α ij (T ) is the temperature dependent electric susceptibility tensor and E j is an electric field component. 13 In stronger fields equations of state can be generalize by including non-linear terms, e.g. P i = α (1) ij (T )E j + α (2) ijk (T )E je k. (4.14) j 10 l is the stretched length and l 0 is the unstretched length 11 E. Guth and and H. M. James, Theory of Elastic Properties of Rubber, J. Chem. Phys. 15, 2941 (1943). 12 Comparing these with Eq.4.10 suggests a physically different origin for rubber elasticity which will be explored in Chapter Is this an external electric field or some average local internal electric field? How do Maxwell s equations fit in here? This will be addressed in an Chapter 12. For the present, we assume that the sample is ellipsoidal in which case the internal and external electric fields are the same. j,k

7 4.1. WORK, HEAT AND THE FIRST LAW 7 5. Non-permanently magnetic materials which are uniformly magnetized in weak magnetic fields are described by the equation of state M = χ M (T ) B (4.15) where M is the magnetization vector, χ M (T ) is a temperature dependent magnetic susceptibility tensor and B is the magnetic induction field prior to insertion of the sample. 14, Examples: I h 2 h 1 Figure 4.1: Quasi-static, reversible work. A piston loaded down with a pile of fine sand. 1. A gas is confined in a cylinder by a frictionless, massless piston of negligible mass and cross-sectional area A. The gas pressure is counteracted by a pile 14 Magnetism and magnetic work, will be discussed in Chapter Solid state physics text books often express Eq.4.15 in terms of an external magnetic field H in which a needle-like sample geometry is assumed in which the long axis is parallel to the external field.

8 8 CHAPTER 4. THE FIRST LAW of fine sand of total mass m resting on the piston [see Figure 4.1] so that it is initially at rest. Neglecting any atmospheric (external) force acting on the piston, the downward force exerted by the sand is, initially, F0 sand = mg. Therefore the initial equilibrium gas pressure, p gas 0, is p gas 0 = mg A. (4.16) Sand is now removed, one grain at a time allowing the gas to expand in infinitesimal steps, continually passing through equilibrium states. Moreover, it is conceivable that at any point a single grain of sand could be returned to the pile and exactly restore the previous equilibrium state. Such a real or idealized infinitesimal process is called quasi-static. A quasi-static process, by definition, takes place so slowly that the system (in this example a gas) always passes through equilibrium states so that an equation of state always relates system pressure, volume, temperature, tension, etc. If, as in this case, the previous, infinitesimally displaced state can be accessed with no dissipation of energy the process is also said to be reversible. 16 Assuming the quasi-static expansion also takes place isothermally at temperature T R, what work is done by the gas in raising the piston from h 1 to h 2? At all stages of the quasi-static expansion equilibrium is maintained with pressure p gas and temperature T R well defined. Therefore incremental work done by the gas (assumed ideal) is d W QS = p gas dv (4.17) = N k B T R V dv. (4.18) Integrating Eq.4.18 along the isothermal path [as pictured in Figure 4.4] gives W QS = N k B T R V 2 =h 2 A V 1 =h 1 A dv V (4.19) = N k B T R ln (h 2 /h 1 ). (4.20) 16 Not all quasi-static processes are reversible but all reversible processes are quasi-static. Reversibility implies no friction or other energy dissipation in the process.

9 4.1. WORK, HEAT AND THE FIRST LAW 9 2. If, instead of a restraining sand pile the piston is now pinned a distance h 1 above the bottom of the cylinder [See Figure 4.2], while at the same time a constant atmospheric pressure p atm exerts an external force F atm on the piston, F atm = p atm A. (4.21) The equilibrium gas pressure p gas 0 initially exerted by the gas on the pinned piston is where F pin is the pinning force, so that p gas 0 > p atm. p gas 0 = F pin A + patm (4.22), atm pin atm Figure 4.2: Irreversible work. A gas filled piston initially pinned at height h 1. Quickly (and frictionlessly) extracting the pin frees the piston allowing the gas to expand rapidly and non-uniformly. After things quiet down the gas reaches an equilibrium state, with the piston having moved upward a distance

10 10 CHAPTER 4. THE FIRST LAW l = h 2 h 1 with a final gas pressure p gas f = p atm. This isothermal process characterized by a finite expansion rate and system (gas) non-uniformity is neither quasi-static nor reversible. 17,18 What isothermal work is done by the rapidly expanding gas in raising the piston a distance l? The gas expands irreversibly. Therefore during expansion a gas law does not apply. But initially the gas has equilibrium pressure p gas 0 with volume V 0 V 0 = Nk BT p gas 0 After the piston has returned to rest (equilibrium), with volume. (4.23) p gas f = p atm (4.24) V f = Nk BT p atm. (4.25) While expanding irreversibly the gas lifts the piston against a constant force F atm (equivalent to raising a weight mg = F atm in the earth s gravitational field.) 19 The incremental, irreversible work done by the confined gas (the system) is which, with constant p atm, is integrated to give With Eqs and 4.25 d W = p atm dv (4.26) W = p atm (V f V 0 ). (4.27) W = Nk B T [1 p atm p gas ]. (4.28) 0 17 All real processes in nature are irreversible. Reversibility is a convenient idealization for thermodynamic modeling. 18 Irreversible as used here, is not intended as an absolute, since surroundings can always be altered to recover the initial equilibrium state. But this is not the same as the sequence of infinitesimal steps associated with reversibility. 19 The piston starts at rest and returns to rest, so there is no change in piston s kinetic energy.

11 4.1. WORK, HEAT AND THE FIRST LAW As in Example 2, the gas is initially confined by a piston exposed to the atmosphere with its initial position secured by a pin. However the piston s movement is restricted by a frictional wedge which, together with the external atmosphere, slows the piston s upward movement, enabling it to be approximated as quasi-static but irreversible. [See Figure 4.3.] Assuming the process is isothermal, what work is done by the expanding gas? The expanding gas (the system) raises the piston against atmospheric force F atm and the force of friction F friction, both taken to have constant value. Therefore the work performed is which after integration becomes d W sys = (F atm + F friction ) A where V 0 is the initial equilibrium volume dv (4.29) W sys = (p atm + F friction /A) (V f V 0 ) (4.30) V 0 = Nk B T p atm (1 + F pin/a p atm ) (4.31) and V f is the final equilibrium volume Therefore V f = Nk BT. (4.32) patm W sys = Nk BT p atm ( F pin A ) (p atm + F friction /A) (p atm + F pin /A) (4.33) The p-v Diagram In performing quasi-static work, a system passes through equilibrium states, so that it is possible to draw a continuous curve on a p V diagram describing the process. An isothermal work path, is represented by the isotherm in Figure 4.4, where the work is the green area under the isotherm.

12 12 CHAPTER 4. THE FIRST LAW Figure 4.3: Quasi-static, irreversible work. A piston slowed by a frictional wedge Examples II 1. A mass 2 m affixed to the end of a vertically stretched massless rubber band, is suspended a distance h 1 (rubber band length l 1 ) above a table. When half the mass is suddenly removed the rubber band rapidly contracts causing the remaining mass to rise. After oscillating a while it finally comes to rest a distance h 2, (h 2 > h 1 ) (rubber band length l 2 ) above the table. [See Figure 4.5.] (a) What is the work done by the rubber band? After returning to rest, the work (not quasi-static) done by the rubber band (work done by the system) in raising the weight m (against gravity) is W = mg l (4.34) where l = l 2 l 1 < 0. (b) What is the change in the rubber band s internal energy?

13 4.1. WORK, HEAT AND THE FIRST LAW 13 p=0 Figure 4.4: p V diagram for an isothermal, quasi-static process. Since the process takes place rapidly it is irreversible. Moreover there is not enough time for the rubber band to lose heat to the surroundings and the process can be regarded as adiabatic. Using the First Law U = +mg l. (4.35) The rubber band loses internal energy. (c) This time, in counteracting rubber elasticity, the total mass 2 m is restrained to rise slowly (quasi-statically) but isothermally. What is the work done by the contracting rubber band? We can safely use an equilibrium tension (an elastic state parameter) which acts to raise the weight. The incremental quasi-static elastic work done by the system is d W QS = τ dχ (4.36) where dχ is the differential of elastic extension. Using the rubber

14 14 CHAPTER 4. THE FIRST LAW l 2 l 1 h 2 h1 Figure 4.5: A loaded rubber band, when released, rapidly contracts. equation of state, Eq.4.12, and integrating over the range of extension χ 0 W QS = K T χ dχ (4.37) l = K T ( l) 2 (4.38) 2. A metal rod, cross sectional area A, is suddenly struck by a sledgehammer, subjecting it to a transient but constant compressive (external) force F which momentarily changes the rod s length by l = l f l 0 < 0.

15 4.1. WORK, HEAT AND THE FIRST LAW 15 (a) What is the work done on the rod during the interval that its length decreases by l? Sudden compression of the rod implies that it does not pass through equilibrium states so an equilibrium internal stress in the bar σ = f/a is not defined. But the mechanical work done on the rod by the constant compressive force is, however W = F l. (4.39) (b) On the other hand, if the rod could be compressed infinitely slowly (perhaps by a slowly cranked vise) the rod s internal stress is a well defined equilibrium property and the quasi-static incremental work done by the system (i.e. the rod) has the standard form d W QS = V σ dɛ (4.40) where V is the rod s volume, σ is the internal stress and dɛ = (dl/l) defines the differential strain. We might now apply an approximate equation of state, such as Eq.4.10, and get W QS = V ɛ f (ɛ/κ) dɛ (4.41) ɛ 0 = (V /2κ) (ɛ 2 f ɛ2 0) (4.42) Heat Capacity Heat capacities have played an important role in exploring microscopic properties of matter. Historically, heat capacity measurements have revealed unexpected behaviors of macroscopic systems which inspired important microscopic quantum discoveries. Equations of state, together with heat capacities, usually offer the basis for a complete thermal picture of matter. An infinitesimal of thermal energy (heat) quasi-statically entering or leaving a system produces an infinitesimal change in temperature. (Exceptions are certain first-order phase transitions, such as boiling of water or melting of ice.) The proportionality constant between the two infinitesimals, together with a list of state variables, α, that

16 16 CHAPTER 4. THE FIRST LAW are fixed during the quasi-static process, is called heat capacity. 20 definition of heat capacity (an extensive quantity) is then The complete d Q QS = C α dt (constant α) (4.43) where C α is the heat capacity at constant α. The units of heat capacity are energy/k. Specific heats are alternative intensive quantities, having units energy/k/mass or energy/k/mole. 1. If heat is transferred at constant volume, from the definition given in Eq.4.43 d Q QS = C V dt (constant volume). (4.44) Heat capacities are equivalently and more usefully expressed in terms of partial derivatives of state parameters. For example, if a gas performs only quasi-static mechanical work, the infinitesimal First Law for the process is du = d Q QS p dv (4.45) where p dv is quasi-static incremental work d W QS done by the gas (system.) Since U can here be expressed in terms of two other thermal parameters, choose U = U (T, V ) so that its total differential is du = ( U V ) T and Eq.4.45 can then be rewritten dv + ( U T ) V dt (4.46) d Q QS = [( U V ) + p] dv + ( U T T ) V dt. (4.47) Thus, for an isochoric (constant volume) path, i.e. dv = 0, and from Eq.4.44 d Q QS = ( U T ) V C V = ( U T ) V dt (4.48) (4.49) 20 The term heat capacity is an anachronistic holdover from early thermodynamics when heat was regarded as a stored quantity. Although we now know better, the language seems not to have similarly evolved.

17 4.1. WORK, HEAT AND THE FIRST LAW Thermodynamic experiments, especially chemical reactions, are often carried out in an open atmosphere where the pressure is constant. In such cases we define d Q QS = C p dt (constant pressure). (4.50) Expressing C p as the partial derivative of a state parameter requires a bit more ingenuity. Here again the infinitesimal First Law description of the process is But this time, using Eq.4.51 can be rewritten as Now impose the definition du = d Q QS dv. (4.51) d (pv ) = p dv + V dp (4.52) d Q QS = d (U + pv ) V dp. (4.53) H = U + pv (4.54) where H also a state function is called Enthalpy, 21 permitting recasting a new differential Law d Q QS = dh V dp. (4.55) Then, expressing H in terms of T and p, i.e. H = H (T, p) so that a total differential is dh = ( H p ) T dp + ( H T ) p Then for an isobaric (constant pressure) path, i.e. dp = 0, d Q QS = ( H T ) p dt. (4.56) dt (4.57) and we have the definition in terms of state variables C p = ( H T ) p (4.58) 21 Enthalpy is one of several defined state variables that are natural to and simplify specific thermodynamic conditions. Others will be acknowledged as need requires.

18 18 CHAPTER 4. THE FIRST LAW Examples III 1. An ideal monatomic gas, volume V 0, temperature T 0, is confined by a frictionless, movable piston to a cylinder with diathermic walls. The external face of the piston is exposed to the atmosphere at pressure P 0. A transient pulse of heat is injected through the cylinder walls causing the gas to expand against the piston until it occupies a final volume V f. (a) What is the change in temperature of the gas? At the beginning and end of the expansion the system is in equilibrium so that an equation of state applies. Initially P 0 V 0 = Nk B T 0 and finally P 0 V f = Nk B T f so that T = P 0 V Nk B (4.59) where T = T f T 0 and V = V f V 0. (b) How much heat Q was in the pulse? The gas will expand turbulently, i.e. irreversibly, against constant atmospheric pressure P 0 so that incremental work done by the expanding gas is d W = P 0 dv. (4.60) Applying the First Law Q = U + P 0 V. (4.61) and using Eqs.4.46, 4.58 and 4.59 gives the internal energy change for this ideal gas Thus the injected heat is U = C V T (4.62) = C V P 0 V Nk B (4.63) Q = ( C V Nk B + 1) P 0 V (4.64) where Eqs.4.61 and 4.63 have been used. Using another ideal gas property 22 C p C V = Nk B (4.65) 22 This is stated, for now, without proof.

19 4.1. WORK, HEAT AND THE FIRST LAW 19 and the definition Eq.4.64 can be condensed to Q = γ = C p /C V, (4.66) γ γ 1 P 0 V. (4.67) L Figure 4.6: A wire held between firmly planted rigid vertical posts. 2. A steel wire, length L and radius ρ is held with zero tension τ between two firmly planted rigid vertical posts, as shown in Figure 4.6. The wire, initially at a temperature T H, is then cooled to a temperature T L. (a) Find an expression for the tension τ in the cooled wire. The thermodynamic state variables used for describing elastic behavior in the steel wire (or rod) are stress σ = τ/a where A is the wire s cross sectional area, and strain ɛ = (l l 0 )/l 0 where l and l 0 are the wire s stretched and unstretched length. These are the same variables referred to in elastic work [see Eq.4.40] and in the elastic equation of state [see Eq.4.15]. In the present situation the rigid posts constrain the length of the wire to remain constant so that a decrease in temperature is expected to increase the strain (tension) in the wire. Referring to Section 4.1.2, consider a functional dependence of strain as ɛ = ɛ(t, σ), which are the physically relevant state variables for the wire, and take the total differential dɛ = ( ɛ T ) σ dt + ( ɛ σ ) T dσ. (4.68)

20 20 CHAPTER 4. THE FIRST LAW The partial derivatives represent physical properties of the system under investigation for which extensive numerical tables can be found in the literature. 23 In Eq.4.68 α L = ( ɛ T ) (4.69) σ where α L is the linear thermal expansivity, and where E T 1 E T = ( ε σ ) T is the isothermal Young s modulus. (4.70) Thus we can write dɛ = α L dt + 1 dσ, (4.71) E T and since the length of the wire is constant, i.e. dɛ = 0, we have 0 = α L dt + 1 E T dσ. (4.72) Finally, assuming α L and E T are average constant values, integration gives and with τ(t H ) = 0 σ = σ(t L ) σ(t H ) = α L E T (T L T H ), (4.73) τ(t L ) = Aα L E T (T L T H ) (4.74) = πr 2 α L E T (T H T L ) (4.75) 3. A gas is allowed to rapidly expand from an initial state (p 0, V 0, T 0 ) to a final state f. What is the final state (p f, V f, T f )? This rapid process can be regarded as taking place adiabatically, Q = 0. Applying the infinitesimal First Law we have, Since U is a state variable the integral d Q = 0 = du + d W. (4.76) U = du (4.77) 23 See for example;

21 4.1. WORK, HEAT AND THE FIRST LAW 21 is path (i.e. process) independent. Therefore the integral d W = U (4.78) depends only on its thermodynamic end points. So we connect these equilibrium states with a unique, quasi-static (maximum work) process and write Then with du as in Eq.4.46 p f,v f,t f 0 = [ du + p dv ]. (4.79) p 0, V 0, T 0 0 = p f,v f,t f p 0, V 0, T 0 {[( U V ) + p] dv + ( U T T ) V dt }. (4.80) So far the result is general (any equation of state). But for simplicity, assume the gas is ideal. Using Eq.4.48 and applying the ideal gas property 24 ( U V ) T = 0 (4.81) together with the ideal gas equation of state [see Eq.4.9] gives which implies 0 = p f,v f,t f p 0, V 0, T 0 {[ Nk BT V ] dv + C V dt } (4.82) [ Nk BT ] dv = C V dt. (4.83) V This differential equation separates quite nicely to give upon integration or Nk B ln ( V f V 0 ) = C V ln ( T 0 T f ) (4.84) ( V Nk B f ) = ( T C V 0 ) (4.85) V 0 T f 24 This useful ideal gas result is assigned as a problem. Note: it only applies to an ideal gas.

22 22 CHAPTER 4. THE FIRST LAW which with Eqs.4.65 and 4.66 becomes ( V γ 1 f ) = ( T 0 ). (4.86) V 0 T f 4. A perfectly insulated, rigid, very narrow necked vessel is evacuated and sealed with a valve. The vessel rests in an atmosphere with pressure P 0 and temperature T 0. [Figure 4.7.] The valve is suddenly opened and air quickly fills the vessel until the pressure inside the vessel also reaches P 0. If the air is treated as an ideal gas, what is the temperature T f of the air inside the vessel immediately after equilibrium is attained? In this problem we introduce the subtle strategy of replacing an essentially open system (the number of molecules inside the vessel is not constant) with a closed system (the gas is conceptually bounded by a movable piston that seals the neck and, ultimately the closed vessel below, as in Figure 4.7) This notional replacement is a common trick in thermodynamics and we take the opportunity to illustrate it here.

23 4.1. WORK, HEAT AND THE FIRST LAW 23 Figure 4.7: Evacuated, insulated vessel in an atmosphere with pressure P 0 and temperature T 0. Physically, the narrow neck separates the air inside the vessel from the air outside the vessel, the latter maintained at pressure P 0 by the infinite atmosphere. But we replace the external atmosphere with a piston that forces outside air across the narrow neck into the vessel by applying the constant external pressure P 0, as shown in Figure 4.8.

24 24 CHAPTER 4. THE FIRST LAW P 0 N T 0 v 0 N T f v f Figure 4.8: Evacuated vessel filling under pressure P 0. Focus attention on a small, fixed amount of air being forced into the vessel by the conjectured piston, say N molecules at pressure P 0 and temperature T 0 occupying a small volume v 0. In forcing these molecules through the valve the piston displaces a volume v 0. The work done on the N molecules by the

25 4.1. WORK, HEAT AND THE FIRST LAW 25 external atmosphere, pressure P 0, is W by the piston = d W (4.87) Therefore the work done by the N molecules of gas is = P 0 v (4.88) = P 0 v 0. (4.89) W by the gas = P 0 v 0. (4.90) Once the N molecules get to the other side of the valve and enter the evacuated, rigid walled vessel, they do NO work. 25 Therefore the total work done by the gas is W total = P 0 v (4.91) Applying the First Law to the N molecules, while approximating this sudden process as adiabatic 26 0 = U + Wtotal (4.92) or Choosing to write U = U (V, T ) 0 = 0 f du = ( U T ) V du P 0 v 0. (4.93) dt + ( U V ) T dv. (4.94) Since ( U T ) V = C V (4.95) and for an ideal gas ( U V ) T 25 This is evident since no weights are raised or lowered. 26 Too fast to allow heat exchange with the external atmosphere. = 0 (4.96)

26 26 CHAPTER 4. THE FIRST LAW we have Then with the ideal gas equation of state which with Eqs.4.65 and 4.66, becomes P 0 v 0 = C V (T f T 0 ). (4.97) Nk B T = C V (T f T 0 ) (4.98) T f = γt 0. (4.99) For an ideal monatomic gas (γ = 5/3) at room temperature (T 0 = 300K ), Eq.4.99 gives T f = 500K, indicating a substantial rise in temperature for the gas streaming into the vessel Concluding Remarks In this hiatus from quantum mechanics 27 we have returned to discuss timeless classical thermodynamics, but with an overriding confidence only quantum mechanics will give in establishing the meaning of state variables such as internal energy. We have also indicated how profoundly U differs from purely classical Work W and Heat Q. In the next chapters we will complete the process of unifying quantum mechanics and thermodynamics. 27 Not really!