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Lecture 2 the First Law of Thermodynamics (Ch.1)

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Lecture 2 the First Law of Thermodynamics (Ch.1) Lecture 2 The First Law of Thermodynamics (Ch.1) Outline: 1. Internal Energy, Work, Heating 2. Energy Conservation – the First Law 3. Quasi-static processes 4. Enthalpy 5. Heat Capacity Internal Energy The internal energy of a system of particles, U, is the sum of the kinetic energy in the reference frame in which the center of mass is at rest and the potential energy arising from the forces of the particles on each other. system Difference between the total energy and the internal energy? boundary system U = kinetic + potential “environment” B The internal energy is a state function – it depends only on P the values of macroparameters (the state of a system), not on the method of preparation of this state (the “path” in the V macroparameter space is irrelevant). T A In equilibrium [ f (P,V,T)=0 ] : U = U (V, T) U depends on the kinetic energy of particles in a system and an average inter-particle distance (~ V-1/3) – interactions. For an ideal gas (no interactions) : U = U (T) - “pure” kinetic Internal Energy of an Ideal Gas f The internal energy of an ideal gas U = Nk T with f degrees of freedom: 2 B f ⇒ 3 (monatomic), 5 (diatomic), 6 (polyatomic) (here we consider only trans.+rotat. degrees of freedom, and neglect the vibrational ones that can be excited at very high temperatures) How does the internal energy of air in this (not-air-tight) room change with T if the external P = const? f ⎡ PV ⎤ f U =Nin room k= T Bin⎢ N room = ⎥ = PV 2 ⎣ kB T⎦ 2 - does not change at all, an increase of the kinetic energy of individual molecules with T is compensated by a decrease of their number. Work and Heating (“Heat”) WORK We are often interested in ΔU , not U. ΔU is due to: Q - energy flow between a system and its environment due to ΔT across a boundary and a finite HEATING thermal conductivity of the boundary – heating (Q >0)/cooling (Q <0) (there is no such physical quantity as “heat”; to emphasize this fact, it is better to use the term “heating” rather than “heat”) W - any other kind of energy transfer across boundary - work Work and Heating are both defined to describe energy transfer across a system boundary. Heating/cooling processes: conduction: the energy transfer by molecular contact – fast-moving molecules transfer energy to slow-moving molecules by collisions; convection: by macroscopic motion of gas or liquid radiation: by emission/absorption of electromagnetic radiation. The First Law The first law of thermodynamics: the internal energy of a system can be changed by doing work on it or by heating/cooling it. ΔU = Q + W conservation of energy. Sign convention: we consider Q and W to be positive if energy flows into the system. P For a cyclic process (Ui = Uf) ⇒ Q = - W. If, in addition, Q = 0 then W = 0 T V An equivalent formulation: Perpetual motion machines of the first type do not exist. Quasi-Static Processes Quasi-static (quasi-equilibrium) processes – sufficiently slow processes, any intermediate state can be considered as an equilibrium state (the macroparamers are well- defined for all intermediate states). Advantage: the state of a system that participates in a quasi-equilibrium process can be described with the same (small) number of macro parameters as for a system in equilibrium (e.g., for an ideal gas in quasi- equilibrium processes, this could be T and P). By contrast, for non- equilibrium processes (e.g. turbulent flow of gas), we need a huge number of macro parameters. Examples of quasi- For quasi-equilibrium processes, P, V, T are equilibrium processes: well-defined – the “path” between two states is a continuous lines in the P, V, T space. isochoric: V = const P 2 isobaric: P = const isothermal: T = const 1 V adiabatic: Q = 0 T Work The work done by an external force on a gas A – the enclosed within a cylinder fitted with a piston: piston area W = (PA) dx = P (Adx) = - PdV force The sign: ifthevolumeisdecreased, W is positive (by Δx compressing gas, we increase its internal energy); if the volume is increased, W is negative (the gas decreases its internal energy by doing some work on the P environment). V2 W1− = 2 P − T(,) V dV ∫V 1 W = - PdV - applies to any shape of system boundary dU = Q – PdV The work is not necessarily associated with the volume changes – e.g., in the Joule’s experiments on determining the “mechanical equivalent of heat”, the system (water) was heated by stirring. W and Q are not State Functions V2 - we can bring the system from state 1 to W1− = 2 P − T(,) V dV ∫V state 2 along infinite # of paths, and for each 1 path P(T,V) will be different. P 2 Since the work done on a system depends not only on the initial and final states, but also on the 1 V intermediate states, it is not a state function. T ΔU = Q + W U is a state function, W - is not ⇒ thus, Q is not a state function either. P A WWWPVVPVVB = + = − ( − )− ( − ) P2 net AB CD 2 2 1 1 1 2 = −PPVV2()() − 1 2 − 1 0 < P1 D C - the work is negative for the “clockwise” cycle; if the cyclic process were carried out in the reverse V1 V2 V order (counterclockwise), the net work done on the gas would be positive. PV diagram Comment on State Functions U, P, T, and V are the state functions, Q and W are not. Specifying an initial and final states of a system does not fix the values of Q and W, we need to know the whole process (the intermediate states). Analogy: in classical mechanics, if a force is not conservative (e.g., friction), the initial and final positions do not determine the work, the entire path must be specified. In math terms, Q and W are not exact differentials of some functions of macroparameters. To emphasize that W and Q are NOT the state U functions, we will use sometimes the curled symbols δ (instead of d) for their increments (δQ and δW). V d U T= d S − P- andV exact differential S z(x ,y ) y 1 1 dz=+ Axy( x,, y) dx A( x y) dy - it is an exact differential if it is the difference between the values of some (state) function z(x ,y ) 2 2 z(x,y) at these points: dz=+ z( x dx,,y +−dy) zx( y) x A∂ ( x, ) yA∂ y ( x, ) y A necessary and sufficient condition for this: x = ∂y ∂x If this condition z∂ ( x, ) y z∂ ( x, ) y ⎛ ∂z ⎞ ⎛ ∂z ⎞ A() x, = y A() x, = y d z= ⎜ ⎟ dx + ⎜ ⎟ dy holds: x x y y ⎜ ⎟ ∂ ∂ ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x ⎛ f T ⎞ - cross derivatives e.g., for an idealQ gas: δ dU= + PdV = B ⎜ dT Nk+ dV ⎟ ⎝ 2 V ⎠ are not equal Problem Imagine that an ideal monatomic gas is taken from its initial state A to state B by an isothermal process, from B to C by an isobaric process, and from C back to its initial state A by an isochoric process. Fill in the signs of Q, W, and ΔU for each step. P, 105 Pa Step QWΔU A 2 A → B +--0 T=const B → C -- + -- B 1 C C → A +0+ 1 2 V, m3 f U = TNk PV= NkB T 2 B Quasistatic Processes in an Ideal Gas isochoric ( V = const ) P 2 W1→ 2= 0 3 PV= NkBT2 Q= Nk( − T () >T0 = ) C Δ T 1 1→ 2 B 2 1 V PV= NkBT1 2 (see the last slide) V1,2 V dU= 1→ Q 2 isobaric ( P = const ) P 2 W P= V −(,) T= dV − − P0 V < V 1→ 2 ∫ ()2 1 2 1 1 PV= NkBT2 5 PV= Nk T Q=1→ 2 NkB( − T2 1 () >T0 =P ) C Δ T B 1 2 V1 V2 V dU=1 W→ 2+ → 1 Q 2 Isothermal Process in an Ideal Gas P isothermal ( T = const ) : PV= Nk T B dU = 0 W V2 V2 dV V2 V VW1→ P= 2 V − (,) T= dV − B Nk=Nk − T B ln T 2 1 V ∫ ∫ V V V1 V1 1 Vi QW1→ 2= − → 1 2 Wi− f = NkB ln T V f Wi-f > 0 if Vi >Vf (compression) Wi-f < 0 if Vi <Vf (expansion) Adiabatic Process in an Ideal Gas adiabatic (thermally isolated system) Q1→ 2= 0 dU= 1→ W 2 the stneeded to change amount of work The system ate of a thermally isolated depends only on the initial and final states and not on the intermediate states. V2 W= P − (,) V T dV 1→ 2 ∫ P V1 2 to calculate W1-2 , we need to know P (V,T) for an adiabatic process PV= Nk T 1 B 2 f f PV= Nk T U =Nk T⇒Nk dU = dT= − PdV B 1 2 B 2 B V 2 V1 V ( f – the # of “unfrozen” degrees of freedom ) 2 PV= NkPdV ⇒ T + VdP= NkPdV+ dT VdP = −PdV ÷ PV B B f V P dV ⎛ 2 ⎞ dP 2 Adiabatic dV dP ⎜1+ ⎟+ =0 ,γ= 1 + γ + =0 V⎜ f⎟ P ∫ f exponent ∫V ∫P ⎝ ⎠ V1 P1 γ ⎛ V ⎞ ⎛ P ⎞ γ ⎜ ⎟ 1 γ ln⎜ ⎟ = ln⎜ PV⎟⇒ P =1 V 1 = const ⎝V1 ⎠ ⎝ P ⎠ Adiabatic Process in an Ideal Gas (cont.) γ γ P PV P=1 V 1 = const 2 An adiabata is “steeper” than an isotherma: PV= NkBT2 in an adiabatic process, the work flowing 1 out of the gas comes at the expense of its PV= NkBT1 thermal energy ⇒ its temperature will V 2 V1 V decrease.
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