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Home , State function

Lecture 2 The First Law of (Ch.1)

Outline: 1. Internal , , Heating 2. Energy Conservation – the First Law 3. Quasi-static processes 4. 5. Capacity The internal energy of a of particles, U, is the sum of the kinetic energy in the reference frame in which the center of 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 (no interactions) : U = U (T) - “pure” kinetic Internal Energy of an Ideal Gas

f The internal energy of an ideal gas U = TNk 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 )

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 = Broomin ⎢NTkN roomin == ⎥ = PV 2 ⎣ BTk ⎦ 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 is increased, W is negative (the gas decreases its internal energy by doing some work on the P environment).

V2 W −21 −= ),( dVVTP ∫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 W −21 −= ),( dVVTP ∫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 B = + = − ( − )− ( −VVPVVPWWW ) P2 net AB CD 211122

()()VVPP 1212 <−−−= 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 −= dVPSdTUd - an 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 ∂ ( , yxA ) ∂ y ( , yxA ) A necessary and sufficient condition for this: x = ∂y ∂x If this condition ∂ ( , yxz ) ∂ ( , yxz ) ⎛ ∂z ⎞ ⎛ ∂z ⎞ (), yxA = (), yxA = zd = ⎜ ⎟ dx + ⎜ ⎟ dy holds: x x y y ⎜ ⎟ ∂ ∂ ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x ⎛ f T ⎞ - cross derivatives e.g., for an ideal gas: δ =+= NkPdVdUQ B ⎜ dT + 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 , from B to C by an , and from C back to its initial state A by an . 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 W →21 = 0 3 PV= NkBT2 ()(0 Δ=>−= TCTTNkQ ) 1 →21 B 12 V PV= NkBT1 2 (see the last slide) V1,2 V = QdU →21

isobaric ( P = const )

P 2 −= ),( ()VVPdVTVPW <−−= 0 →21 ∫ 12 2 1 1 PV= NkBT2 5 PV= Nk T →21 B ()(12 0 PΔ=>−= TCTTNkQ ) B 1 2 V1 V2 V = → +QWdU →2121 Isothermal Process in an Ideal Gas

P isothermal ( T = const ) :

PV= Nk T B dU = 0

W V2 V2 dV V2 V V →21 −= ),( −= BTNkdVTVPW −= BTNk ln 2 1 V ∫ ∫ V V V1 V1 1

Vi → = −WQ →2121 − fi = BTNkW ln V f

Wi-f > 0 if Vi >Vf (compression)

Wi-f < 0 if Vi

adiabatic (thermally ) Q →21 = 0 =WdU →21 The amount of work needed to change the state of a thermally isolated system depends only on the initial and final states and not on the intermediate states.

V2 −= ),( dVTVPW →21 ∫ P V1 2 to calculate W1-2 , we need to know P (V,T) for an PV= Nk T 1 B 2 f f PV= Nk T U = dUTNk =⇒ −= PdVdTNk B 1 2 B 2 B V 2 V1 V ( f – the # of “unfrozen” degrees of freedom ) 2 TNkPV ⇒= + = dTNkVdPPdV VdPPdV −=+ PdV ÷ PV B B f V P dV ⎛ 2 ⎞ dP 2 Adiabatic dV dP ⎜1+ ⎟ =+ 0 γ 1, += γ =+ 0 ⎜ fV ⎟ P ∫ f exponent V ∫∫ P ⎝ ⎠ V1 P1 γ ⎛ V ⎞ ⎛ P ⎞ γ ⎜ ⎟ 1 γ ⎜ ⎟ = lnln ⎜ ⎟ 11 ==⇒ constVPPV ⎝V1 ⎠ ⎝ P ⎠ Adiabatic Process in an Ideal Gas (cont.)

γ γ P 11 == constVPPV 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 ⇒ its will V 2 V1 V decrease.

V VV22PV γ 1 2 W=− P(,) V T dV=−11 dV =− PVγγ V −+1 12→ ∫∫V γ 11−+γ 1 VV11 V1

γ 11⎛⎞ 1 =−PV11 ⎜⎟γγ−−11 γ −1⎝⎠VV21 γ ⇒ 1+2/3≈1.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 ≈1.33 (polyatomic) (again, neglecting the vibrational degrees of freedom) ff Prove WPVNkTU= Δ=() Δ=Δ 12→ 22B Summary of quasi-static processes of ideal gas

ΔUU≡−f Ui Quasi-Static Ideal gas ΔU Q W process law fff + 2 isobaric Vi V f Δ=UNkTPVB Δ= Δ PVΔ −PVΔ = 222 (ΔP=0) TTi f P P isochoric fff i = f Δ=UNkTPVB Δ=() Δ ()ΔPV 0 (ΔV=0) 222 TTi f V isothermal f PV= P V 0 −W −NkB T ln ii ff (ΔT=0) Vi adiabatic ff γ γ PVii = Pff V Δ=UNkTB Δ=Δ() PV 0 ΔU (Q=0) 22 Problem

Imagine that we rapidly compress a sample of air whose initial is 105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature? Rapid compression – approx. adiabatic, no time for the energy exchange with the environment due to thermal conductivity

11 = BTNkVP 1 γ −1 VP γ VP γ VP ⎛ V ⎞ T = TNkVP P = 11 11 TNk == 11 T ⇒ ⎜ 1 ⎟ = 2 22 B 2 2 V γ V γ −1 B 2 T 2 ⎜V ⎟ T γ γ 2 2 1 ⎝ 2 ⎠ 1 = VPVP 2211 For adiabatic processes: γ −1 γ −1 11 VTVT 22 == const also −1 / γγ= constTP γ −1 ⎛ V ⎞ ⎜ 1 ⎟ 4.0 = TT 12 ⎜ ⎟ K K ≈×≈×= 51474.12954295 K ⎝V2 ⎠ - poor approx. for a bike pump, works better for diesel engines Non-equilibrium Adiabatic Processes Free expansion 1. TV γ −1 = const V – increases ⇒ T – decreases (cooling)

2. On the other hand, ΔU = Q + W = 0 U ~ T ⇒ T – unchanged (agrees with experimental finding)

Contradiction – because approach #1 cannot be justified – violent expansion of gas is not a quasi- static process. T must remain the same.

γ −1 TV = const - applies only to quasi-equilibrium processes !!! The Enthalpy

Isobaric processes (P = const):

dU = Q - PΔV = Q -Δ(PV) ⇒ Q = Δ U + Δ(PV)

⇒ H ≡ U + PV - the enthalpy The enthalpy is a state function, because U, P, and V are state functions. In isobaric processes, the energy received by a system by heating equals to the change in enthalpy. in both cases, Q isochoric: Q = Δ U does not depend on the isobaric: Q = Δ H path from 1 to 2.

Consequence: the energy released (absorbed) in chemical reactions at constant volume (pressure) depends only on the initial and final states of a system. f ⎛ f ⎞ The enthalpy of an ideal gas: PVUH B BTNkTNk ⎜ +=+=+= 1⎟ BTNk (depends on T only) 2 ⎝ 2 ⎠

The heat capacity of a system - the amount of energy δQ transfer due to heating required to produce a unit C ≡ temperature rise in that system ΔT T f1 f2 f3 C is NOT a state function (since Q is not a T1+dT state function) – it depends on the path T between two states of a system ⇒ 1 i V ( isothermic – C = ∞, adiabatic – C = 0 )

C The specific heat capacity c ≡ m CV and CP

⎛ ∂U ⎞ C = ⎜ ⎟ the heat capacity at V ∂T δQ + PdVdU ⎝ ⎠V constant volume C == dT dT ⎛ ∂H ⎞ the heat capacity at C = ⎜ ⎟ P constant pressure ⎝ ∂T ⎠P

To find CP and CV, we need f (P,V,T) = 0 and U = U (V,T) f ⎛⎞f For an ideal gas U = TNk H =+⎜⎟1 NkB T 2 B ⎝⎠2 f f ⎛⎞f CV NkB == nR CnRP =+⎜⎟1 22 ⎝⎠2

# of moles 3 5 For one mole of a = = RCRC monatomic ideal gas: V 2 P 2 Another Problem P P During the ascent of a meteorological helium-gas filled balloon, i 3 3 its volume increases from Vi =1m to Vf =1.8m, and the pressure inside the balloon decreases from 1 bar (=105 N/m2)to Pf 0.5 bar. Assume that the pressure changes linearly with volume

between Vi and Vf. (a) If the initial T is 300K, what is the final T? (b) How much work is done by the gas in the balloon? Vi Vf V (c) How much “heat” does the gas absorb, if any? ()VP −= bar/m625.0 3 V +× bar625.1

3 (a) PV VP ff ×1.8mbar5.0 = B TTNkPV = TT if == K300 3 = K270 NkB VP ii ×1mbar1 V f V f

(b) δ ON −= )( dVVPW - work done on a system δ = )( dVVPW - work done by a system ∫ BY ∫ V i Vi V f δ −= δWW δ = dVVPW ()8.05.0)( mbar 3 4.05.0 mbar 3 6.0 mbar 3 ⋅=⋅=⋅×+⋅×= 106 4 J ON BY BY ∫ Vi

(c) δ +=Δ δWQU ON 3 3 ⎛T ⎞ ()⎜ f ⎟ 5 ()4 4 δδON =−Δ= ONifB =−− VPWTTNkWUQ ii ⎜ −1⎟ δWBY J ⋅=⋅+−×⋅=+ 105.41061.0105.1 JJ 2 2 ⎝ Ti ⎠