PPT 107 PHYSICAL CHEMISTRY

PPT 107 PHYSICAL CHEMISTRY Semester 2 Academic session 2012/2013

Chapter 3The Second Law of Thermodynamics

Content 1. The Second Law of Thermodynamics 2. Heat Engines. 3. Entropy. 4. Calculation of Entropy Changes. 5. Entropy, Reversibility, and Irreversibility.

A major application of thermodynamics to chemistry is to provide information about equilibrium in chemical systems. If we mix nitrogen and hydrogen gases together with a catalyst, portions of each gas react to form ammonia. The first law assures us that the total energy of system plus surroundings remains constant during the reaction, but the first law cannot say what the final equilibrium concentrations will be. We shall see that the second law provides such information. The second law leads to the existence of the state function entropy S, which possesses the property that for an isolated system the equilibrium position corresponds to maximum entropy.

1st Law: energy is conserved • But is that enough ? • Object drops converting KE to heat but never see the opposite • H2 and O2 react to form H2O when ignited at room temperature but not the reverse • 1st Law would permit the reverse but 2nd Law does not.

Energy is both a molecular and a macroscopic property and plays a key role in both quantum chemistry and thermodynamics. Entropy is a macroscopic property but is not a molecular property. A single molecule does not have an entropy. Only a collection of a large number of molecules can be assigned an entropy.

THE SECOND LAW OF THERMODYNAMICS For a heat engine to produce continuous mechanical work, it must exchange heat with two bodies at different temperatures, absorbing heat from the hot body and discarding heat to the cold body “Carnot principle”. Without a cold body for the discard of heat, the engine cannot function continuously. This is the essential idea of one form of the second law of thermodynamics.

Kelvin-Planck Statement It is impossiblefor a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of anequivalent amount of work by the system on the surroundings.

Heat (q) Cyclic machine (machine) Heat Reservoir A system that violates the second law of thermodynamics but not the first law. Work done by the system = q

The Second Law Follows the consequence system of certain assumption about the molecule level. Failure of anyone to construct a machine (cyclic machine) Leads to many conclusion have been verified eg heat vaporization, volume change in vaporization etc.

First Law Second Law Impossible to have a cyclic machine that completely converted the random molecular energy of heat flow into the ordered motion of mechanical work Work output cannot be produced by a cyclic machine Without an equivalent amount of energy balance.

Clausius Statement It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a cold reservoir and the flow of an equal amount of heat out of the system into a hot reservoir.

Direction of Thermodynamic Process

Reversible • Internally reversible process The internally reversible process which is once having taken place, can bereversedand in so doing leave no change in the system. This says nothing about what happens to the surroundings about the system. • Totally or externally reversible process The externally reversible process which is, once having taken place, can be reversedand in so doing leave no change in the system or surroundings.

Irreversible • All real processes are irreversible. • Irreversible processes occur because of the following: • Friction • Unrestrained expansion of gases • Heat transfer through a finite temperature difference • Mixing of two different substances

Heat Engines We shall use the second law to deduce theorems about the efficiency of heat engines. The study of the efficiency of heat engines is related to the basic question of what limitations exist on the conversion of heat to work. A heat engine converts some of the random molecular energy of heat flow into macroscopic mechanical energy (work).

Heat Engine Device that transform heat partly into work or mechanical energy Matters inside the engine undergoes inflow and outflow of heat, expansion, compression and sometimes changes of phase. The simplest process is cyclic process

The working substance (for example, steam in a steam engine) is heated in a cylinder, and its expansion moves a piston, thereby doing mechanical work. If the engine is to operate continuously, the working substance has to be cooled back to its original state and the piston has to return to its original position before we can heat the working substance again and get another work-producing expansion. Hence the working substance undergoes a cyclic process.

The essentials of the cycle are: The absorption of heat qH by the working substance from a hot body (for example, the boiler). The performance of work -w by the working substance on the surroundings. The emission of heat -qC by the working substance to a cold body (for example, the condenser)

Our convention is that w is work done on the system. • Work done by the system is -w. • qh means the heat flowing into the system. • -qC is the heat that flows fromthe system to the cold body in one cycle. • For a heat engine, qH>0, -w>0, and –qC>0, so w<0and qC<0. • The quantity w is negative for a heat engine because the engine does positive work on its surroundings; qC is negative for a heat engine as positive heat flows out of the system to the cold body.

e = efficiency of heat engine ∆U=0=q+w=qh+qc+w For a cycle of operation The carnot efficiency Since qC is negative and qH is positive, the efficiency is less than 1. To further simplify the analysis, we assume that the heat qH is absorbed from a hot reservoir and that -qC is emitted to a cold reservoir, each reservoir being large enough to ensure that its temperature is unchanged by interaction with the engine.

The Carnot’s Principle No heat engine can be more efficient than a reversible heat engine when both engine work between the same pair of temperature THand TC.

Hot Reservoir Revers Reversible Engine Super engine Net Work output Cold Reservoir

Now consider two reversible heat engines, A and B, that work between the same two temperatures with efficiencies erev,A and erev,B. If we replace the superengine in the above reasoning with engine A running forward and use engine B running backward as the heat pump, the same reasoning that led to previous result as it gives erev,A ≤ erev,B. If we now interchange A and B, running B forward and A backward, the same reasoning gives erev,B ≤ erev,A. These two relations can hold only if erev,B = erev,A.

We have shown that (1) all reversible heat engines operating between the same two temperatures have the same efficiency erev, and (2) this reversible efficiency is the maximum possible for any heat engine that operates between these temperatures, so: eirrev ≤ erev

Calculation of erev Since the efficiency of any reversible engine working between the temperatures τH and τC is the same, this efficiency erev can depend only on τH and τC: erev = ʄ(τH, τC)

Consider first the nature of the cycle we used to derive the previous relation. The first step involves absorption of heat qH from a reservoir whose temperature remains at TH. Since we are considering a reversible engine, the gas also must remain at temperature TH throughout the heat absorption from the reservoir. (Heat flow between two bodies with a finite difference in temperature is an irreversible process.) Thus the first step of the cycle is an isothermalprocess. Moreover, since ∆U=0for an isothermal process in a perfect gas, it follows that, to maintain U as constant, the gas must expand and do work on the surroundings equal to the heat absorbed in the first step. The first step of the cycle is thus a reversible isothermal expansion, as shown by the line from state 1 to state 2 in the Fig Similarly, when the gas gives up heat at TC, we have a reversible isothermal compression at temperature TC. The TC isotherm lies below the TH isotherm and is the line from state 3 to state 4 in the Fig. To have a complete cycle, we must have steps that connect states 2 and 3 and states 4 and 1. We assumed that heat is transferred only at TH and TC. Therefore the two isotherms in the Fig. must be connected by two steps with no heat transfer, that is, by two reversible adiabats.

Carnot cycleis defined as a reversible cycle that consists of two isothermal steps at different temperatures and two adiabatic steps.

The second integral on the right side of the general relation must also vanish. This is because dV/V is the differential of a state function (namely, lnV), and its line integral is therefore zero for a cyclic process.

ENTROPY Clausiusdiscovered the state function S in 1854 and called it the transformation content (Verwandlungsinhalt). Later, he renamed it entropy, from the Greek word trope, meaning “transformation,” since S is related to the transformation of heat to work. Unit used: (J / K)

We now extend the result of [(dqrev / T) is equal to zero] to an arbitrary reversible cycle, removing the constraint that heat be exchanged with the surroundings only at TH and TC. This will then show that (dqrev / T) is the differential of a state function (S).

* The curve in the Fig. depicts an arbitrary reversible cyclic process. * We draw reversible adiabats (shown as dashed lines) that divide the cycle into adjacent strips. * Consider one such strip, bounded by curves ab at top and cd at the bottom. * We draw the reversible isotherm mn such that the area under the zigzag curve amnb equals the area under the smooth curve ab. * Since these areas give the work w: we have wamnb= wab, where ab is the process along the smooth curve and amnb is the zigzag process along the two adiabats and the isotherm.

* ∆U is independent of the path from a to b. so ∆Uamnb = ∆Uab. * As ∆U = q + w, this will lead to qamnb = qab. * Since am and nb are adiabats, we have qamnb = qmn. Hence qmn = qab. * Similarly, we draw the reversible isotherm rs such that qrs = qcd. Since mn and rs. are reversible isotherms and ns and rm are reversible adiabats, we could use these four curves to carry out a Carnot cycle; (qmn/Tmn) + (qsr/Tsr) = 0 and (qab/Tmn) + (qdc/Tsr) = 0

Now consider the limit as we draw the adiabats closer and closer together in The Fig., ultimately dividing the cycle into an infinite number of infinitesimally narrow strips, in each of which we draw the zigzags at the top and bottom. As the adiabatbd comes closer to the adiabatac. This will lead to :Tmn = Tab and Tsr = Tdc (qab/Tab) + (qdc/Tdc) = 0

The sum of the infinitesimals is a line integral around the cycle, and we get: Since the integral of dqrev/T around any reversible cycle is zero, so that the value of the line integral ∫dqrev/T is independent of the path between states 1 and 2 and depends only on the initial and final states. Hence dqrev/T is the differential of a state function. This state function is called the entropyS

CALCULATION OF ENTROPY CHANGES to calculate ∆S for any process: Identify the initial and final states 1and 2. Devise a convenient reversible path from 1 to 2. Calculate ∆S from [∆S = ∫ dqrev / T ]

Let us calculate ∆S for some processes. Note that, as before, all state functions refer to the system, and ∆S means ∆Ssyst , and does not include any entropy changes that may occur in the surroundings. Cyclic process. Since S is a state function, ∆S = 0 for every cyclic process. Reversible adiabatic process. Here dqrev = 0; therefore ∆S = 0rev. ad. proc.

Reversible phase change at constant T and P: where, since T is constant, we took 1/T outside the integral. qrev is the latent heat of the transition. Since P is constant, qrev = qP= ∆H Reversible isothermal process. Here T is constant:

Constant-pressure heating with no phase change.First, suppose the heating is done reversibly. At constant pressure (provided no phase change occurs), dqrev = dqP = CPdT. The relation [∆S = ∫ dqrev / T] becomes: If CP is essentially constant over the temperature range, then ∆S=CPln(T2/T1) Reversible change of state of a perfect gas. From the first law we have for a reversible process in a perfect gas

Summary To calculate ∆S ≡ S2 - S1, we devise a reversible path from state 1 to state 2 and we use ∆S = ∫ (1/T)dqrev. If T is constant, then ∆S = qrev/T. If T is not constant, we use an expression for dqrev to evaluate the integral; for example, dqrev= CPdT for a constant-pressure process or dqrev= dU - dwrev = CVdT + (nRT/V)dVfor a perfect gas.

ENTROPY, REVERSIBILITY, AND IRREVERSIBILITY The total entropy change that occurs in any process; is the sum of the entropy changes in the system and the surroundings: ∆Ssyst+ ∆Ssurr We call this sum the entropy change of the universe ∆Suniv = ∆Ssyst + ∆Ssurr

Reversible Processes In a reversible process, any heat flow between system and surroundings must occur with no finite temperature difference; otherwise the heat flow would be irreversible. Let dqrev be the heat flow into the system from the surroundings during an infinitesimal part of the reversible process. The corresponding heat flow into the surroundings is -dqrev . We have Integration gives ∆Suniv = 0 rev. proc. Although Ssyst and Ssurr may both change in a reversible process, Ssyst+ Ssurr = Sunivis unchanged in a reversible process.

Irreversible Processes We first consider the special case of an adiabatic irreversibleprocess in a closed system. Let the system go from state 1 to state 2 in an irreversible adiabatic process. The disconnected arrowheads from 1 to 2 indicate the irreversibility and the fact that an irreversible process cannot in general be plotted on a P-V diagram since it usually involves non-equilibrium states.

PPT 107 PHYSICAL CHEMISTRY