P2P2P2P2 PfPfPfPf P=nRT/V Work in an Irreversible Isothermal Expansion [T constant] V2V2V2V2 VfVfVfVf Pressure Volume PiPiPiPi Path 2: w2= -P2 [V2-Vi] -Pf [Vf-V2] ViViViVi Path 2: PiP2Pf stoping at an intermediate pressure P2 with ViV2Vf Path 1: PiPf at start with ViVf Path 1: w1= -Pf [Vf-Vi] Clearly, w2 > w1  wwww w2 = w1 +w -w 1

PiPiPiPi PfPfPfPf P=nRT/V Work in an Isothermal Expansion: Approaching Reversibility ViViViVi VfVfVfVf Pressure Volume Path 1: PiPf at start with ViVf Path 1: w1= -Pf [Vf-Vi] Path 2: PiP2P3 …….  Pf with ViV2V3   …….  Vf Clearly, w2 > w1  Path 2: w2= -P2 [V2-Vi] -P3 [V3-V2] ………... -Pf [Vf-Vf-1]

PiPiPiPi PfPfPfPf P=nRT/V Work in an Isothermal Expansion: Approaching Reversibility ViViViVi VfVfVfVf Pressure Volume Path 1: PiPf at start with ViVf Path 1: w1= -Pf [Vf-Vi] Path 2: PiP2P3 …….  Pf with ViV2V3   …….  Vf Clearly, w2 > w1  Path 2: w2= -P2 [V2-Vi] -P3 [V3-V2] ………... -Pf [Vf-Vf-1]

“True” Reversible, Isothermal Expansion of an Ideal Gas w rev = - nRT ln p gas = p ext p gas = - p ext = p gas = nRT/V Maximum possible work is done by a gas in a reversible expansion. -w rev

P ex = P 2 and suddenly drop pressure to: so fast that gas doesn’t expand  nRT v2v2v2v2 GasGas P1P1P1P1 P P2P2P2P2 P Then expansion takes place with Pex = P2 = Quantitative comparison of reversible and irreversible work: Reversible work:

Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

Preliminary statement: There exists a state function called entropy whose change is given by ∆S = = dq rev Here dqrev is an infinitesemal change in the heat added for a reversible process, T = Kelvin temperature. (1) Choose a reversible path (3) Divide dqrev by T at each point Entropy and the Second Law

If qrev were an ordinary well defined variable (state function) like P, V could think of it this way: q rev State 1 State 2 1T (graph is misleading because q is not an ordinary variable!) Thus, over any path between the same initial and final states (reversible or irreversible) ∆S has the same value. 1/T ∆S for specific reversible path in qrev space

The Second Law The simplest way to state the second law is for an isolated system. This is a system for which dq = 0, and dw = 0. ∆S = 0for a reversible change ∆S > 0 for an irreversible change (corollary : ∆S < 0 is impossible for an isolated system).

The word spontaneous is used interchangeably with irreversible to describe processes that are observed in nature to "proceed on their own". Since we can always combine the system together with surroundings to form an isolated system which is the universe, we can restate the 2nd Law: ∆S (universe) < 0 impossible (never observed)

Entropy Calculations 1) Cardinal Rule: Can only calculate entropy change if choose a reversible path. 3) ∆S for isothermal process (reversible) T = const

dqrev = sum of heat changes over path. Total is qrev or heat added going from state 1 to 2 over reversible path. For an ideal monatomic gas E = (3/2) RT ∆E = q + w = 0  w = - q For an ideal gas,P = In a reversible process, P=P ext and V is not constant.

∆S = nR ln (V2 / V1 )Ideal gas isothermal change. Note this last case is NOT inconsistent with the 2nd law! System entropy can decrease as long as entropy of universe either increases or stays the same. If system entropy decreases, entropy of surroundings must increase. As we shall see later, the fact that we need to include both the system and surroundings when considering the Second Law makes the ∆S function “inconvenient”.

Temperature Dependence of Entropy For constant pressure process For infinitesimal change dqrev = = = = nCp dT [Now assume Cp independent of T] ∆Sp = = = = nCp ln (T2 / T1) Assumes Cp, Cv are constant over range T1  T2 ∆S =

For an ideal gasCp = (3/2)R + R = (5/2)R Absolute Entropies and the Third Law In same way can set up an arbitrary scale for ∆Sf. As it turns out, can also set up an absolute scale for entropy. Note that ∆Sp, ∆Sv are not the same. Cannot be between same initial and final states! (T1, T2 can be same but P1, P2 and V1, V2 must be different).

Third Law of Thermodynamics S298°  Standard absolute entropy: TF = 298 K and P = 1 atm. But S =  [dqrev/T] ∆S = - S 0 = - 0 = Note: Really cannot take Cp independent of T here because are going from absolute zero to some finite T. Most substances go from solid to liquid to gas over this range! dqrev=CpdT