Chem 340 - Fall 2013 – Lecture Notes 2 - Thermodynamic Concepts

Thermodynamics – macro scale, seems old fashioned, but practical Build a chemical plant, want to know if get product, if yield will make profit Develop a drug, want to know if it will bind to target enzyme Study metabolism, concerned with how much energy to make a process go, e.g. ATP driven ion transport through membranes, helps you think in class!

Statistical Thermodynamics, does this on microscale, atoms and molecules moving and interacting, and then averages/analyzes many steps (trajectories) to get properties of macro system, but more complex than simpler goals above, macro scale general

What do we study? - a system and its surroundings, in principle that is everything Open system – exchange matter with surrounding, but Closed cannot exchange Isolated system cannot exchange matter or energy with surroundings (e.g. sealed thermos container) Interface is a boundary, often a barrier or wall System could be the earth, an island or a cell, depends on interest, various boundaries, each system (atmosphere; solid-water; membrane) State of a system defined by its properties, readily observed, for us most often temperature, pressure, volume, number of particles and components, T,P,V,n,. . .

Equilibrium means there is no dynamic change in a property value with time and across the system, typically consider pressure, temperature and density: P, T, n/V e.g. soap bubble, P inside and out is same, expand to match, but n/V is not Temperature indirect, must measure impact on something else and calibrate it Ideal gas law (book terms this perfect gas): PV = nRT or P = RT where  = n/V  T = P/R Measuring pressure and density can gives a scale for T, and provides definition of thermal equilibrium, same pressure and density, or ratio of P/(since R constant)

While we use SI, need to be aware of other units to communicate and read reports

Containers affect equilibrium Adiabatic boundary allows no (heat) energy flow, bring two systems together, temperatures not equilibrate (if same , different P– values do not change, Imagine contacting styrofoam cups, hot and cold coffee, each stays same) Diathermal boundary lets heat flow, temperatures come to equilibrium (now put coffee in metal cup, and set it on cold table - cools)

Adiabatic Diathermal

Zeroeth law of Thermodynamics: Two systems that are separately in thermal equilibrium with a third system are also in thermal equilibrium with one another (logically before other three, but came after, can measure temperature, thermometer)

General observable, can sense relative hotness or coldness of objects, and experience shows if bring them together can come to the same level (equilibrium)

Temperature scales Thermometer best if some property changes linearly with temperature: T = f(P,V) e.g. Pressure ideal gas or volume of mercury (viewed in capillary)

Centigrade scale developed by letting 0oC be freezing point of water, 100oC, boiling pt Celsius a bit refined, 0.01oC is triple point water (ice, vapor and liquid in equilibrium) Water boil at 99.975oC under 1 atm – but values pretty arbitrary Absolute (thermodynamic) temperature scale defined based on ideal gas behavior Measure P vs. T at constant V and n (closed system - , extrapolate to P = 0

Ideal gas extrapolate to P = 0 at T = 0K Real gases deviate at Ttp with P > 0

Experimentally, all extrapolated plots (different V) intersect at a low temp, linear: . P = a + b T - where a and b determined from fit of data to linear function, T- oC Intersection defines 0 (zero) for new scale, degree step chosen same as Celsius o Kelvin scale: T(K) = T( C) + 273.15 for gas thermometer: T(K) = 273.16 P/Ptp

Ideal gas Equation of State (text uses p for P) – text uses “perfect gas” no interactions pV = nRT summarizes observables named a. Boyle’s Law: pV = const, fixed n, T b. Charles’ Law: V/T = const, fix n, p and p/T = const, fix n, V (or ) c. Avogadro’s principle:  = const, fix T, p

R – gas constant, is proportionality for p,V,T at fixed n, note nR = Nk, where N = nNA R = 0.08206 L atm mol-1 K-1 units for gas law, also energy R = 8.314 J mol-1 K-1

Isotherms plot p vs. V for constant T, n get set of hyperbolas: p = const/V - line across, shows T and V for const p - vertical line, shows p and T for const V p and T independent of amount of gas Intensive variable n and V depend on how much gas Extensive variable Ratio them:  = n/V, density ( ) is intensive Only two independent intensive variables p = RT may confuse, use: P = RT

Alternate (old) units 1 J = 4.184 cal so R = 1.987 cal mol-1 K-1

Example: 1 L blood can carry the equivalent of

0.20 L O2 measured at T=273K and p=1.0 atm. The average person has 5 L blood and it circulates in about 1 min. How many moles of O2 can be transported per min. (max)?

-1 -1 n = pV/RT = 1.0 atm x 0.20 L (x 5)/(0.082 L atm mol K x 273K) = 0.045 mole O2/min

Example: imagine a bicycle pump, starts at 1 atm and 1 L compress to 0.2 L, T = 25oC

Final pressure: p2 = p1V1/V2 = (1 atm x 1 L)/0.2 L = 5 atm (Boyle Law method) o If friction heated to 27 C: p2 = (p1V1/V2)(T2/T1) = 5 atm(300 K/298 K) = 5.03 atm How much gas? n = pV/RT = (1atm 1L)/(0.082 L.atm.mol-1K-1 298K) = 0.042 mol

Ideal mixed gases – no interaction, contributions to pV independent (sum over i) -1 pV =  niRT or for one V, partial pressure (pi): p = V  niRT =  pi = p1+p2+p3 . . .

Mole fraction (xi) related to partial pressure of ideal gases (V and T set):

pi/p = (niRT/V)/(nRT/V) = ni/n = xi where n =  ni

Example: Consider system above, at T=298 K, compute final pressure and partial pressures of each gas if barriers removed, assume ideal gas and barrier has no volume: -1 -1 nHe = pV/RT = (1.50 bar x 2.00 L)/(8.314 L bar K mol x 298 K) = 0.121 mol -1 -1 nNe = pV/RT = (2.50 bar x 3.00 L)/(8.314 L bar K mol x 298 K) = 0.303 mol -1 -1 nXe = pV/RT = (1.00 bar x 1.00 L)/(8.314 L bar K mol x 298 K) = 0.0403 mol sum these for total moles: n = nHe + nNe + nXe = 0.464 , remove walls, total pressure, pt: -1 -1 pt = (nHe + nNe + nXe)RT/V = (0.464 mol x 8.314 L bar mol K x 298 K)/6.0 L = 1.92 bar and mole fractions: and partial pressures:

xHe = nHe/n = 0.261 pHe = nHe p = 0.261 x 1.92 bar = 0.501 bar xNe = nNe/n = 0.653 pNe = nNe p = 0.653 x 1.92 bar = 1.25 bar xXe = nXe/n = 0.086 pXe = nXe p = 0.086 x 1.92 bar = 0.165 bar p. 28-29, Text has nice discussion of atmosphere as mix of gases, have a read

Real gases – interact and have molecular volume Compare energy - potential – V(r ) – very close, repel, overlap non-zero volumes - Separate further attract, van der Waals - V(r) has minimum (distance for max attract) - large separations, approach ideal (p ~ 0) - big V - molecule sizes become negligible - large distance, electrons don’t “feel” other

- Transition to ideal at |V(rtrans)| ~ kT

Attraction due to electron response to electrons in other molecules, fluctuating dipole moments Polarizability property, varies by molecule Effect: pressure is lowered at attractive distances and raised at very short (repulsive) distance Result—low T, high p, get high   gives minimum V (liq,sol), but if ideal V  0 van der Waals equation of state: p = nRT/(V-nb) – a(n2/V2) where parameters are for: a, attractive interaction, and b, finite molecular size

5 a,b vary with molecular properties, see Table 1.6 (text), fit to experimental observable e.g. He much less than Ar, Xe due to fewer (less polarizable) electrons and smaller

-1 Example: Calculate pressure of CO2, T = 300 K for molar volumes of 250, 0.10 L mol -1 3 (recall  = n/V, so  = molar vol, and L = dm , for problems in molar vol, neff = 1) p = RT/(-1-b) – a(2) = (0.0802 atm mol-1 K-1 x 300 K)/(0.1 – 0.0429) L mol-1 - 3.610 atm L2 mol-2 /(0.1 L mol-1)2 = 421 – 360 = 61 atm p = RT/(-1-b) – a(2) = (0.0802 atm mol-1 K-1 x 300 K)/(250 – 0.0429) L mol-1 - 3.610 atm L2 mol-2 /(250 L mol-1)2 = 0.09624 – 5.8x10-5 = 0.0962 atm Ideal: p=RT – 0.1 L mol-1 : 0.0802 atm mol-1K-1x 300 K/0.1 Lmol-1 = 241 atm (higher) 250 Lmol-1: 0.0802 atm mol-1K-1x 300 K/250 Lmol-1 = 0.09624 atm (same)

Related topics Compressibility: if attractive easier to compress, repulsive, harder to compress

Compression factor, Z, compare molar volume (Vm) to o o ideal, V m : Z = Vm/V m = pVm/RT

Virial coefficients, alternate expression gas law: pV = RTZ = RT(1 + B’p + C’p2 +. . .) 2 pV = RT(1 + B/Vm +C/Vm + . . . ) in PChem, often expand real relationship in form like ideal measure by compare to compressibility

Condensation: at some p,T sudden volume loss with no increase in p, liquid drops form (CDE in diagram 1.15) C—all vapor, E - all liquid, D – mix, separate phases

Critical constants: at Tc, cannot distinguish liquid – vapor isotherm touches blue (condensed phase) area Top of blue area, where two phase lines converge

Practice - see attached example problems, web site – partial pressure, van der Waals