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V ∝ 1/P, Boyle's Law V ∝ T, Charles's Law V ∝ N, Avogadro's Law Therefore, V ∝ Nt/P Or PV ∝ Nt

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V ∝ 1/P, Boyle's Law V ∝ T, Charles's Law V ∝ N, Avogadro's Law Therefore, V ∝ Nt/P Or PV ∝ Nt Ideal Gas Law Deduced from Combination of Gas Relationships: V ∝ 1/P, Boyle's Law V ∝ T, Charles's Law V ∝ n, Avogadro's Law Therefore, V ∝ nT/P or PV ∝ nT PV = nRT where R = universal gas constant The empirical Equation of State for an Ideal Gas Boyle’s Law (experimental) PV = constant (hyperbolae) T4 Is T4 > or < T1 ?? Temp T1 Ideal Gas Equation of State Ideal Gas Law PV = nRT where R = universal gas constant R = PV/nT R = 0.0821 atm L mol–1 K–1 R = 0.0821 atm dm3 mol–1 K–1 R = 8.314 J mol–1 K–1 (SI unit) Standard molar volume = 22.4 L mol–1 at 0°C and 1 atm Real gases approach ideal gas behavior at low P & high T General Principle!! Energy is distributed among accessible configurations in a random process. The ergodic hypothesis → Consider fixed total energy with multiple particles and various possible energies for the particles. → Determine the distribution that occupies the largest portion of the available “Phase Space.” That is the observed distribution. Energy Randomness is the basis* of an exponential distribution of occupied energy levels n(E) ∝ A exp[-E/<E>] Average Energy <E> ~ kBT n(E) ∝ A exp[-E/kBT] This energy distribution is known as the Boltzmann Distribution. * Shown later when we study statistical mechanics. Maxwell Speed Distribution Law 32 1 dN m −mu2 2k T = 4π ue2 B N du 2π kB T 1 dN is the fraction of molecules per unit speed interval N du Maxwell Speed Distribution Law dN 2kT Most probable speed, ump = 0 for u = u u = du mp mp m ∞ 8kT Average speed, <u> or ū u= ∫ u N( u) du = 0 π m ∞ 22 3kT Mean squared speed, <u2> u= ∫ u N( u) du = 0 m 2 3kT Root mean square speed uu= = rms m Distinguish between System & Surroundings Internal Energy Internal Energy (U) is the sum of all potential and kinetic energy for all particles in a system U is a state function Depends only on current state, not on path ∆U = Ufinal - Uinitial Internal Energy, Heat, and Work If heat (q) is absorbed by the system, and work (w) is done on the system, the increase in internal energy (U) is given by: ∆U = q (heat absorbed by the system) + w (work done on the system) Reversible and Irreversible Work Reversible or Irreversible Processes Reversible: carried out through a sequence of equilibrium states Irreversible: anything else! Internal Energy (2) U is a state function It depends only on state, not on path to get there ∆U = Ufinal - Uinitial This means mathematically that dU is an f exact differential: ∆=U∫ dU i For now, consider a system of constant composition. U can then be regarded as a function of V, T and P. Because there is an equation of state relating them, any two are sufficient to characterize U. So we could have U(P,V), U(P,T) or U(V,T). Enthalpy Defined Enthalpy, H ≡ U + PV At Constant P, ∆H = ∆U + P∆V ∆U = q + w q= qP = ∆U - w, w = -P∆V qP = ∆U + P∆V= ∆H At constant V, q = ∆U = ∆H Comparing ∆H and ∆U at constant P ∆H = ∆U + P∆V 1. Reactions that do not involve gases ∆V ≅ 0 and ∆H ≅ ∆U 2. Reactions in which ∆ngas = 0 ∆V ≅ 0 and ∆H ≅ ∆U 3. Reactions in which ∆ngas ≠ 0 ∆V ≠ 0 and ∆H ≠ ∆U Heat Capacity at Constant Volume or Pressure CV = dqV/dT = (∂U/∂T)V → Partial derivative of internal energy with respect to T at constant V CP = dqP/dT = (∂H/∂T)P → Partial derivative of enthalpy with respect to T at constant P → Ideal Gas: CP = CV + nR Heat Capacity, C Heat Capacity (J K-1 ) Heat needed to raise T of system by 1 K q = C∆T Specific Heat Capacity (J K-1 kg-1 ) Heat needed to raise T of 1 kg by 1 K q = CSm∆T Molar Heat Capacity (J K-1 mol-1 ) Heat needed to raise T of 1 mole by 1 K q = Cmn∆T Endothermic & Exothermic Processes ∆H = Hfinal - Hinitial ∆H Positive Positive amount of heat absorbed by the system → Endothermic Process ∆H Negative Negative amount of heat absorbed (i.e. heat released by the system) → Exothermic Process Thermochemical Equations CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) (a combustion reaction) ∆ cH = – 890 kJ Reaction must be balanced Phases must be specified ∆H is an extensive property Sign of ∆H changes when reaction is reversed Standard State The Standard State of an element is defined to be the form in which it is most stable at 25 °C and 1 bar pressure Some Standard States of elements: Hg (l) O2 (g) Cl2 (g) Ag (s) C (graphite) The standard enthalpy of formation (∆fH°) of an element in its standard state is defined to be zero. Enthalpies of Formation The standard enthalpy of formation (∆ f H°) of a compound is the enthalpy change for the formation of one mole of compound from the elements in their standard state. Designated by superscript o, Ø, or o: H°, . For example, CO2: C (graphite) + O2 (g) → CO2 (g) ∆rxnH° = -393.5 kJ/mol Table 2.7 ∆ f H° CO2 (g) = -393.5 kJ/mol Enthalpies of Reaction The enthalpy of reaction can be calculated from the enthalpies of formation of the reactants and products O O ∆rxnH = ∑ ∆fH (Products) O - ∑ ∆fH (Reactants) Example: Find ∆ rxn H° (using Standard Enthalpies of Formation) CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) ∆ f H° (from Tables 2.4&5, text): CH4 (g) -74.6 kJ/mol O2 (g) 0 CO2 (g) -393.5 H2O (l) -285.8 ∆rxnH° = -393.5 -2 (285.8) – 0 – (-74.6) kJ/mol Therefore, ∆rxnH° = -890.5 kJ/mol Exam 1 ~ 5 problems (weights given) – budget your time Closed book Don’t memorize formulas/constants You will be given things you need I would assume that you could derive PV work in an isothermal process. I would not assume that you could derive the Maxwellian distribution. Exam will not be heavily numeric, but will emphasize concepts If a problem seems lengthy, do another problem & come back later Understanding homework will be useful EXAM held in JILA Auditorium, Oct 1, 10 AM.
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