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Transcript
Week 1 Wednesday September 26, 2012 page 1 Thermodynamic properties: composition, internal energy, volume, pressure, temperature, entropy, etc Properties: 1. Extensive a. Depends on size b. Larger system means larger value c. Volume is an example 2. Intensive a. Density is an example b. More important than extensive c. Pressure and temperature are examples Always think about whether a quantity is intensive or extensive. Systems: 1. Homogeneous a. Uniform, 1 phase b. Refers to mixture (more than one substance) 2. Heterogeneous a. More than one phase (like oil and water) State: We need at least two intensive properties to define the state of a pure substance. Temperature: ο· ο· Most important intensive quantity Canβt measure directly (thermometer shows volume, not temperature) We can skip section 1.3 Gas Laws Equation of state for ideal gas (perfect gas): PV=nRT πΏ ππ‘π πππ π½ ππ3 ππ‘π π = .08206 = 1.987 = 8.314 = 82.06 πππ πΎ πππ πΎ πππ πΎ πππ πΎ So L atm is a unit of energy. R is universal gas constant. PV=nRT is true at low pressure and high temperature only. ο· Ignores intermolecular forces since molecules are so far apart ο· ο· Assumes molecules take up no space Have to memorize Van der Waals Equation (chapter 8) π= ππ π ππ2 β 2 π β ππ π a and b are both gas specific constants. The pressure correction (a term) is subtracted because intermolecular forces are always attractions, which reduces the pressure on the wall. The b term is due to the size of the molecules. (π + ππ2 ) (π β ππ) = ππ π π2 We donβt have to memorize the Van der Waals equation, but itβs important. Symbols can have more than one use or meaning to pay attention to context. We donβt have to memorize every equation. PVT behavior in condense phases (solid and liquid) For any substance, V=f(T,P) assuming n is constant. dV is change in volume d is differential (infinitesimal change) ππ = ( ππ ππ ) ππ + ( ) ππ ππ P ππ π dV is a total differential The P in the first term means constant pressure. The T in the second term means constant temperature. Divide that equation by V. ππ 1 ππ 1 ππ = ( ) ππ + ( ) ππ π π ππ π π ππ π The equation is now intensive. The ratio of two extensive quantities is intensive. π= π π£ m is extensive, v is extensive, π is intensive 1 ππ πΌ = π (ππ ) π alpha 1 ππ βπΎ = π (ππ) kappa π πΌ is the coefficient of thermal expansion or thermal expansivity, or cubic expansion coefficient. K is isothermal compressibility. Isothermal means constant temperature. πΎ= β1 ππ ( ) π ππ π Constant temperature 1 ππ πΌ = ( ) constant pressure π ππ π ππ = πΎ= π = πππππ π£πππ’ππ π β1 πππ 1 πππ ( ) πππ πΌ = ( ) ππ ππ π ππ ππ π πππ‘πππ ππ£π = πππ‘πππ ππ£π πππ‘πππ ππ£π πΌ, πΎ are both intensive. πΌ: 10-5 to 10-4 K-1 for solids 10-3.5 to 10-3 K-1 for liquids 10-3 to 10-2 K-1 for gasses K: 10-6 to 10-5 atm-1 for solids 10-4 to ? atm-1 for liquids .1 to 1 atm-1 for gasses Liquid is the most complicated phase. K=f(T,P) πΌ = π(π, π) Example: Calculate the volume change that occurs when 50cm3 of argon, treated as an ideal gas, is heated though 5.0K at 298K. ππ ππ ππ = ( ) ππ + ( ) ππ π ππ ππ π ππ πβπ ( ) ππ π‘πππ = 0 π ππππ π ππ ππππ π‘πππ‘. ππ π βV is a finite change. ππ βπ β ( ) βπ β πΌππ ππ π What is Ξ± for an ideal gas? πΌ= 1 ππ 1 π ππ π 1 ππ 1 ( ) = ( ( )) = = π π π π ππ π π ππ π π βπ β End of chapter 1 1 50ππ3 β 5πΎ πβπ β β .84ππ3 π 298πΎ