Potential Concept Questions

Transcription

1 Potential Concept Questions Phase Diagrams Equations of State Reversible Processes Molecular Basis of Heat Capacity State Functions & Variables Temperature Joules Experiment for conversion of one form of energy into another Equations of State P = F(V,T) V = f(p,t) Ideal gas law van der Waals Equation They relate the P,V,T coordinates of a thermodynamic system Use to predict U,H, S, G & A. 1

2 Ideal Gases 1662 Boyles Law PV = Constant at fixed T 1787 Charles LawΔV T at fixed P 1801 Dalton s Law Pt = Pa + Pb + Pc 1822 Discovery of the Critical State 1834 Discovery of the Ideal Gas Law PV = nrt or PV = nr(t ) Ideal Gas Equation of State PV = nrt Dilute, Volume of Molecules is ~ Zero No Attractions Between Molecules Temperature must be in Absolute Units, K or R R is the Gas Constant a Universal Constant R = cal/(mol K) or L bar/(mol K) Ideal Gas Limit Defines Absolute Temperature scale and the Gas Constant T(K) = t(c) & T(R) - t(f) Defines limit of real world equations of state Useful for many industrial gases (air, methane, noble gases, etc. Molecules take up no volume No interactions between molecules Limit of PV RT as P 0 Plot PV vs P at constant T to get R van der Waals Equation of State P = RT V b a V 2 V 3 - b - RT P V 2 + a P V - ab P = 0 Where a and b can be evaluated from the partials given below : P = 0 & 2 P 2 V m V m = 0 Tc Tc 2

3 van der Waals EOS Cubic Analytic solutions Defined Roots Simple Attractive Forces are represented by (a) Repulsive Forces represented be (b) Predicts a liquid phase Serves as the model for modern EOS What Makes up Heat Capacity? Molecular translation Molecular rotation Molecular vibration Energy stored in bonds Energy stored in energy levels of elections T temperature P pressure V volume X or n composition State Variables These are the coordinates that define a thermodynamic state State Functions for Pure Fluids U = U(T,V) H = H(T,P) S = S(T,V) or S(T,P) G = G(T,P) A = A(T,V) or T = T(S,A)???? System is over specified 3

4 State Functions They are path independent in PVT space ie. They only depend on the initial and final values of PV&T. They are continuous functions in PVT space therefore we can write them as total derivatives. Because they are path independent we can pick any path we desire going from inlet to outlet State Functions H(T, P) dh = dp + H(T, P) dt Abstract P T H(T, P) T H(T, P) P P T T = C P = V - T V T dh = C P dt + V - T V dp Real World T P P P Reversible Processes ΔS = 0, reversible ΔS > 0, irreversible ΔS < 0, impossible for an isolated system Isolated system - no mass or heat transfer Closed system - no mass transfer Open system - both heat and mass transfer allowed ΔS Q Rev T 4

5 Temperature (T) Related to molecular motion Related to phase changes (freezing point, boiling point, triple point, change of crystalline structure) Calibration required, usually between fix reference with the assumption of linearity. Temperature Measurement Expansion of a liquid (Mercury) Change of metal - metal emf (thermocouple) Change of electrical resistance (PRT) All use F, C, R or K temperature scales Joules Experiment & Q as a Form of Energy Describe the experimental system Demonstrates the conversion of mechanical work to heat. The change in T is proportional to the amount of mechanical work ΔU = U initial -U final ΔU = w for an adiabatic process 5

6 Example The vessel in the last slide contains 1.00 kg of water and is heated 1.82 C by a 100 kg weight falling in a Gravitational field. C pwater = 4,184 J/(kg K), how far the The weight have to fall? ΔU & ΔH ΔU = Q - W - ΔE p - ΔE K ΔU = Q - ΔE P - ΔE k -W SH -W pv ΔH = Q - ΔE P - ΔE k -W SH Comparison of ΔH & ΔU Analytical Problems At 1 atm Air C Water 20-30C Δ(pv) = ΔH - ΔU 2,850 J/kg 0.1 J/kg ΔU 7,170 J/kg 41,840 J/kg ΔH 10,020 J/kg 41,840 J/kg Applying Equations of State Conversion of temperature scales Thermal expansion of a solid or volumetric expansion of a liquid The flow of heat Thermal resistance Iron C J/kg 4,494 J/kg 4,494 J/kg 6

7 Thermal Expansion α 1 dl L dt α is the coefficient of thermal expansion 1 dv β V dt β is the coefficient of volume expansion Heat Flow Thermal Resistance Δ T = T 1 - T 2 Consider a beam of length L and crossection A Δ Q A Δ T Δ t L If we look at a small length of beam x dq dt = -κ A dt dx κ = the thermal conductivity kw ( ) m K R L κ A high R value means good insulation w ft 2 h o F or m2 K BTU W ΔQ Δt = 1 R AΔT 7

8 1 R eff = Thermal Resistance R eff = R 1 + R 2 + etc. 1 A 1 + A 2 A 1 + A 2 R 1 R 2 Example A 300 cm 3 glass is filled with 100 g of ice and 200g of water at 25 C. A) characterize the contents of the glass after equilibrium has been established. B) Repeat the calculation for 50 g of ice and 250 g of water. Solution Example Latent heat of ice is 80 cal/g, if all the ice melts it will require 8,000 cal. If water goes to the freezing point it will require 25 x 200 = 5,000 cal therefore we have an ice/water mixture at 0C, with 3,000/80 = 37.5 g of Ice remaining. For part B 50 x x (T - 0) = 250 (25 -T) 300T = 6,250-4,000 T = 7.5 C A container holds a gas at a pressure of 1.0 atm and a temperature of 300K. Half of the gas leaks out while the temperature is being raised to 340 K. What is the pressure in the container? For the two states of the gas in the container we can write p 1 V = n 1 RT 1 and p 2 V = n 2 RT 2, which can be combined to give (p 2 /p 1 ) = (n 2 /n 1 )(T 2 /T 1 ); (p 2 /1 atm) = (0.5)(340 K/300 K), which gives p 2 = 0.57 atm. 8