Download presentation
Presentation is loading. Please wait.
Published byJeffry Matthews Modified over 9 years ago
1
1 What is Thermodynamics 1. Understanding why things happens 2. Concerning heat, work, related temperature, pressure, volume and equilibrium 3. Equations relate macroscopic properties
2
2 The laws of thermodynamics Number BasisProperty Zeroth Law Thermal Equilibrium Temperature First LawRelation between work, energy and heat Internal Energy U Second Law Spontaneous process Entropy Third LawAbsolute Zero Degree of Temperature Entropy S 0 as T 0 Kelvin
3
3 Study of heat engines Being studied by all students in physical science and engineering
4
4
5
5 Concept of State
6
6 From Avogadro’s hypothesis the volume per mole of all ideal gases at 0 o C and 1atm pressure is 22.414 litres.
7
7 For n mole gas PV=nRT
8
8 For water vapour, the number of moles for Kg water is obtained by
9
9 Thermodynamics Process Work and Energy Heat
10
10
11
11 1.Open system- material and energy exchange 2.Closed system- energy exchange only 3.Isolated system- no material and energy exchange
12
12 What we learn from this module? 1.Internal energy U and entropy S 2.Combining U and S with P, T and V gives enthalpy H=U + PV and Gibbs energy G=H-TS 3. H is related to heat adsorption or release at constant pressure 4. G controls the position of equilibrium in closed systems at constant temperature and pressure.
13
13 Why is Thermodynamics useful? 1.Qualitative explanation of materials behaviour 2.Quantitatively understanding of materials status. 3. Physical significance of thermodynamic functions.
14
14 Applications of Thermodynamics 1.Extraction, refining 2.Corrosion 3.Phase transformation-phase diagram calculation 4.Materials processing 5.Design of new materials.
15
15 The First Law of Thermodynamics Conservation of Energy Principle Same principle in mechanics, physics and chemistry
16
16
17
17
18
18
19
19 Work done in an Expansion (or contraction) against an External Pressure
20
20 Expansion against a constant external pressure
21
21 W 12 Q 12 Reversible process
22
22 Reversible process and Maximum Work Reversible process for a closed system W system-environment =W environment-system Q environment-system =Q system-environment
23
23
24
24 For an ideal gas For isothermal process, ie. T=constant
25
25 Questions*: 1.How to calculate W for a constant pressure process? 2.How to calculate W for an isothermal reversible process? 3.Is a constant pressure process an reversible process? Explain why? * All of these questions are concerned with ideal gas systems.
26
26 Example: Gas compress during quasi-equilibrium processing, with PV=constant. The system is the gas P 1 =101325 N/m 2, V 1 =0.01m 3, V 2 =0.005m 3 Find W W=-702J
27
27 Enthalpy U=q-w · Q Heat is transferred due to the presence of a temperature difference. · Work here is considered as the work of expansion. · U results from the oscillation of atoms or ions in solid and movement of the particles in gas and liquid. · Q and w depend on how the change is carried out where the difference between them does not. · At constant volume, w=0 and U=q
28
28 The Enthalpy Function When P=constant
29
29 At constant pressure, the change in enthalpy is equal to the heat The change of enthalpy is independent of path. Q: Does q or W depend on path? For the change involving solids and liquids, H U, but for gases, H U Q:explain why?
30
30 Example 1: Given p=constant=101.3 kPa V 1 =1m 3, V 2 =2m 3 Q 12 =200kJ Find a) U b) an expression for Q 12 in terms of thermodynamics properties for a quasi- equilibrium process.
31
31 Example 2. Given T=T 1 =T 2 =constant for the process P 1 =200 kPa, T 1 =300K V 1 =2m 3, V 2 =4m 3 Find a) W and b) Q W=277KJ
32
32 Heat Capacities (C p and C v ) a) Under constant volume conditions Cv- all heat supplied increases energy of sample b) Under constant pressure conditions Cp- Heat supplied increases energy of sample and provides energy for work performed.
33
33 Relation between Cv and U The 1 st Law When V=constant Therefore
34
34 For n moles of materials over small ranges in temperature Cv constant
35
35 Relation between Cp and H At constant pressure Over small range of T for n moles of materials
36
36 Summary At constant volume At constant pressure Molar heat capacity at constant pressure Molar heat capacity at constant volume
37
37 Questions: 1.For a constant temperature process of an ideal gas, prove H= U. 2.For a gas system, explain why Cp is larger than Cv? 3.For a solid/liquid system, explain why Cp is close to Cv? 4.What are the equations for calculating change of enthalpy and internal energy due to temperature change?
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.