The First Law of Thermodynamics. Lecture 5

Transcription

1 The First Law of Thermodynamics Lecture 5

2 First Law of Thermodynamics Overlooks the fine microscopic details (which in many cases are irrelevant). Describes the conversion of one form of energy (heat) into other forms Important Consequences: Conservation of energy Allows estimates of maximum efficiency of energy conversion Completely General! (not just ideal gas) 1 st Law

3 The Effect of Heat Distributed Energy 1 click

4 Statement of 1 st Law (1842) E int = Q + W E int E int represents the distributed energy in a closed system. It can contain contributions from a variety of sources like thermal, chemical, etc. Sign conventionl

5 The system Universal Appeal! The system can be a collection of matter in any form, separated from its surroundings by real or imaginary boundaries. The system may be a biological cell, the contents of a test tube, the gas in a piston, a thin film, or a cup of coffee. How you define a system depends on the problem you want to solve.

6 When using the 1 st Law, it s important to determine Sign Conventions the proper signs Work (W) has a sign associated with it: If work is done ON a system, it is assigned a positive sign If work is done BY a system (external work or work on the environment), it is assigned a negative sign Note: This sign convention is consistent with definition in mechanics in which work done on a system (eg lifting a box) increases the mechanical (potential) energy of the system. By analogy, positive work on a system (eg work on a volume of gas) increases the internal energy of the system. Heat (Q) also has a sign associated with it: If system GAINS heat, Q is positive If system LOSES heat, Q is negative Note: Q and W are considered NEGATIVE if heat energy or work energy flows OUT of the system An analogy

7 Work is a Signed Number! Work done on system Work done by system System = gas in airtight piston compression expansion negative positive 0 Work

8 A Simple Analogy....Measuring the Depth of a Lake stream rain evaporation Snow-capped mountains stream lake river river Depth of water is like internal energy. Rainfall is like heat going into the system (Q>0). Evaporation is like heat leaving the system (Q<0). Streams flowing into lake are like work done on the system (W>0). Rivers flowing out of lake are like work done by the system (W<0). Defining internal energy

9 What is Internal Energy? E int = Q in + W on Notation emphasizes positive quantities! Internal energy is an intrinsic, distributed property of a system Internal energy increases if work is done ON the system (W on is positive work) on Internal energy increases if heat is absorbed BY the system (Q in is positive heat) For an ideal gas, E int KE =3/2 NkT = 3/2 nrt For an Ideal Gas, the Internal Energy does not depend on P or V! Defining the systeml

10 The same situation described by defining two different SYSTEMS Heat (-) SYSTEM SYSTEM Heat (+) OR Heat (-) Work (+) Note: In thermodynamics, the word WORK is used to describe any energy other than HEAT that crosses a boundary between two objects. Sign conventions

11 Dual Sign Conventions A Source of Confusion? Many textbooks write the 1 st Law as E = Q - W These textbooks focus on the Work done by the system rather than the Work done on the system. The above form of the 1 st Law reflects the original interest in understanding steam engines, where the primary interest is focused on the work EXTRACTED from the engine. Thinking thru a problem

12 General Case A Few Examples No heat lost Constant T W(+) W(+) W(+) coolant good insulation Q(-) T i T f Q(-) Q = 0 P coolant E int = -Q + W W = E int E int = 0 Q = -W joggingl

13 Applying the 1 st Law While jogging, you use 4.3 x 10 5 J of energy(work) and you give off 3.8 x 10 5 J of heat. What is the change in your internal energy? You are the system! The system loses heat, so Q < 0 The system does work, so W < 0 E int = Q in + W on Application to gas

14 Now, focus on ideal gas We need a general formula for W

15 volume of gas decreases, so V<0 An example of the 1 st Law: Compression of a Gas W on = F d = (PA) d = - P V Q out Negative sign makes W on positive Initial Conditions: P 1, V 1, T 1 Final Conditions: P 2, V 2, T 2 Path is importnat

16 The Problem: There are many ways to go from (P 1,V 1,T 1 ) to (P 2,V 2,T 2 ) and each path requires a different Q and W P Final State: (P 2, V 2, T 2 ) Initial State: (P 1, V 1, T 1 ) V There is only ONE quantity ( E int ) that is the same for all these processes E int = Q in + W on Work/ideal gas

17 Calculating the Work from the PV Diagram for an Ideal Gas Constrained Processes: General case has both P and V inside integral P adiabatic isochoric isothermal T 1 isobaric T 2 T3 T 1 >T 2 >T 3 V f V i V consequences

18 Constrained Processes for an Ideal Gas Process Restriction Consequence Const. Volume (Isochoric) Const. Temperature (Isothermal) Const. Pressure (Isobaric) V=0 E int = Q T=0 E int = 0 P=0 Free Expansion Q=W=0 E int = 0 No Heat Loss (Adiabatic) Q=0 E int = W Closed Cycle E int = 0 Q= W W P dv = Important to know

19 Why are these constraints important to know? Most heat engines require an analysis of a sequence of gas processes often called a closed cycle. Example of a Closed Cycle A B C A E int = Q in + W on P A? Q in C?? Q in B Step Process Q in W on E int A B B C C A isothermal isobaric isochoric or clicks Q in V Closed cycle Σ(ΔE int ) = 0 Isothermal process

20 An important case Work done by isothermally compressing an ideal gas; E int =0 W on = PdV PV = nrt P = dv Won = nrt V nrt V Area Isothermal example

21 A gas initially at atmospheric pressure in a box 10 cm on a side is isothermally compressed to half its volume. How much work (energy) is required? W = nrt ln(v i /V f ) If T is constant, then nrt is a constant Assume ideal gas, use PV=nRT=constant W = PV ln[v i /(0.5 V i )] = 1x10 5 Pa (10 cm) 3 (1m/10 2 cm) 3 ln[2] = 69 J Adiabatic example

22 In an automobile engine, 0.1 mole of hot gas expands against a piston in about 10 msec. During this expansion, the temperature decreases from 1200 K to 400K. Estimate the work done by the gas on the piston? volume of fuel mixture increases mixture of gas vapor ( C 8 H 18 ) and air ideal gas? Initial Conditions: P 1, V 1, 1200K Final Conditions: P 2, V 2, 400K solution

23 Solution: use First Law plus constraints 10 msec very fast for heat flow adiabatic process W on gas = E int Q in = E int since Q in = 0 Treat the fuel/gas mixture as an ideal gas (E int =3/2 nrt): E int = E final E initial = 3/2 nr T = 3/2 nr(t f T i ) where n=0.1 mole; R=8.314 J/(mole K) E int = -997 J W by gas on piston = -W on gas = 997 J Closed cycle

24 For a cyclic process: Work = area enclosed