Chapter 19pThe first law of thermodynamics19.1 Thermodynamic systems19.2 Work done during volume changesg g19.3 Paths between thermodynamic states19 4 Internal energy and first law of19.4 Internal energy and first law of thermodynamics19 5 Kinds of thermodynamic processes19.5 Kinds of thermodynamic processes19.6 Internal energy of an ideal gas19 7 Heat capacities of an ideal gas19.7 Heat capacities of an ideal gas19.8 Adiabatic processes for an ideal gas

ThermodynamicsThermodynamics

• The study of relationships involving heat,The study of relationships involving heat, mechanical work, and other aspects of energy and energy transfer

• Name few example of thermodynamics processes

• First law of thermodynamics deals with conversion of energy in a thermodynamic system

• Name an example of mechanical energy transfer• Name an example of heat transfer• First law will combine two forms of energy

Thermodynamic systemsThermodynamic systems• Thermodynamic systems: Any collection of objects

that is convenient to regard as a unit and that maythat is convenient to regard as a unit, and that may have potential to exchange energy with its surroundingsg

• Thermodynamic processes: any process during which there are changes in the state of the thermodynamic system

• State of a thermodynamic system is defined by its state variables such as T, P, V

• Thermodynamics does not depend on details of the structure of matter so both microscopic and macroscopic treatment lead to the same result.

Sign convention for exchange of energy between a thermodynamic system and environmenta thermodynamic system and environment

Thermodynamics: work done by the system is + Mechanics: work done by the forces acting on the body is +Sign of work in thermodynamics and mechanics are opposite

Sign of the work done during a volume change (whoever does the work has the positive W)(whoever does the work has the positive W)

AContraction: W(molecules)<0

Expansion: pW(molecules)>0

Work done in a volume changeF AF pAdW Fdx pAdxAd dV

== =

V

Adx dVdW pdV

==

2

1

( )V

V

W p V dV= ∫

1 2Always is initial state is finalIf is a variable (not a constant)

V Vp

we need to know its functionality with to solve the integral. V

Name an example of a PV relationship

Work done in constant p

2 1

ressure: ( )W p V V= −

Calculating work using PV diagramsWork is the area under the curve that represents pressure change as a function of volume change

Area under the curvei ii

W p V pV= Δ =∑

• Verify sign of the work during an expansion and contraction process with the convention set before

Example:Isothermal expansion of an ideal gas

• An ideal gas undergoes an isothermalg g(constant-temperature) expansion at T, during which its volume changes from V1g g 1to V2. How much work does the gas do?

• How much is the volume expansion of anHow much is the volume expansion of an ideal gas if the work is 2nRT?

Paths between thermodynamic states: workstates: work• Calculate the work done by the system to get from P1V1 to

P2V2 on the paths 132, 142,12. D th k d d th th?• Does the work depend on the path?

Paths between thermodynamic states: heatstates: heatGradual expansion and moving the piston up

Rapid removal of the wall and free expansion

Q1W1

Q2=0W2=0

• How much is the amount of the work done by the system in a and b for one mole of an ideal gas

W2 0ΔT=0

in a and b for one mole of an ideal gas

Stop and thinkStop and think

• What do you think about “the• What do you think about the amount of work contained in a system”? Can it be defined clearly?clearly?

• Can you name a system that the work done by it does not depend on the path?on the path?

Internal energy and the first law of thermodynamics: a generalizationthermodynamics: a generalization

of the conservation of energyInternal energ U of a s stem is s m of the kinetic energies of all of its• Internal energy, U, of a system is sum of the kinetic energies of all of its constituents plus sum of the potential energy of the interaction among these particles.

• When heat added to a system its internal energy increases ΔU=QWhen heat added to a system, its internal energy increases ΔU Q• When work is done by a system its internal energy decreases by the

amount of work by ΔU=-W• Total change in internal energy of a system is ΔU=Q-WTotal change in internal energy of a system is ΔU Q W

How we determine the internal f ?energy of a system?

• We can’t determine absolute value of the internal energy of a system (similar to the potential energy).

• We choose an state of a system as referenceWe choose an state of a system as reference and give an arbitrary value to its internal energy.

• Then we can measure change in internal energy i th h t h d k d b /using the heat exchange and work done by/on

the system ΔU=Q-W • While dQ and dW both depend on the path, it e dQ a d d bot depe d o t e pat , t

turns out that ΔU does not depend on the path. ΔU only depends on the state of a system which is a function of its state variables (p V T)which is a function of its state variables (p, V, T)

Change in internal energy of human body in 24 hourshuman body in 24 hours

Work done during a change of sate

Summary

Summary: The first law of h d ithermodynamics

Summary: Internal energy

ExampleExample

• You intake 900 calorie and run up a cliffYou intake 900 calorie and run up a cliff. How high do you have to climb to burn all calories consumed? (your mass is 60 kg)calories consumed? (your mass is 60 kg) (ans. 6410 m)

• Evaluate your answer• Evaluate your answer.

A cyclic processA cyclic process• State of a system starts and y

ends at a. The total work is -500 J.

Why is the work negative?– Why is the work negative? – Find the change in internal

energy and the heat added to the system in this processthe system in this process.

• For pV diagrams total work is positive for clockwise processes and negative for counterclockwise. Can you contradict it?contradict it?

Comparing thermodynamic processes

Kinds of thermodynamic processesKinds of thermodynamic processes

• Adiabatic: no heat transferAdiabatic: no heat transfer• Isobaric: constant pressure

I h i t t l• Isochoric: constant volume • Isothermal: constant temperature

Adiabatic processes and the first lawAdiabatic processes and the first law

• No heat transfer: Q=0 can be achieved in– Isolated systems – Fast processes

• First Law: ΔU=-WFirst Law: ΔU W• In adiabatic expansion system does positive work and

hence looses energy. W>0 and ΔU<0I di b ti i t d ti k• In adiabatic compression system does negative work and hence gains energy. W<0 and ΔU>0

• Example: compression stroke in an internal combustion, p pexpansion of the fuel during the power stoke (temperature rise or drop happens during the process)

Isochoric processes and the first lawIsochoric processes and the first law

• No change of volume: W=0 is zeroNo change of volume: W 0 is zero• First Law: ΔU=Q; • All the energy added to the system remains in itAll the energy added to the system remains in it

and cusses temperature change• Example: heating gas in a cylinder (don’t try it atExample: heating gas in a cylinder (don t try it at

home)• Although there are other kinds of work that do t oug t e e a e ot e ds o o t at do

not involve volume change but “isochoric” is used for the systems that no work is done at all.

Isobaric processes and the first lawIsobaric processes and the first law

• No change of pressure: none of the ΔU QNo change of pressure: none of the ΔU, Q, or W is zero

• First Law: ΔU=Q W;• First Law: ΔU=Q-W; • Work is calculated easily W=p(V2-V1)

Isothermal processes and the first lawIsothermal processes and the first law

• No change of temperature: process has toNo change of temperature: process has to occur slowly and none of the ΔU, Q, or W is zerois zero

• First Law: ΔU=Q-W; I i l ΔT 0 ΔU 0• In special cases ΔT=0 means ΔU=0 (Example: ideal gas) but generally ΔU is

tnot zero.

Thermodynamic processes on PV didiagrams

Internal energy of an ideal gasInternal energy of an ideal gas• Example: adiabatic expansion of a

gas If U depended on V or p thengas. If U depended on V or p then we would have observed a change in T.

• Experiment shows no change in T• Experiment shows no change in T for expansion of low density gasses.I t l f id l• Internal energy of an ideal gas depends only on its temperature, not on its pressure or volume.

• For a non-ideal gas adiabatic expansion is accompanied with drop in temperature.p p

• Does internal energy of a solid depend on its volume?

Heat capacities of an ideal gasHeat capacities of an ideal gas• Cp molar heat capacity p p y

under constant pressure (usually for gasses)C l h t it• Cv molar heat capacity under constant volume (for solids or liquids)

• For a given system (ideal gas) why these should be different? Which one isdifferent? Which one is greater?

Cp=Cv+R molar heat capacities of an ideal gasUsually Cv=.4Cp For water Cv<Cp between 0 and 40C

More about heat capacitiesMore about heat capacities

• Ratio of heat capacities γ=C /CRatio of heat capacities γ=Cp/Cv

• For an ideal gas internal energy change only depends on T so ΔU=nC ΔT no matter whatdepends on T so ΔU=nCvΔT no matter what the volume change is.E l 19 6 Fi d th h i i t l• Example 19.6: Find the change in internal energy of a dorm room containing 2400

l f i h it i l d f 23 90C tmoles of air when it is cooled from 23.90C to 11.60C at a constant pressure of 1.00 atm. T t th i id l ith 1 4Treat the air as an ideal gas with γ=1.4.

Adiabatic Process for an ideal gas• Q=0 and ΔU=-W<0• An adiabatic curve on PVAn adiabatic curve on PV

diagram is always steeper than an isotherm.

• Adiabatic process has to beAdiabatic process has to be fast for not letting heat exchange to happen.

• If we use ideal gas law to gdeduce any results the process has to be slow to allow equilibrium condition to b i t i dbe maintained.

• In reality it is hard to meet both of these conditions simultaneously So whateversimultaneously. So whatever we discuss here is an approximation.

Adiabatic Process for an ideal gas

• For an adiabatic process in an ideal gas:

• TVγ-1=constant so T V γ-1= T V γ-1T1V1

γ-1= T2V2γ-1

• PVγ=constant so P1V1

γ= P2V2γP1V1 P2V2

• W=-ΔU=nCv (T1-T2)• Use PV=nRTUse PV nRT• W=Cv (P1V1-P2V2) /R• W= (P1V1-P2V2) /(γ-1) ( 1 1 2 2) (γ )

Summary

SummarySummary

Summary

Summary

Example: 19 7Example: 19.7• The compression ratio of

a diesel engine is 15 to 1a diesel engine is 15 to 1. If the initial pressure is 1.01x105pa and the initial t t i 270C fi dtemperature is 270C, find the final pressure and the temperature aftertemperature after compression. Air is a mixture of mostly diatomic

d it th toxygen and nitrogen that is an ideal gas with γ=1.40. (T2=886K, γ ( 2 ,p2=44.8x105Pa=44 atm)

Work done in an adiabatic processWork done in an adiabatic process

• In the previous example how much work isIn the previous example how much work is done by the gas do during the compression if the initial volume of the cylinder is 1.00 L=1.00x10-3m3? Assume that CV for air is 20.8 J/mol.K and γ=1.40