CHAPTER 16 : CHAPTER 16 : THERMODYNAMICSTHERMODYNAMICS

16.1 Work 16.1 Work HeatHeat

Form of energy which is transferred from one Form of energy which is transferred from one body to another body at lower temperature, by body to another body at lower temperature, by virtue of the temperature difference between the virtue of the temperature difference between the bodies. bodies.

Internal energyInternal energyTotal energy content of a system. It is the sum of Total energy content of a system. It is the sum of all forms of energy possessed by the atoms and all forms of energy possessed by the atoms and molecules of the system.molecules of the system.

16.1 Work16.1 Work

SystemSystemThe collection of matter within prescribed The collection of matter within prescribed and identifiable boundaries. and identifiable boundaries.

SurroundingsSurroundingsEverything else in the environment. Everything else in the environment.

Thermodynamics systemThermodynamics systemSystem that can interact with its System that can interact with its surroundings or environment in at least surroundings or environment in at least two ways, one of which is heat transfer.two ways, one of which is heat transfer.

16.1 Work16.1 Work

WorkWork Assume a system in a piston that contains Assume a system in a piston that contains fluid, the work done by the system as its fluid, the work done by the system as its volume changes.volume changes.

pA

dx

A

16.1 Work16.1 Work

When the piston moves out an infinitesimal When the piston moves out an infinitesimal distance dx, the work dW done by this force is distance dx, the work dW done by this force is

dw = Fdx = pAdxdw = Fdx = pAdx

A : cross-sectional area of the cylinderA : cross-sectional area of the cylinder

p : pressure exerted by the system at the p : pressure exerted by the system at the

surface of the piston. surface of the piston.

F : total force exerted by the system on F : total force exerted by the system on

the pistonthe piston

16.1 Work16.1 Work

But dV=Adx, therefore But dV=Adx, therefore dW=pdV. dW=pdV.

The work done in a The work done in a volume change from volume change from VV11 to V to V22,,

AdxdV

2

1

1

2

P

PnRT

V

VnRTW lnln

Graph p versus V Graph p versus V pp

(a) p vs V at changes pressure

V

(a) p vs V at constant pressure

V

16.2 First Law of 16.2 First Law of ThermodynamicsThermodynamics

The internal energy of The internal energy of a system changes a system changes from an initial value from an initial value U1 to a final value U2 U1 to a final value U2 due to heat, Q and due to heat, Q and work, W:work, W:

WQUUU 12

Cyclic processCyclic process

The change in internal energy depends on the The change in internal energy depends on the gas temperature.gas temperature.

It also depends on the initial and final state.It also depends on the initial and final state.

Cyclic process :Cyclic process :

Initial state = Final stateInitial state = Final state

U1 = U2 U1 = U2

Q = WQ = W

Example:Example:

Two moles of the monatomic argon gas Two moles of the monatomic argon gas expand isothermally at 298 K, from an expand isothermally at 298 K, from an initial volume of 0.025 m3 to a final volume initial volume of 0.025 m3 to a final volume of 0.050 m3 (Argon is an ideal gas), find of 0.050 m3 (Argon is an ideal gas), find (a) work done by the gas (a) work done by the gas

(b) the heat supplied to the gas. (b) the heat supplied to the gas.

(a) W=+3400 J, (a) W=+3400 J,

(b) Q =+3400 J(b) Q =+3400 J

16.3 Thermodynamic 16.3 Thermodynamic ProcessesProcesses

Isothermal processIsothermal processIsochoric processIsochoric processIsobaric processIsobaric process

Adiabatic processAdiabatic process

Graph Graph Isothermal processIsothermal process

The work done by the system is the area The work done by the system is the area under the curve of a pV-diagram as a under the curve of a pV-diagram as a graph shown below..graph shown below..

p

VVi Vf

Work,W

pi

pf

Isothermal processIsothermal process

2

1

1

2 lnlnP

PnRT

V

VnRTW

Isochoric processIsochoric process

a constant-volume process, where there is a constant-volume process, where there is no work done in the system, W = 0.no work done in the system, W = 0.

The first law of thermodynamic will be,The first law of thermodynamic will be,

U2-U1 = ΔU = Q.U2-U1 = ΔU = Q.

Graph Graph Isochoric processIsochoric process

The graph pV of an isochoric process is The graph pV of an isochoric process is shown below.shown below.

pp

vv

Isothermal processIsothermal process

a constant-temperature process. a constant-temperature process. Heat transfer and work in the system must not Heat transfer and work in the system must not change and the internal energy will be zero, change and the internal energy will be zero, ΔU=0.ΔU=0.

The changes in pressure,p and volume,V will The changes in pressure,p and volume,V will take place. take place.

Isobaric processIsobaric process

Isobaric process is a constant-pressure Isobaric process is a constant-pressure process. process.

W = p(V2 – V1) W = p(V2 – V1)

The first law of thermodynamic will be,The first law of thermodynamic will be,

Q= ΔU +W = ΔU + p( V2-V1)Q= ΔU +W = ΔU + p( V2-V1)

Graph Isobaric processGraph Isobaric processp

V

Isobaric process carried out at a constant pressure

Adiabatic processAdiabatic process

a process with no a process with no heat transfer to or heat transfer to or from a system, Q=0.from a system, Q=0.

ΔU = -WΔU = -W

ExampleExample

The temperature of three moles of a monatomic The temperature of three moles of a monatomic ideal gas is reduced from Ti = 540 K to Tf = 350 ideal gas is reduced from Ti = 540 K to Tf = 350 K by two different methods. In the first method K by two different methods. In the first method 5500 J of heat flows into the gas, while in the 5500 J of heat flows into the gas, while in the second method, 1500 J of heat flows into it. In second method, 1500 J of heat flows into it. In each case, find (a) the change in the internal each case, find (a) the change in the internal energy of the gas and (b) the work done by the energy of the gas and (b) the work done by the gas.gas.

16.4 Molar Specific Heat Capacities At Constant 16.4 Molar Specific Heat Capacities At Constant

Pressure And VolumePressure And Volume Molar specific heat capacities of constant pressure, CpMolar specific heat capacities of constant pressure, CpWhen an n mole of gas undergoes a heating process, When an n mole of gas undergoes a heating process, the equation that relates Q and the temperature changes, the equation that relates Q and the temperature changes,

dT dT is written as ;is written as ; Q = nCdT ……………..(1)Q = nCdT ……………..(1) Whereas Q = heatWhereas Q = heat n = number of molen = number of mole C = molar specific heat capacitiesC = molar specific heat capacities dT = temperature changesdT = temperature changes

When the gas is heated at constant When the gas is heated at constant pressure, the (1) equation will be;pressure, the (1) equation will be;

Q = nCpdTQ = nCpdT

Or Cp = Or Cp =

Where Cp = molar specific heat capacities at Where Cp = molar specific heat capacities at constant constant

pressure.pressure.

ndT

Q

Molar specific heat capacities of constant Molar specific heat capacities of constant volume, Cvvolume, Cv

Hence, when an n mole of gas is heated at Hence, when an n mole of gas is heated at constant volume, the (1) equation will be;constant volume, the (1) equation will be;

Q= nCvdTQ= nCvdT

Or Or

Where Cv = molar specific heat capacities at Where Cv = molar specific heat capacities at constant volumeconstant volume

The universal gas constant, RThe universal gas constant, R

the substraction between Cp and Cv that the substraction between Cp and Cv that is;is;

Cp-Cv = RCp-Cv = R

Where R = 8314 J mol-1 KWhere R = 8314 J mol-1 K

ExampleExample

A certain perfect gas has specific heat A certain perfect gas has specific heat capacities as follows:capacities as follows:

Cp = 0.84 kJ / mol K and Cp = 0.84 kJ / mol K and

Cv = 0.657 kJ / mol KCv = 0.657 kJ / mol K

Calculate the gas constant. Calculate the gas constant.

16.5 Specific Heat At Constant Pressure And 16.5 Specific Heat At Constant Pressure And VolumeVolume

ExampleExample

A 2.00 mol sample of an ideal gas with γ = A 2.00 mol sample of an ideal gas with γ = 1.40 expands slowly and adiabatically 1.40 expands slowly and adiabatically from a pressure of 5.00 atm and a volume from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L. of 12.0 L to a final volume of 30.0 L.

(a) What is the final pressure of the gas?(a) What is the final pressure of the gas?

(b) What are the initial and final (b) What are the initial and final temperature?temperature?

THERMODYNAMICS THERMODYNAMICS

THE ENDTHE END