Gas Processes

7/31/2019 Gas Processes

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Heat and Work

Ideal-Gas Processes

Specific Heat of Gases

Ideal Gas Processes

Unit 4 - Lecture 4

Charles W Fay IV

August 22, 2011

Charles W Fay IV U4-04: Gas Processes


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Heat and Work

Ideal-Gas Processes


First Law of Thermodynamics

PV Diagrams

Work done on a Gas


Q = U + WG (1)

The heat transfered to a system is equal to the change in

internal energy plus the work done. A system is a definite quantity of material surrounded by a

boundary (real or imagined)

The state of a system is defined by its state variables

state variables are the macroscopic variables that express

physical quantities of a system, (P,V,N,T). A particular set of state variables define a definite state.

the equation of state is,

PV = N kBT



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Heat and Work

Ideal-Gas Processes



PV Diagrams

Work done on a Gas

Thermodynamics

The study of the transfer of energy. Heat: the transfer of energy due to a temperature difference.

Q = U + WG

Temperature: measure of thermal energy U =3

2N kBT. Heat is measured in Joules.

If there is not transfer of heat energy the objects are in

thermal equilibrium.

Ta = Tb = Tc (2)



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Heat and Work

Ideal-Gas Processes



PV Diagrams

Work done on a Gas

Heat

Heat is a method of transferring energy into or out of a system. Heat is energy transferred through a thermal interaction.

ETH transferred from fast (hot) atoms to slow (cold) atoms.

Transfer continues until the system reaches thermal

equilibrium.T1f = T2f = Tf

The heat gained by one object is the heat lost by the other

object.

Q2 = Q1



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H d W k Fi L f Th d i


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Heat and Work

Ideal-Gas Processes



PV Diagrams

Work done on a Gas

Example: Mechanical equivalence of heat

Given: m1

= m2

= 50kg mw = 1kg Vw = 1LT = 1CHow far do the weights fall?


H t d W k Fi t L f Th d i


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Heat and Work

Ideal-Gas Processes



PV Diagrams

Work done on a Gas

Example: Mechanical equivalence of heat

Given:m1 = m2 = 50kg m

w= 1kg V

w= 1L

T = 1CHow far do the weights fall?

Q = 4186J

Q = mTgh = 2m1ghh =

Q

2m1g= 4.27m


Heat and Work First Law of Thermodynamics


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Heat and Work

Ideal-Gas Processes



PV Diagrams

Work done on a Gas


For system where only the thermal energy of a system changes,we can write the conservation of energy as,

ETH = U = Q + W

The thermal energy is sometimes called the internal energy andgiven the symbol U.

W > 0 work done on the system E

W < 0 work done by the system E




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Heat and Work

Ideal-Gas Processes



PV Diagrams

Work done on a Gas

Processes

A process is the method of changing from one state to another,the processes can be,

1. irreversible - A process for which intermediate steps are not

equilibrium states. As such the path cannot be retraced.

2. reversible - The change of state is so slow that each

intermediate step is an equilibrium state (U = 0?)




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Heat and Work

Ideal-Gas Processes



PV Diagrams

Work done on a Gas

Ideal-Gas Processes

Quantity of a gas is fixed (N=constant) Well defined initial state P1, V1, and T1.

Well defined final state P2, V2, and T2.

in a sealed container (N=Constant)

PV

T= nR = constant

P1V1T1

=P2V2

T2(3)




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Heat and Work

Ideal-Gas Processes



PV Diagrams

Work done on a Gas

PV Diagrams

It is useful to represent ideal-gas processes on a graph, PVdiagram.

Each point on a graph represents a single, unique state of the

gas.

each point represents (P, V, T) specifying the state.

The path from one ideal-gas state to another ideal-gas state is

called a trajectory.




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Ideal-Gas Processes


y

PV Diagrams

Work done on a Gas

Work done on a Gas

suppose a gas expands x = xf xi The work is then related to,

W = Fx = PAx = PV (4)

Work done on a gas is the area under the PV curve

Figure: Work done on a gas




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Ideal-Gas Processes


y

PV Diagrams

Work done on a Gas

Work done on a Gas

In order for the gas to do work, the volume must change. The work WG = PV is positive V > 0.

The work WG = PV is negative V < 0.

The pressure in Pa, volume in m3 gives work in Joules.

WG is not the work that appears in the 1st law ofthermodynamics

W is the word done on the system.

WG is the work done by the system.

W = WG (5)

U = Q WG

Q = U + WG (6)




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Ideal-Gas Processes


PV Diagrams

Work done on a Gas

Work done on a Gas

Work done on a gas is the area under the PV curve Work is dependent upon the path taken.

Heat also depends upon the path taken.

The change in internal energy U is only dependent upon the

end point temperatures, so it is independent of the path. In order for the gas to do work, the volume must change.

The work W = PV is positive V > 0. The work W = PV is negative V < 0.


Heat and Work Isometric Process


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Ideal-Gas Processes


Isobaric Process

Isothermal Process

Adiabatic Process

Isometric process

An isometric process is defined by the volume being constant.Since V2 = V1, WG = 0.

Q = U + WG

Q = U


Heat and Work Isometric Process


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Ideal-Gas Processes


Isobaric Process

Isothermal Process

Adiabatic Process

Example: isometric process

An ideal gas inside a can is heated the volume remains constant.What is its new pressure?

Given:P = 1atm = 1.01 105kP aT1 = 20

C = 293K T2 = 200C = 473K



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Heat and Work

Ideal Gas Processes

Isometric Process

Isobaric Process


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Ideal-Gas Processes


Isobaric Process

Isothermal Process

Adiabatic Process

Isobaric Process

An isobaric process is defined by the pressure being constant.

Pf = Pi

PV = N kBT

P = N kBT

VV

T= constant

internal energy increases as the gas expands.

Wisobaric = PV (7)

Q = U + PV = U + N kBT


Heat and Work

Ideal Gas Processes

Isometric Process

Isobaric Process


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Ideal-Gas Processes


Isobaric Process

Isothermal Process

Adiabatic Process

Isothermal process

Figure: Isotherms

Temperature is constant.

T = 0 U = 0.

The graph of P vs V for a specific T is

known as an isotherm.

PV = N kBT = constant

P 1

V

Q = W

WIT = N kBT lnV2

V1

(8)


Heat and WorkIdeal-Gas Processes

Isometric ProcessIsobaric Process


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Ideal-Gas Processes


Isobaric Process

Isothermal Process

Adiabatic Process

Example: Compare isothermal and isobaric

2 moles of an ideal gas initially at 0

C and 1atm is expanded totwice its original volume. Is more work done if it is expandisothermally or isobaricly.

Given:P = 1atm = 1.01 105P a T1 = 0

C = 273Kn = 2.00mol V 2 = 2V1





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Ideal Gas Processes


Isobaric Process

Isothermal Process

Adiabatic Process




Given:P = 1atm = 1.01 105P a T1 = 0

C = 273Kn = 2.00mol V 2 = 2V1

WIT = nRT lnV2

V1

= (2mol)(8.31J/molK)(273K) ln2V1

V1

= 3.14 103J





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Ideal Gas Processes


Isobaric Process

Isothermal Process

Adiabatic Process




Given:P = 1atm = 1.01 105P a T1 = 0

C = 273Kn = 2.00mol V 2 = 2V1

V =nRT

PWIB = P = P(nRTP)

= P(V2 V1) = P(2V1 V1) = PV1

= PnRT1

P= nRT1

= (2.00mol)(8.31J/molK)(273K) = 4.53 103J

More work is done if the gas is expanded isobarically.





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Specific Heat of Gases Isothermal Process

Adiabatic Process

Adiabatic Process

An adiabatic process is defined by the heatQ

being constant.

Q = U + W = 0

U = Wadiabatic (9)

Adiabatic compression raises T.

Adiabatic expansion lowers T.

The pressure and volume are related by,

P2V2

= P1V1





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Specific Heat of Gases Isothermal Process

Adiabatic Process

Adiabatic Process

This leads to the work for an adiabatic process being,

WA =P1V1 P2V2

1

where,

= cPcV

cP is the specific heat of a gas at constant pressure, cV is thespecific heat of a gas at constant volume.



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Now lets calculate cP

,

QP = cPT

QP = U + W = U + PV

cPT = cVT + PV

cP = cV + PVT

cP = cV + N kB (10)



S ifi H f G


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Specific Heats of Gases

Thus for a monatomic gas,

cV =3

2N kB (11)

cP = 32N kB + N kB =5

2N kB

and,

m =cPcV

=5

2N kB

3

2N kB

=5

3(12)

For diatomic gases,

cV =5

2N kB, cP =

7

2N kB, d =

7

5(13)



S ifi H t f G


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Example: Adiabatic process

a monatomic gas initially at 20C expands adiabatically fromP = 1atm to 3 times its original volume. How much work is done?Given:

n = 2.00mol V 2 = 3V1 P1 = 1atm = 101.3 103P a

m = 5/3 T = 20C = 293K





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Example: Adiabatic process

a monatomic gas initially at 20C expands adiabatically fromP = 1atm to 3 times its original volume. How much work is done?Given:

n = 2.00mol V 2 = 3V1 P1 = 1atm = 101.3 103P a

m = 5/3 T = 20C = 293K

P2V2 = P1V1

P2 = P1

V1V2

= P1

13

= 1.62 104N/m2

V1 =nRT

P1= 4.82 102m3

V2 = 3V1 = 1.45 101m3

Wadiabatic =P1V1 P2V2

1= 3.76 103J





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