Ideal-Gas Processes 21st July 2014 - Physics & Astronomyphysics.unm.edu/Courses/morgan-tracy/151Summer/Slides/151-07-21-… · The Ideal-Gas Equation Ideal-Gas Processes 21st July

July 21, Week 8

Ideal-Gas Processes 21st July 2014

Today: Ideal-Gas Processes, Chapter 12

Final Homework #7 due today at 5:00. Office hours, 1:00-5:00

Final Exam, Thursday. 9:00-10:30 or 11:00-12:15

Pressure


When the molecules in a gas or liquid collide with its container, theyexert a force on the container

Pressure



Pressure



Collisions cause an outward forceover the entirety of the container’ssurface area

Pressure




Pressure = “average” force per area

p =F

Aunit: N/m2 = Pa (Pascal)

Pressure





p =F


Other pressure units: atmospheres (atm), millimeters of mercury(mmHg = torr), and pounds per square inch (psi)

Pressure





p =F


Other pressure units: atmospheres (atm), millimeters of mercury(mmHg = torr), and pounds per square inch (psi)

1 atm = 101300Pa = 101.3 kPa = 760mmHg = 14.7 psi

The Ideal-Gas Equation


In an ideal gas, the relationship between the pressure (p), volume(V ) and temperature (T ) of the gas is given by:




pV = NkBTN = number of molecules

kB = 1.38× 10−23 J/K = Boltzmann’s constant






If forced to use moles:






If forced to use moles: n = number of moles: N = n×NA

Avogadro’s Number: NA = 6.02× 1023








pV = nRT R = kB ×NA = 8.31 J/mole ·K









When a gas is in a sealed container, the total number of particles mustremain constant









When a gas is in a sealed container, the total number of particles mustremain constant

pV

T= NkB = constant ⇒

piVi

Ti

=pfVf

Tf

Ideal-Gas Exercise I


A gas is sealed in a container that has a moveable piston on oneside (so the volume can change). If the volume is cut in half, bywhat ratio does the temperature change?

Moveable Piston




Moveable Piston




Moveable Piston

(a) The temperature will double.




Moveable Piston


(b) The temperature will stay the same.




Moveable Piston



(c) The temperature will be cut in half.




Moveable Piston




(d) There is not enough information todetermine.




Moveable Piston





(e) Intentionally left blank.




Moveable Piston









Moveable Piston


Even when N is fixed, pV = nkBT is a function of three variables: p, Vand T . In order to figure out how one of them is changing, we need toknow something about the other two.

Ideal-Gas Exercise II


A gas is sealed in a container that has a moveable piston on oneside (so the volume can change). If the volume is cut in half whilethe pressure is kept constant, by what ratio does the temperaturechange?

Moveable Piston




Moveable Piston




Moveable Piston





Moveable Piston






Moveable Piston







Moveable Piston








Moveable Piston









Moveable Piston









Moveable Piston(c) The temperature will be cut in half.

With N and p are fixed, V =NkBp

T ⇒ V = constant× T ⇒ cutting V

in half, cuts T in half.

Work done by a Gas


Volume changes occur when the gas exerts a large enough forceon its container to move it by some distance

Work done by a Gas



Initial−→

Fgas−→

Fext

Work done by a Gas



Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext

Work done by a Gas



Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext

For a constant force: W = Fd

d

Work done by a Gas



Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext


dWgas =

(

Fgas

A

)

(dA)

Area = A

Work done by a Gas



Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext


dWgas =

(

Fgas

A

)

(dA)

Area = A

Wgas = p∆V Constant Pressure

Work done by a Gas



Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext


dWgas =

(

Fgas

A

)

(dA)

Area = A


V

p i

f

When pressure changes,we use a “pV diagram ⇒

pressure on the y-axis andvolume on the x-axis.

Work done by a Gas



Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext


dWgas =

(

Fgas

A

)

(dA)

Area = A


V

p i

f

Wgas is the area under thepV diagramWhen pressure changes,

we use a “pV diagram ⇒

pressure on the y-axis andvolume on the x-axis.

Processes Exercise I


Which of the following pV diagrams corresponds to an Isobaric =constant pressure process?




(a)

V

p i f




(a)

V

p i f

(b)

V

p i

f




(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f




(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f




(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f

(e) Intentionallyleft blank




(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f





(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f


For an Isobaric process, Wgas = p (Vf − Vi)

Processes Exercise II


Which of the following pV diagrams corresponds to an Isothermal =constant temperature process?




(a)

V

p i f




(a)

V

p i f

(b)

V

p i

f




(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f




(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f




(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f





(a)

V

p i f

(b)

V

p i

f

pV = NkBT⇒ pV = constant⇒ hyperbola

(c)

V

p i

f

(d)

V

p i

f





(a)

V

p i f

(b)

V

p i

f


(c)

V

p i

f

(d)

V

p i

f


For an Isothermal process, Wgas = NkBT ln

(

Vf

Vi

)




(a)

V

p i f

(b)

V

p i

f


(c)

V

p i

f

(d)

V

p i

f


For an Isothermal process, Wgas = NkBT ln

(

Vf

Vi

)

(c) is an Isochoric process, ⇒ constant volume ⇒ Wgas = 0

First Law for Ideal Gasses


First Law of Thermodynamics: Q+W = ∆Eth




Wgas is the work done by the system ⇒ W = −Wgas





For an Ideal Gas: ∆Eth =3

2NkB∆T






2NkB∆T

Q−Wgas =3

2NkB∆T First Law for an Ideal Gas






2NkB∆T

Q−Wgas =3

2NkB∆T First Law for an Ideal Gas

If forced to at gunpoint: Q−Wgas =3

2nR∆T

First-Law Exercise I


An ideal gas expands isobarically as shown. If the number ofmolecules in the gas is such that NkB = 2 J/K, how much heatwas involved in this process? Hint: Given the initial pressure andvolume values, the gas’s initial temperature was 2.5K.

V

p




V (m3)

p(Pa)

i f

1 3

5




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(a) Q = 5 J




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(a) Q = 5 J

(b) Q = 7.5 J




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(a) Q = 5 J

(b) Q = 7.5 J

(c) Q = 15 J




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(a) Q = 5 J

(b) Q = 7.5 J

(c) Q = 15 J

(d) Q = 17.5 J




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(a) Q = 5 J

(b) Q = 7.5 J

(c) Q = 15 J

(d) Q = 17.5 J

(e) Q = 25 J




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(a) Q = 5 J

(b) Q = 7.5 J

(c) Q = 15 J

(d) Q = 17.5 J

(e) Q = 25 J




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(e) Q = 25 J

Isobaric ⇒ Wgas = p∆V




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(e) Q = 25 J


Wgas = (5Pa)(

3m3− 1m3

)

= 10J




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(e) Q = 25 J


Wgas = (5Pa)(

3m3− 1m3

)

= 10J

Constant pressure ⇒ tripling thevolume also triples the temperature




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(e) Q = 25 J


Wgas = (5Pa)(

3m3− 1m3

)

= 10J


⇒ Tf = 7.5K and ∆T = 5 k




V (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(e) Q = 25 J


Wgas = (5Pa)(

3m3− 1m3

)

= 10J


⇒ Tf = 7.5K and ∆T = 5 k

First Law: Q− 10J = 15J

First-Law Exercise II


The pV diagram shown is for an Adiabatic expansion. (Note: Theshape is NOT a hyperbola.) An adiabatic process is one in whichQ = 0. Which of the following statements about this expansionmust be correct?

V

p i

f




V

p i

f

(a) The gas’s temperature musthave decreased during theexpansion.




V

p i

f


(b) The gas’s temperature musthave stayed constant during theexpansion.




V

p i

f



(c) The gas’s temperature musthave increased during theexpansion.




V

p i

f




(d) and (e) Intentionally left blank.




V

p i

f




(d) and (e) Intentionally left blank.




V

p i

f


First Law: Q−Wgas =3

2NkB∆T




V

p i

f



2NkB∆T

Adiabatic ⇒ Q = 0 ⇒ −Wgas =3

2NkB∆T




V

p i

f



2NkB∆T

Adiabatic ⇒ Q = 0 ⇒ −Wgas =3

2NkB∆T

In an expansion, Vf > Vi ⇒ Wgas is positive ⇒ ∆T is negative

Phase Transitions


Thermal energy, Eth, is NOT temperature!

Phase Transitions



Thermal energy is the combination of the molecules’ kinetic energyand their potential energy. (Translational Kinetic energy is whatdetermines temperature.)

Phase Transitions




During a phase transition (melting, freezing, evaporating,condensing), the thermal energy of the substance changesbecause of a change in potential energy only. (Energy is releasedwhile making bonds during freezing and condensing. It requiresenergy to break bonds during melting and evaporation.)

Phase Transitions




During a phase transition (melting, freezing, evaporating,condensing), the thermal energy of the substance changesbecause of a change in potential energy only. (Energy is releasedwhile making bonds during freezing and condensing. It requiresenergy to break bonds during melting and evaporation.)

Since the kinetic energy doesn’t change,There is no change in temperature during a phase transition

Documents

Ideal-Gas Processes 21st July 2014 - Physics & Astronomyphysics.unm.edu/Courses/morgan-tracy/151Summer/Slides/151-07-21-… · The Ideal-Gas Equation Ideal-Gas Processes 21st July