Ideal-Gas Processes 21st July 2014 - Physics &...

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

Ideal-Gas Processes 21st July 2014

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

Pressure

Ideal-Gas Processes 21st July 2014

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

Pressure

Ideal-Gas Processes 21st July 2014

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

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

Pressure

Ideal-Gas Processes 21st July 2014

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

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

Pressure = “average” force per area

p =F

Aunit: N/m2 = Pa (Pascal)

Pressure

Ideal-Gas Processes 21st July 2014

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

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

Pressure = “average” force per area

p =F

Aunit: N/m2 = Pa (Pascal)

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

Pressure

Ideal-Gas Processes 21st July 2014

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

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

Pressure = “average” force per area

p =F

Aunit: N/m2 = Pa (Pascal)

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

Ideal-Gas Processes 21st July 2014

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

The Ideal-Gas Equation

Ideal-Gas Processes 21st July 2014

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

The Ideal-Gas Equation

Ideal-Gas Processes 21st July 2014

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:

The Ideal-Gas Equation

Ideal-Gas Processes 21st July 2014

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: n = number of moles: N = n×NA

Avogadro’s Number: NA = 6.02× 1023

The Ideal-Gas Equation

Ideal-Gas Processes 21st July 2014

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: n = number of moles: N = n×NA

Avogadro’s Number: NA = 6.02× 1023

pV = nRT R = kB ×NA = 8.31 J/mole ·K

The Ideal-Gas Equation

Ideal-Gas Processes 21st July 2014

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: 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

The Ideal-Gas Equation

Ideal-Gas Processes 21st July 2014

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: 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

pV

T= NkB = constant ⇒

piVi

Ti

=pfVf

Tf

Ideal-Gas Exercise I

Ideal-Gas Processes 21st July 2014

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

Ideal-Gas Exercise I

Ideal-Gas Processes 21st July 2014

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

Ideal-Gas Exercise I

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

Ideal-Gas Exercise I

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

Ideal-Gas Exercise I

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

(c) The temperature will be cut in half.

Ideal-Gas Exercise I

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

(c) The temperature will be cut in half.

(d) There is not enough information todetermine.

Ideal-Gas Exercise I

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

(c) The temperature will be cut in half.

(d) There is not enough information todetermine.

(e) Intentionally left blank.

Ideal-Gas Exercise I

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

(c) The temperature will be cut in half.

(d) There is not enough information todetermine.

(e) Intentionally left blank.

Ideal-Gas Exercise I

Ideal-Gas Processes 21st July 2014

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

(d) There is not enough information todetermine.

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

Ideal-Gas Processes 21st July 2014

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

Ideal-Gas Exercise II

Ideal-Gas Processes 21st July 2014

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

Ideal-Gas Exercise II

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

Ideal-Gas Exercise II

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

Ideal-Gas Exercise II

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

(c) The temperature will be cut in half.

Ideal-Gas Exercise II

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

(c) The temperature will be cut in half.

(d) There is not enough information todetermine.

Ideal-Gas Exercise II

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

(c) The temperature will be cut in half.

(d) There is not enough information todetermine.

(e) Intentionally left blank.

Ideal-Gas Exercise II

Ideal-Gas Processes 21st July 2014

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

(a) The temperature will double.

(b) The temperature will stay the same.

(c) The temperature will be cut in half.

(d) There is not enough information todetermine.

(e) Intentionally left blank.

Ideal-Gas Exercise II

Ideal-Gas Processes 21st July 2014

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(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

Ideal-Gas Processes 21st July 2014

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

Work done by a Gas

Ideal-Gas Processes 21st July 2014

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

Initial−→

Fgas−→

Fext

Work done by a Gas

Ideal-Gas Processes 21st July 2014

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

Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext

Work done by a Gas

Ideal-Gas Processes 21st July 2014

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

Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext

For a constant force: W = Fd

d

Work done by a Gas

Ideal-Gas Processes 21st July 2014

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

Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext

For a constant force: W = Fd

dWgas =

(

Fgas

A

)

(dA)

Area = A

Work done by a Gas

Ideal-Gas Processes 21st July 2014

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

Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext

For a constant force: W = Fd

dWgas =

(

Fgas

A

)

(dA)

Area = A

Wgas = p∆V Constant Pressure

Work done by a Gas

Ideal-Gas Processes 21st July 2014

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

Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext

For a constant force: W = Fd

dWgas =

(

Fgas

A

)

(dA)

Area = A

Wgas = p∆V Constant Pressure

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

Ideal-Gas Processes 21st July 2014

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

Initial−→

Fgas−→

Fext

Final−→

Fgas−→

Fext

For a constant force: W = Fd

dWgas =

(

Fgas

A

)

(dA)

Area = A

Wgas = p∆V Constant Pressure

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

Ideal-Gas Processes 21st July 2014

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

Processes Exercise I

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

Processes Exercise I

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

Processes Exercise I

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

Processes Exercise I

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f

Processes Exercise I

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f

(e) Intentionallyleft blank

Processes Exercise I

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f

(e) Intentionallyleft blank

Processes Exercise I

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f

(e) Intentionallyleft blank

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

Processes Exercise II

Ideal-Gas Processes 21st July 2014

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

Processes Exercise II

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

Processes Exercise II

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

Processes Exercise II

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

Processes Exercise II

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f

Processes Exercise II

Ideal-Gas Processes 21st July 2014

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

(a)

V

p i f

(b)

V

p i

f

(c)

V

p i

f

(d)

V

p i

f

(e) Intentionallyleft blank

Processes Exercise II

Ideal-Gas Processes 21st July 2014

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

(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

(e) Intentionallyleft blank

Processes Exercise II

Ideal-Gas Processes 21st July 2014

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

(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

(e) Intentionallyleft blank

For an Isothermal process, Wgas = NkBT ln

(

Vf

Vi

)

Processes Exercise II

Ideal-Gas Processes 21st July 2014

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

(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

(e) Intentionallyleft blank

For an Isothermal process, Wgas = NkBT ln

(

Vf

Vi

)

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

First Law for Ideal Gasses

Ideal-Gas Processes 21st July 2014

First Law of Thermodynamics: Q+W = ∆Eth

First Law for Ideal Gasses

Ideal-Gas Processes 21st July 2014

First Law of Thermodynamics: Q+W = ∆Eth

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

First Law for Ideal Gasses

Ideal-Gas Processes 21st July 2014

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

First Law for Ideal Gasses

Ideal-Gas Processes 21st July 2014

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

Q−Wgas =3

2NkB∆T First Law for an Ideal Gas

First Law for Ideal Gasses

Ideal-Gas Processes 21st July 2014

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

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

Ideal-Gas Processes 21st July 2014

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

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (m3)

p(Pa)

i f

1 3

5

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (m3)

p(Pa)

i f

1 3

5

Q−Wgas =3

2NkB∆T ⇒

Q−Wgas =3

2(2)∆T

(a) Q = 5 J

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

Wgas = (5Pa)(

3m3− 1m3

)

= 10J

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

Wgas = (5Pa)(

3m3− 1m3

)

= 10J

Constant pressure ⇒ tripling thevolume also triples the temperature

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

Wgas = (5Pa)(

3m3− 1m3

)

= 10J

Constant pressure ⇒ tripling thevolume also triples the temperature

⇒ Tf = 7.5K and ∆T = 5 k

First-Law Exercise I

Ideal-Gas Processes 21st July 2014

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 (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

Wgas = (5Pa)(

3m3− 1m3

)

= 10J

Constant pressure ⇒ tripling thevolume also triples the temperature

⇒ Tf = 7.5K and ∆T = 5 k

First Law: Q− 10J = 15J

First-Law Exercise II

Ideal-Gas Processes 21st July 2014

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

First-Law Exercise II

Ideal-Gas Processes 21st July 2014

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

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

First-Law Exercise II

Ideal-Gas Processes 21st July 2014

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

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

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

First-Law Exercise II

Ideal-Gas Processes 21st July 2014

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

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

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

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

First-Law Exercise II

Ideal-Gas Processes 21st July 2014

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

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

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

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

(d) and (e) Intentionally left blank.

First-Law Exercise II

Ideal-Gas Processes 21st July 2014

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

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

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

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

(d) and (e) Intentionally left blank.

First-Law Exercise II

Ideal-Gas Processes 21st July 2014

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

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

First Law: Q−Wgas =3

2NkB∆T

First-Law Exercise II

Ideal-Gas Processes 21st July 2014

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

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

First Law: Q−Wgas =3

2NkB∆T

Adiabatic ⇒ Q = 0 ⇒ −Wgas =3

2NkB∆T

First-Law Exercise II

Ideal-Gas Processes 21st July 2014

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

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

First Law: Q−Wgas =3

2NkB∆T

Adiabatic ⇒ Q = 0 ⇒ −Wgas =3

2NkB∆T

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

Phase Transitions

Ideal-Gas Processes 21st July 2014

Thermal energy, Eth, is NOT temperature!

Phase Transitions

Ideal-Gas Processes 21st July 2014

Thermal energy, Eth, is NOT temperature!

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

Phase Transitions

Ideal-Gas Processes 21st July 2014

Thermal energy, Eth, is NOT temperature!

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

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

Ideal-Gas Processes 21st July 2014

Thermal energy, Eth, is NOT temperature!

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

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