Reversible Process

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# Reversible Process - PowerPoint PPT Presentation

Grains of sand. In limit of infinitesmal changes, system moves through equilibrium states during transition--&gt;reversible process. Reversible Process. Remove one grain of sand at a time. Allow system to equilibrate. Pressure decreases by very small amount.

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Grains of sand

In limit of infinitesmal changes, system moves through

equilibrium states during transition-->reversible process

Reversible Process

Remove one grain of sand at a time.

Allow system to equilibrate

Pressure decreases by very small amount

Volume increases by very small amount.

A reversible process is one where every step of the path is at equilibrium with its surroundings

Examples of Reversible and Irreversible
• Jumping into a pool? (from the perspective of the water)
• Getting into a pool slowly?
• Free expansion of a gas
• Isothermal expansion (piston must move very slowly for bath)
• Moving a glass slowly

Isothermal Reversible Process: T constant: Ideal gas

For an ideal gas U= q + w =0

-w = q

Isothermal Reversible Process: T constant: Ideal gas

For an ideal gas E=q+w=0

w=-q

In rev process Pext=Pgas=P=nRT/V

w=-P1V -P2V- …

Free Expansion

the same final state

the same initial state

Internal energy of an ideal gas

Partition is broken and gas is

Allowed to expand in an adiabatic

(no heat transfer container)

Controlled expansion

(Isothermal Reversible)

Work done

No work done

Different paths

Heat exchanged

No heat exchanged

Is DU is the same for both processes?

Different paths in the Expansion of a gas

Consider the expansion of an ideal gas with various coupling between it (the system) and the surroundings

• Free expansion (No external Pressure)
• Isothermal (Constant Temperature)
• Isobaric (Constant pressure)
• Isochoric (No work)

Thermal reservoir

Change in State at Constant Volume

1)

2)

Substitute in 1) into 2)

What is this at constant volume?

How do we measure heat flow? (calorimeter)

What is

for an Ideal Gas?

Free expansion

Joule originally did a free expansion experiment and found that virtually no heat was transferred between the system and its surroundings.

• U = KE + PE
• U = q + w
• w= pextdV , pext=0, w=0
• No temperature change in surroundings so q=0
• Therefore DU =0, the internal energy is independent of pressure and volume and is only dependent on Temperature

This is a calorimeter

Joule originally did a free expansion experiment and concluded that virtually no heat was transferred.

However, for non-ideal gas some temperature change occur.

The internal energy U is the sum of the kinetic and potential

energies for all the particles that make up the system. Non-ideal

gas have attractive intermolecular forces. So if the internal energy

is constant, the kinetic energies generally decrease (think of the vanderwaals potential). Therefore as the temperature is directly related to molecular kinetic energy, for a non-ideal dilute gas a free expansion results in a drop in temperature.

U(r)

Internal energy of Real Gas

r

van der Waals

interaction

What about for a dense gas?

Joule did his experiments with relatively dilute gases so that their behavior was similar to an ideal gas. How would you expect a denser gas to behave?

Would they become more or less favorable?

If the expansion was done in an adiabatic container what would happen? Would the final temperature be higher, lower or remain the same?

If the Temperature was controlled with a bath would heat be transferred into or out of the system?

Free Expansion

the same final state

the same initial state

Internal energy of an ideal gas

Partition is broken and gas is

Allowed to expand in an adiabatic

(no heat transfer container)

Controlled expansion

(Isothermal Reversible)

Work done

No work done

Different paths

Heat exchanged

No heat exchanged

Is DU is the same for both processes?

Problem

Imagine that an ideal monatomic gas is taken from its initial state A to state B by an isothermal process, from B to C by an isobaric process, and from C back to its initial state A by an isochoric process. Fill in the signs of Q, W, and ¢U for each step.

P,

A

2

0

T=const

B

1

C

1

2

V

Problem

Imagine that an ideal monatomic gas is taken from its initial state A to state B by an isothermal process, from B to C by an isobaric process, and from C back to its initial state A by an isochoric process. Fill in the signs of Q, W, and ¢U for each step.

P

A

2

+ -- 0

T=const

-- + --

B

1

C

+ 0 +

1

2

V

Is this an engine?

heat QI

heat QII

C) QI > QII

A) QI < QII

B) QI = QII

Two other Systems
• Consider the two systems shown to the right. In Case I, the gas is heated at constant volume; in Case II, the gas is heated at constant pressure.

Compare QI , the amount of heat needed to raise the temperature 1K (= 1ºC) insystem I to QII, the amount of heat needed to raise the temperature 1ºC insystem II.

TexPoint fonts used in EMF.

Read the TexPoint manual before you delete this box.: AAAA

heat QI

heat QII

C) QI > QII

A) QI < QII

B) QI = QII

U = Q - Wby

ACT 3 Solution
• Consider the two systems shown to the right. In Case I, the gas is heated at constant volume; in Case II, the gas is heated atconstant pressure.

Compare QI , the amount of heat needed to raise the temperature 1ºC insystem I to QII, the amount of heat needed to raise the temperature 1ºC insystem II.

• Apply the First Law of Thermodynamics to these situations!
• In Case I, NO WORK IS DONE, since the volume does not change.
• In Case II, POSITIVE WORK IS DONE by the system, since the volume is expanding, therefore Wby > 0
• In each case, temperature rise is the same, and so is DU. Therefore, more heat is required in Case II!
Change in State at Constant Volume

1)

2)

Substitute in 1) into 2)

What is this at constant volume?

What do each of these terms mean?

How do we measure heat flow? (calorimeter)

What is

for an Ideal Gas? For a Lennard Jones Gas?

Heat Capacity at Constant Volume

The heat capacity at constant volume CV is defined as:

It takes more energy to increase the temperature of a material that has a higher heat capacity.

Heat and Heat Capacity are both extensive properties. That is, they depend on the amount of material present. We therefore define

Molar Heat Capacity = heat capacity per mole; Cm = C/N

Specific Heat = heat capacity per gram; c = C/mass.

For an ideal gas,

The internal energy for an ideal gas is just dependent on the Temperature so both states at T2 have the same Energy

At constant-volume (no work is done)

At constant-pressure (work is also done)

Heat capacity of an ideal gas at constant Pressure

### The Enthalpy

Nice to have a quantity for Isobaric processes (P = const):

H  U + PV - the enthalpy

This is a Legendre Transform of the Energy

At constant Pressure the change in enthalpy is equal to the heat transferred during the process:

### The Enthalpy

Nice to have a quantity for Isobaric processes (P = const):

H  U + PV - the enthalpy

The enthalpy is a state function, because U, P, and V are state functions. In isobaric processes, the energy received by a system by heating equals to the change in enthalpy.

in both cases, Q does not depend on the path from 1 to 2.

isochoric:

q =  U

isobaric:

q =  H

Consequence: the energy released (absorbed) in chemical reactions at constant volume (pressure) depends only on the initial and final states of a system.

The enthalpy of an ideal gas:

(depends on T only)

CV and CP

the heat capacity at constant volume

V = const

P = const

the heat capacity at constant pressure

To find CP and CV, we need f (P,V,T) = 0 and U = U (V,T)

CV and CP

the heat capacity at constant volume

V = const

P = const

the heat capacity at constant pressure

To find CP and CV, we need f (P,V,T) = 0 and U = U (V,T)

=

U

3

Nk

T

Internal Energy of a Classical ideal gas

• “Classical” means Equipartition Principle applies: each molecule has average energy ½ kT per quadratic modein thermal equilibrium.

At room temperature, for most gases:

• monatomic gas (He, Ne, Ar, …)
• 3 translational modes (x, y, z)

U= 3/2 NkT

diatomic molecules (N2, O2, CO, …)

3 translational modes (x, y, z)

+ 2 rotational modes (wx, wy)

U= 5/2 NkT

• non-linear molecules (H2O, NH3, …)
• 3 translational modes (x, y, z)
• + 3 rotational modes (wx, wy, wz)
• For an ideal gas, the internal energy only depends on the Temperature: U=® NkT = ® nRT
• ( f depends on the type of molecule)
CV and CPforan Ideal Gas

For an ideal gas

CV = dU/dT

CVof one mole of H2

7/2NkB

Vibration

5/2NkB

Rotation

3/2NkB

Translation

# of moles

100

1000

10

T, K

0

( for one mole )

For one mole of a monatomic ideal gas:

E depends only on T

Constant Volume Heat Capacity for monoatomic ideal gas

E depends only on T

Constant Volume Heat Capacity for Diatomic Ideal Gas

Quasistatic Processes in an Ideal Gas
• isochoric ( V = const)

P

2

PV= NkBT2

1

PV= NkBT1

V1,2

V

• isobaric ( P = const )

P

2

1

PV= NkBT2

PV= NkBT1

V1

V2

V

Grains of sand

In limit of infinitesmal changes, system moves through

equilibrium states during transition-->reversible process

Reversible Process

Remove one grain of sand at a time.

Allow system to equilibrate

Pressure decreases by very small amount

Volume increases by very small amount.

A reversible process is one where every step of the path is at equilibrium with its surroundings

Thermodynamic processes of an ideal gas(DU = q - wby )

2

p

Q

Q

1

Temperature changes

FLT:

V

p

• Isobaric (constantpressure)

1

2

p

FLT:

Temperature and volume change

V

• Isochoric (constant volume)

1

2

p

Q

Thermal Reservoir

T

FLT:

V

Volume and pressure change

• Adiabatic (isolated - no heat input)

1

2

p

FLT:

Volume, pressure and temperature change

V

( FLT: DU = q + w )

• Isothermal (constant temperature)
Details of an Adiabatic process (Q = 0)

U, W for an ideal gas:

Know this

derivation.

Integrate:

TaV = constant

Using pV = NkT, we can also write this in the form:

where  = (+1)/

pVg = constant

isotherm : pV = constant

• In an adiabatic process, p, V, and T all change

monatomic gas g =(5/2)/(3/2) = 5/3 = 1.67

diatomic gas g =(7/2)/(5/2) = 7/5 = 1.4

• Adiabatic and isothermal processes are reversible -- no heat flow between objects at different temperatures.
The p-V curve for an Adiabatic process

 = (+1)/

p

V

Reversible Processes

A reversible process is at equilibrium (mechanical and thermal) with its surroundings at all times. The path is always on the surface described by the equation of state.

Examples of Reversible and Irreversible
• Jumping into a pool? (from the perspective of the water)
• Getting into a pool slowly?
• Free expansion of a gas
• Isothermal expansion (piston must move very slowly for bath)
• Moving a glass slowly

Isothermal Reversible Process: T constant: Ideal gas

For an ideal gas \Delta U= q + w =0

-w = q

Isothermal Reversible Process: T constant: Ideal gas

For an ideal gas E=q+w=0

w=-q

In rev process Pext=Pgas=P=nRT/V

w=-P1V -P2V- …

Isothermal Process in an Ideal Gas

P

• isothermal ( T = const ) :

PV= NkBT

W

V2

V1

V

Wi-f > 0 if Vi>Vf (compression)

Wi-f < 0 if Vi<Vf (expansion)

Adiabatic Process in an Ideal Gas

The amount of work needed to change the state of a thermally isolated system depends only on the initial and final states and not on the intermediate states.

P

2

to calculate W1-2 , we need to know P(V,T)

PV= NkBT2

1

PV= NkBT1

V2

V1

V

( f – the # of “unfrozen” degrees of freedom )

Adiabatic Process in an Ideal Gas (cont.)

P

2

An adiabata is “steeper” than an isotherma: in an adiabatic process, the work flowing out of the gas comes at the expense of its thermal energy  its temperature will decrease.

PV= NkBT2

1

PV= NkBT1

V2

V1

V

 1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic)

(again, neglecting the vibrational degrees of freedom)

Problem

Imagine that we rapidly compress a sample of air whose initial pressure is 105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature?

Rapid compression – approx. adiabatic, no time for the energy exchange with the environment due to thermal conductivity

also

- poor approx. for a bike pump, works better for diesel engines

Isothermal Reversible Change
• In an isothermal reversible expansion, temperature does not change
• T is not a function of V and can be removed from the integral
• For ideal gas, w = -nRT ln (Vf/Vi)
• If final volume > initial (expansion), w < 0
• System has done work on surroundings internal energy(U) decreased
• If final volume < initial (compression), w > 0
• System has work done on it, U increased
• Note that as T increases |w| increases
Reversible vs. Irreversible Change
• Irreversible expansion: w= = -pexDV
• Reversible (isothermal) expansion: w=-nRTln{Vf/Vi}
• Which is greater?
• Consider indicator diagram
• Work(rev.) = Area under curve
• Work (irrev.) = Area under rectangle
• Work (rev.) > W (irrev.)
• Reversible work is the maximum which can be done
• True of PV work
• True of all work
Heat Changes - Calorimetry
• Recall: dU = dq + dw
• If expansion work is dwexp and dwother is other work done (electrical, magnetic, etc.), then dw = dwexp + dwother
• dU = dq + dwexp + dwother
• If V is constant, dwexp is zero since no PV work can be done
• Assume dwother is also zero, then dU = dq = qv
• qv is the change in q at constant volume
• Thus, measuring the heat change in a system at constant volume gives a measure of the change in internal energy
• This process of measuring heat change is known as calorimetry
• One constant volume container for measuring heat change is a bomb calorimeter
• Experiment 1 in lab
• Typically, substance is burned in calorimeter and temperature rise is measured (dV is constant, but P does change in bomb)
• In a bomb calorimeter, DT a qv
• This proportionality is quantified by calibration, typically by combusting a known substance
Heat Changes - Heat Capacity
• If V constant, U increases as temperature increases
• The rate of change of U at any temperature, (dU/dT)V is called the heat capacity, CV
• CV(A) is not equal to Cv(B) generally smaller if TA < TB
• Note volume is const. If that changes CV(T may change)
• CV is an extensive property (2x the amt gives twice the heat capacity)
• Molar heat capacity, CVm, is the
• Rate of change of CV with T typically small around room temperature, you can assume it’s constant
• This means that dU = qV= CV dT ≈ CVDT
• You can estimate CV by determining the amount of heat supplied to a sample
• Because qV ≈ CVDT, for a given amount of heat the larger CV, the smaller DT
• At a phase transition (boiling point) DT = 0 so CV = ∞

Grains of sand

In limit of infinitesmal changes, system moves through

equilibrium states during transition-->reversible process

Reversible Process

Remove one grain of sand at a time.

Allow system to equilibrate

Pressure decreases by very small amount

Volume increases by very small amount.

A reversible process is one where every step of the path is at equilibrium with its surroundings

Examples of Reversible and Irreversible
• Jumping into a pool? (from the perspective of the water)
• Getting into a pool slowly?
• Free expansion of a gas
• Isothermal expansion (piston must move very slowly for bath)
• Moving a glass slowly

Isothermal Reversible Process: T constant: Ideal gas

For an ideal gas \Delta U= q + w =0

-w = q

Isothermal Reversible Process: T constant: Ideal gas

For an ideal gas E=q+w=0

w=-q

In rev process Pext=Pgas=P=nRT/V

w=-P1V -P2V- …

¢ U depends only on ¢T

Constant Volume Heat Capacity for monoatomic ideal gas

E depends only on T

Constant Volume Heat Capacity for Diatomic Ideal Gas

P

Another Problem

Pi

During the ascent of a meteorological helium-gas filled balloon, its volume increases from Vi = 1 m3 to Vf = 1.8 m3, and the pressure inside the balloon decreases from 1 bar (=105 N/m2) to 0.5 bar. Assume that the pressure changes linearly with volume between Vi and Vf.

(a) If the initial T is 300K, what is the final T?

(b) How much work is done by the gas in the balloon?

(c) How much “heat” does the gas absorb, if any?

Pf

Vi

Vf

V

(a)

(b)

- work done on a system

- work done by a system

(c)

Quasistatic Processes in an Ideal Gas
• isochoric ( V = const)

P

2

PV= NkBT2

1

PV= NkBT1

(see the last slide)

V1,2

V

• isobaric ( P = const )

P

2

1

PV= NkBT2

PV= NkBT1

V1

V2

V

Isothermal Process in an Ideal Gas

P

• isothermal ( T = const ) :

PV= NkBT

W

V2

V1

V

Wi-f > 0 if Vi>Vf (compression)

Wi-f < 0 if Vi<Vf (expansion)

Adiabatic Process in an Ideal Gas

The amount of work needed to change the state of a thermally isolated system depends only on the initial and final states and not on the intermediate states.

P

2

to calculate W1-2 , we need to know P(V,T)

PV= NkBT2

1

PV= NkBT1

V2

V1

V

( f – the # of “unfrozen” degrees of freedom )

Adiabatic Process in an Ideal Gas (cont.)

P

2

An adiabata is “steeper” than an isotherma: in an adiabatic process, the work flowing out of the gas comes at the expense of its thermal energy  its temperature will decrease.

PV= NkBT2

1

PV= NkBT1

V2

V1

V

 1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic)

(again, neglecting the vibrational degrees of freedom)

Problem

Imagine that we rapidly compress a sample of air whose initial pressure is 105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature?

Rapid compression – approx. adiabatic, no time for the energy exchange with the environment due to thermal conductivity

also

- poor approx. for a bike pump, works better for diesel engines

Isothermal Reversible Change
• In an isothermal reversible expansion, temperature does not change
• T is not a function of V and can be removed from the integral
• For ideal gas, w = -nRT ln (Vf/Vi)
• If final volume > initial (expansion), w < 0
• System has done work on surroundings internal energy(U) decreased
• If final volume < initial (compression), w > 0
• System has work done on it, U increased
• Note that as T increases |w| increases
Reversible vs. Irreversible Change
• Irreversible expansion: w= = -pexDV
• Reversible (isothermal) expansion: w=-nRTln{Vf/Vi}
• Which is greater?
• Consider indicator diagram
• Work(rev.) = Area under curve
• Work (irrev.) = Area under rectangle
• Work (rev.) > W (irrev.)
• Reversible work is the maximum which can be done
• True of PV work
• True of all work
Heat Changes - Calorimetry
• Recall: dU = dq + dw
• If expansion work is dwexp and dwother is other work done (electrical, magnetic, etc.), then dw = dwexp + dwother
• dU = dq + dwexp + dwother
• If V is constant, dwexp is zero since no PV work can be done
• Assume dwother is also zero, then dU = dq = qv
• qv is the change in q at constant volume
• Thus, measuring the heat change in a system at constant volume gives a measure of the change in internal energy
• This process of measuring heat change is known as calorimetry
• One constant volume container for measuring heat change is a bomb calorimeter
• Experiment 1 in lab
• Typically, substance is burned in calorimeter and temperature rise is measured (dV is constant, but P does change in bomb)
• In a bomb calorimeter, DT a qv
• This proportionality is quantified by calibration, typically by combusting a known substance
Heat Changes - Heat Capacity
• If V constant, U increases as temperature increases
• The rate of change of U at any temperature, (dU/dT)V is called the heat capacity, CV
• CV(A) is not equal to Cv(B) generally smaller if TA < TB
• Note volume is const. If that changes CV(T may change)
• CV is an extensive property (2x the amt gives twice the heat capacity)
• Molar heat capacity, CVm, is the
• Rate of change of CV with T typically small around room temperature, you can assume it’s constant
• This means that dU = qV= CV dT ≈ CVDT
• You can estimate CV by determining the amount of heat supplied to a sample
• Because qV ≈ CVDT, for a given amount of heat the larger CV, the smaller DT
• At a phase transition (boiling point) DT = 0 so CV = ∞
Heat Changes - Enthalpy
• If PV work can be done by a system, at constant pressure, V changes
• Some of the internal energy lost through work
• U=q-pV or q = U+ pV
• dU < dq, if work is done; dU=dq, if no work is done
• Define new variable, H, such that H= U + pV
• H is called enthalpy
• Like U, H is a state function
• At constant p, dH = dU +pdV
• This is an important function because most experiments are carried out at constant pressure, e.g., atmospheric pressure
• Enthalpy is the heat supplied at constant pressure
• If pdV=0, dH=dq or DH=qP
• As with Cv, Cp is defined as dqp/dt = (dH/dT)p
Measuring Enthalpy Changes
• Isobaric calorimeter measures heat changes at constant pressure
• Unlike bomb calorimeter in which volume constrained
• You can use a bomb calorimeter to measure enthalpy by converting DU to DH
• See Experiment 1
• Assume molar enthalpy and molar internal energy are nearly the same
• True for most solids and liquids
• If process involves only solids and liquids DU ≈ DH
• Calculate pV term based on density differences (Example 2.2)
• Differential scanning calorimeter (see Atkins, p. 46)
• Themogram: plot of Cp vs. temperature
• Peaks correspond to changes in enthalpy
• Area in the peak = DH
• Typically small samples (mg) and high temperatures reached

exothermic

Glass transition

melting

crystallization

endothermic

Typical DSC

Thermogram

Cp(dH/dt) mJ/s

Temperature (K)

Source: www.chem.neu.edu/Courses/U331/CHM1222DSC.ppt

Enthalpy Calculations
• Ideal Gas
• Since PV term for ideal gas is nRT, DH = DU +pV = DU +nRT
• If a reaction changes the amount of gas, PV term becomes (nfinal -ninitial)RT or

DH = DU +DngRT, where Dng= nfinal -ninitial

DH - DU = DngRT

• Note: sign Dng of is important
• Enthalpy changes can be directly related to heat input

Example 2.3: Calculate molar internal energy change and enthalpy changes at 373.15 when 0.798 g of water is vaporized by passing 0.5A @ 12V through a resistance in contact with the water for 300 seconds

Change of Enthalpy with Temperature (Cp)
• The instantaneous slope of the plot of enthalpy vs temperature is the heat capacity at constant pressure
• Extensive property
• Cp,m, the molar heat capacity is an intensive property
• As with U and Cv, dH = Cvdt, or for measurable intervals DH=CvDT
• Also, qp = CvDT, because at constant pressure DH= qp
• This assumes Cv is constant over the temperature range
• True if range is small, esp. for noble gases
• More generally Cv = a + bT +(c/T2)

a, b,and c are constants depending on the substance

• If Cv is not constant with T, DH ≠ CvDT

Now what happens????

Variation of Cp with Temperature
• If Cp varies with temperature, you must integrate over the temperature limits of interest.
• Example: Assume Cv = a + bT +(c/T2)
• Try Self Test 2.5
• In most cases Cp > Cv
• For ideal gases, Cp = Cv +nR
• This amounts to ~8 JK-1mol-1 difference
Adiabatic expansion of an ideal gas
• Work is done, U decreases, hence T decreases
• To calculate the final state (Tf and Vf) assume a two step process
• Remember U is a state function, it doesn’t matter how you get there
• Step 1: Volume is changed and temperature held constant
• Internal energy change = 0 since it’s independent of volume the molecules occupy
• Step 2: Temperature changed from Ti to Tf
• Adiabatic so q = 0
• If Cv is independent of temperature, adiabatic work = wad= CvDT
• DU = q + wad = 0 + CvDT= CvDT
• This says that for an adiabatic expansion the work done is only proportional to the initial and final temperatures
• The relationship between initial and final volumes can be derived using what we’ve learned about reversible adiabatic expansions and perfect gases
• Given dq = 0 (adiabatic) and dw = -pdV (reversible expansion)
• dU = dq + dw =-pdV [1]
• For perfect gas dU = CvdT [2]
• Thus, combining [1] and [2], CvdT = -pdV [3]
• For an ideal gas, pV = nRT, so [3] becomes CvdT = -(nRT/V)dV [4]

or, rearranging, (Cv/T)dT = -(nR/V)dV [5]

• To obtain the relationship between Ti, Vi and Ti, Vi [5] must be integrated
• Ti corresponds to Vi
• Cv independent of T
• The adiabatic work, wad= CvDT, can thus be calculated once this relationship is found
• This relationship means that the product, pVg doesn’t change as you go through an adiabatic expansion
• For ideal gas, g,the ratio of the heat capacities is >1 since Cp,m = Cv,m +R

g= Cp,m /Cv,m = (Cv,m+R) /Cv,m

g= 1 +(R /Cv,m)

• For monotomic gas, Cv,m= 3/2 R, so g= 5/2
• For monotomic gas, Cv,m= 3 R, so g= 4/3
• Plot of P vs V for an adiabatic change is called an adiabat
• p a Vg, where g>1
• Recall isotherm, p a V
• Adiabats fall more steeply than isotherms
The First Law and Chemistry
• The study of heat produced from chemical reactions is called thermochemistry
• We can apply what we learned from our 1st law considerations to the system of reactants and products in a chemical reaction
• In a reaction, heat either flows into the system (endothermic reactions) or out of the system (exothermic reactions)
• At constant pressure (reactions done in a beaker for example), the first law tells us for
• Endothermic reactions DH>0 {heat flows in}
• Exothermic reactions DH<0 {heat flows out}
• Standard enthalpy change,DHø, is the change in enthalpy for substances in their standard state
• Standard state is the form of the substance at 1bar (~1atm) and specified temperature
• DHø298 is the standard enthalypy change at 1bar and 298K (~RT)
Enthalpies of Physical Change
• Changes in state (liquid to gas, solid to liquid, solid to gas, etc.) have enthalpies associated with the transition, DtransHø
• Liquid-to-Gas, vaporization: DvapHø
• Solid-to-Liquid, fusion: DfusHø
• Solid-to-Gas, sublimation: DsubHø
• Table 2.4 gives symbols for a variety of transitions
• Because transition temperature is an important point, transition temperatures are often listed at this point
• May like to know them at convenient temperature as well
• Recall that enthalpy is a state function. Thus, value is path independent.
• Can construct information if you don’t have it by making a path and summing the steps
• Example: You need the enthalpy of sublimation, but can’t find it

Path:

• Step 1 - solid to liquid (DfusHø)
• Step 2 - liquid to gas (DvapHø)
• Step 3 - sum of steps1 & 2
• Reverse of process involves only change of sign
Enthalpies of Chemical Change
• Like state transitions reactions have enthalpies associated with them
• Can write a thermochemical equation
• Standard chemical equation plus enthalpy change
• Reactants and products in standard states ® enthalpy is standard enthalpy of reaction, DrHø
• Typically, you must calculate DrHø
• Again because enthapy is a state function you create a path using enthalpies you know or can find and sum them
• Usually a table of standard free energies of formation of components are available
• Given per mole of substance, DrHøm
• Recall that decomposition is the same as formation except sign is reversed
• For aA + bB ® cC + dD,

DrHøm = cDrHøm(C) + dDrHøm(C) - aDrHøm(A) - bDrHøm(D)

• Generally,

Remember state is important!

• More generally, the standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which it may be divided - Hess’s Law
• Consequence of the fact that enthalpy is a state function
• Component reactions could be hypothetical
Standard Enthalpies of Formation , DfHø
• As above, the std enthalpy of formation refers to compound in standard state (1bar, typically 298K although other temperature could be specified)
• Enthalpies of reaction can be determined from enthalpies of formation as before
• Enthalpies of formation can be estimated by considering contributions from individual groups and summing
• Molecular modeling can be used to modify this to consider steric effects
• Not quantitatively reliable but qualitatively o.k.
Temperature Dependence of DrHøKirkoff’s Law
• To be accurate DfHø needs to be measured at temperature of interest
• If not available, can be estimated as follows:
• This equation is known as Kirkoff’s Law
• Assumes Cpø is independent of temperature (good approx. esp. since differences are being considered
• Can use functional dependence of Cp, if known (we showed this earlier)
• Best to use T1 as close to T2 as you can get
System & Surroundings

surroundings

system

• System - volume of interest (reaction vessel, test tube, biological cell, atmosphere, etc.)
• Surroundings volume outside system
• Open system - matter can pass between system & surroundings
• Closed system - matter cannot pass between system & surroundings
• Isolated system - Neither matter nor energy can pass between system & surroundings
• No mechanical or thermal contact

Energy

Matter

Open System

surroundings

system

Energy

Closed System

surroundings

system

Isolated System

TexPoint fonts used in EMF.

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System and surroundings

The "system" is the part of the universe you are looking at in an experiment.

The "surroundings" is the rest of the universe.

Energy is conserved for "system+surroundings", not for the "system" separately.

TexPoint fonts used in EMF.

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System and Surroundings

Adiabatic - The orange walls insulate the system completely (no heat can pass through)

Isochoric - Literally means no work. For our purposes it means the Volume is constant.

Isothermal - The thermal reservoir keeps the system at a constant temperature

Free Expansion - The piston offers no resistance (there is vacuum above the brown piston)

Isobaric - The force on the piston from above is constant (for example atmospheric pressure)

Thermal reservoir

### The First Law of Thermodynamics

Signs for heat (q) and work (w)

Thermodynamic transformations

q is + if heat is added to system

q is – if heat is lost by system

w is + if work is done on the system.

w is – if work is done by the system.

The classical expression of the First Law

Work done by the system decreases its internal energy.

Heat into the system increases its internal energy.

Work is “usable energy” – the capacity to push a piston or a turbine, or to exert a force, F, over a distance, dl.

Where the force is pressure x area.

Increasing volume lowers Eint.

The classical expression of the First Law

The First Law of thermodynamics is most often written in terms of the change in the internal energy of a system as

Work done by the system decreases its internal energy.

Heat into the system increases its internal energy.

Where the force is pressure x area.

Increasing volume lowers Eint.

Internal energy

Tentative definition:

The internal energy U of a system is the sum of the kinetic

energies of all of its constituent particles, plus the sum of

all the potential energies of interactions among these

particles.

• The first law ofthermodynamics

What is Energy

• Infinitesimal changes in state
What is Heat?
• Up to mid-1800’s heat was considered a fluid that flowed from hot objects to cold objects.
• A better definition is that it is the energy that flows from hot bodies to cold bodies in order to establish thermal equilibrium. This is called thermal energy.
• The term Heat (Q) is properly used to describe energy in transit, that is thermal energy transferred into or out of a system. Heat only makes sense in the context of the process of the transformation from one state to another. It is not a state or equilibrium property of the system.

f

2

i

1

p

V

c) < W1

b) = W1

a) > W1

f

a

2

i

p

V

Act 4: Work along different paths
• Consider the two paths, 1 and 2, connecting points i and f on the pV diagram.
• The work W2 done by the system in going from i to f along path 2 is:

Consider the first part of path 2 (i to a): the work done by the system is negative (V is decreasing) and equal in magnitude to the area under the curve.

Now consider the second part of path 2 (a to f): the work done by the system is positive (V is increasing) and equal in magnitude to the area under the curve

The sum of the work is just the area between the curves (i to a) and (a to f)

The corresponding area for path 1 is larger so |W2| < |W1|.

Thermodynamic processes of an ideal gas(DU = q - wby )

2

p

Q

Q

1

Temperature changes

FLT:

V

p

• Isobaric (constantpressure)

1

2

p

FLT:

Temperature and volume change

V

• Isochoric (constant volume)

1

2

p

Q

Thermal Reservoir

T

FLT:

V

Volume and pressure change

• Adiabatic (isolated - no heat input)

1

2

p

FLT:

Volume, pressure and temperature change

V

( FLT: DU = q + w )

• Isothermal (constant temperature)
Details of an Adiabatic process (Q = 0)

U, W for an ideal gas:

Know this

derivation.

Integrate:

TaV = constant

Using pV = NkT, we can also write this in the form:

where  = (+1)/

pVg = constant

isotherm : pV = constant

• In an adiabatic process, p, V, and T all change

monatomic gas g =(5/2)/(3/2) = 5/3 = 1.67

diatomic gas g =(7/2)/(5/2) = 7/5 = 1.4

• Adiabatic and isothermal processes are reversible -- no heat flow between objects at different temperatures.
The p-V curve for an Adiabatic process

 = (+1)/

p

V

Example problem 1: Kinetic Energy of a Gas

A small room at room temperature (T=300 K) and atmospheric pressure measures 3.0 m x 2.4 m x 5.2 m.

1) Estimate both the total translational kinetic energy associated with the molecules in the room, and the number of molecules in the room.

2) If we assume these molecules are primarily nitrogen (N2) molecules, what is their average speed?

Example problem 1: Kinetic Energy of a Gas

A small room at room temperature (T=300 K) and atmospheric pressure measures 3.0 m x 2.4 m x 5.2 m.

1) Estimate both the total translational kinetic energy associated with the molecules in the room, and the number of molecules in the room.

2) If we assume these molecules are primarily nitrogen (N2) molecules, what is their average speed?

Example problem 1: Kinetic Energy of a Gas

A small room at room temperature (T=300 K) and atmospheric pressure measures 3.0 m x 2.4 m x 5.2 m.

1) Estimate both the total translational kinetic energy associated with the molecules in the room, and the number of molecules in the room.

2) If we assume these molecules are primarily nitrogen (N2) molecules, what is their average speed?

How many molecules are in the room?

If these are N2 molecules, what is their average speed?

Example problem 1: Kinetic Energy of a Gas

A small room at room temperature (T=300 K) and atmospheric pressure measures 3.0 m x 2.4 m x 5.2 m.

1) Estimate both the total translational kinetic energy associated with the molecules in the room, and the number of molecules in the room.

2) If we assume these molecules are primarily nitrogen (N2) molecules, what is their average speed?

How many molecules are in the room?

If these are N2 molecules, what is their average speed?

<v> = 515 m/s

Equipartition of translational kinetic energy. The rotational energy of nitrogen molecules is not important here.

Helium doesn’t form molecules.

KE = mv2/2 so the speed ratio is the inverse square root of the mass ratio.

The actual ratio of sound speeds is 2.76.

<KE>He = <KE>N2

mHe/mN2 = 4/28

vHe/vN2 = (28/4)1/2 = 2.65

You expect the ratio of sound speeds to be approximately the same as this ratio.

Example Problem 2

The speed of sound in a gas is roughly equal to the average velocity of the particles. Compare the speed of a typical helium atom (mHe = 4amu) to that of a typical nitrogen molecule (mN2 = 2 x 14amu) in a gas mixture in thermal equilibrium. What do you expect the ratio of sound speeds in pure helium and nitrogen to be?

Solution:

A Joule is a Newton×meter, which is the same as a Pascal×meter3.

Convert Pascals to atmospheres,

and meter3 to liters.

Arrange the factors so the units cancel correctly.

1 Atm = 1.013´105 Pa

1 liter = 0.001 m3

R = 8.314 Pa×m3/ mol ×K/(0.001 m3/ liter) / (1.013´105 Pa/Atm) = 0.0821 liter ×atm/ mol ×K

Example Problem 3

Convert the gas constant, R = 8.31 J/mol×K, to units where volume is measured in liters, and pressure is measured in atmospheres. That is, if you write R = x liter×atm/mol×K, calculate x.

Solution:

Simplify using GMEm/rE2 = g = 9.8 m/s2. Use

rE = 6.4´106 m (4000 mi), and mN2 = 4.7´10-26 kg.

Equipartition tells us that <KE> = 3kT/2. That’s hot!

The mass of a helium atom is smaller by a factor of 4/28, so both KE and T are reduced by the same factor.

That’s still hot, but it turns out (for reasons you’ll learn later in the course) to be low enough that all the helium has escaped from the Earth’s atmosphere.

KE = GMEm/rE

= gmrE

= 2.9´10-18 J

TN2 = 2<KE>/3k = 1.4´105 K

THe = 2´104 K.

Example Problem 4

How much kinetic energy must a nitrogen molecule have in order to escape from the Earth’s gravity, starting at the surface? Ignore collisions with other air molecules. How about a helium atom? At what temperatures will the average molecule of each kind have enough energy to escape?

Solution:

P

Another Problem

Pi

During the ascent of a meteorological helium-gas filled balloon, its volume increases from Vi = 1 m3 to Vf = 1.8 m3, and the pressure inside the balloon decreases from 1 bar (=105 N/m2) to 0.5 bar. Assume that the pressure changes linearly with volume between Vi and Vf.

(a) If the initial T is 300K, what is the final T?

(b) How much work is done by the gas in the balloon?

(c) How much “heat” does the gas absorb, if any?

Pf

Vi

Vf

V

(a)

(b)

- work done on a system

- work done by a system

(c)