Idea Transcript
• Temperature ~ Average KE of each particle • Particles have different speeds • Gas Particles are in constant RANDOM motion • Average KE of each particle is: 3/2 kT • Pressure is due to momentum transfer Speed ‘Distribution’ at CONSTANT Temperature is given by the Maxwell Boltzmann Speed Distribution
Mean Free Path A single molecule follows a zig-zag path through a gas as it collides with other molecules. The average distance between the collisions is called the mean free path:
(N/V) is the number density of the gas in m−3. r is the the radius of the molecules when modeled as hard spheres; for many common gases r ≈ 10−10 m. © 2013 Pearson Education, Inc.
Slide 18-20
QuickCheck 18.1
The temperature of a rigid container of oxygen gas (O2) is lowered from 300°C to 0°C. As a result, the mean free path of oxygen molecules A. Increases. B. Is unchanged. C. Decreases.
© 2013 Pearson Education, Inc.
Slide 18-21
QuickCheck 18.1
The temperature of a rigid container of oxygen gas (O2) is lowered from 300°C to 0°C. As a result, the mean free path of oxygen molecules A. Increases. B. Is unchanged. C. Decreases.
© 2013 Pearson Education, Inc.
λ depends only on N/V, not T.
Slide 18-22
Pressure and Kinetic Energy • •
•
•
•
•
Assume a container is a cube with edges d. Look at the motion of the molecule in terms of its velocity components and momentum and the average force Pressure is proportional to the number of molecules per unit volume (N/V) and to the average translational kinetic energy of the molecules. This equation also relates the macroscopic quantity of pressure with a microscopic quantity of the average value of the square of the molecular speed One way to increase the pressure is to increase the number of molecules per unit volume The pressure can also be increased by increasing the speed (kinetic energy) of the molecules
___ 2 N 1 2 P = mo v 3 V 2
Molecular Interpretation of Temperature • We can take the pressure as it relates to the kinetic energy and compare it to the pressure from the equation of state for an ideal gas 2 N 1 ___2 = = NkBT P v nRT m= 3 V 2
• Temperature is a direct measure of the average molecular kinetic energy
Molecular Interpretation of Temperature • Simplifying the equation relating temperature and kinetic energy gives ___ 1 3 2 mo v = kBT 2 2
• This can be applied to each direction, 1 ___2 1 m v x = kBT 2 2
– with similar expressions for vy and vz
Total Kinetic Energy • The total kinetic energy is just N times the kinetic energy of each molecule 1 ___2 3 3 K = N = mv = NkBT nRT tot trans 2 2 2 • If we have a gas with only translational energy, this is the internal energy of the gas • This tells us that the internal energy of an ideal gas depends only on the temperature
QuickCheck 18.4
A rigid container holds both hydrogen gas (H2) and nitrogen gas (N2) at 100°C. Which statement describes the average translational kinetic energies of the molecules? A. єavg of H2 < єavg of N2. B. єavg of H2 = єavg of N2. C. єavg of H2 > єavg of N2.
© 2013 Pearson Education, Inc.
Slide 18-37
QuickCheck 18.4
A rigid container holds both hydrogen gas (H2) and nitrogen gas (N2) at 100°C. Which statement describes the average translational kinetic energies of the molecules? A. єavg of H2 < єavg of N2. B. єavg of H2 = єavg of N2. C. єavg of H2 > єavg of N2.
© 2013 Pearson Education, Inc.
Slide 18-38
Kinetic Theory Problem A 5.00-L vessel contains nitrogen gas at 27.0°C and 3.00 atm. Find (a) the total translational kinetic energy of the gas molecules and (b) the average kinetic energy per molecule.
Hot Question Suppose you apply a flame to 1 liter of water for a certain time and its temperature rises by 10 degrees C. If you apply the same flame for the same time to 2 liters of water, by how much will its temperature rise? a) 1 degree
b) 5 degrees
c) 10 degrees
d) zero degrees
Ludwig Boltzmann or Dean Gooch? • 1844 – 1906 • Austrian physicist • Contributed to – Kinetic Theory of Gases – Electromagnetism – Thermodynamics
• Pioneer in statistical mechanics
Distribution of Molecular Speeds •
• •
The observed speed distribution of gas molecules in thermal equilibrium is shown at right NV is called the Maxwell-Boltzmann speed distribution function mo is the mass of a gas molecule, kB is Boltzmann’s constant and T is the absolute temperature
mo NV = 4π N k T π 2 B
3/2 2
v e
− mv 2 / 2 kBT
Molecular Speeds and Collisions
Speed Summary • Root mean square speed
= v rms
kT 3kBT = 1.73 B mo mo
• The average speed is somewhat lower than the rms speed 8k T kT = v avg
B = 1.60 B mo π mo
• The most probable speed, vmp is the speed at which the distribution curve reaches a peak • vrms > vavg > vmp
= v mp
2kBT kT = 1.41 B m m
Some Example vrms Values
At a given temperature, lighter molecules move faster, on the average, than heavier molecules
Speed Distribution • The peak shifts to the right as T increases – This shows that the average speed increases with increasing temperature
• The asymmetric shape occurs because the lowest possible speed is 0 and the highest is infinity
The Kelvin Temperature of an ideal gas is a measure of the average translational kinetic energy per particle:
3 / 2kT = KE = 1/ 2mv
2 rms
k =1.38 x 10-23 J/K Boltzmann’s Constant Root-mean-square speed:
v= rms
= v 2
3kT m
Kinetic Theory Problem Calculate the RMS speed of an oxygen molecule in the air if the temperature is 5.00 °C. The mass of an oxygen molecule is 32.00 u (k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)
vrms
3kT = m
What is m?
m is the mass of one oxygen molecule in kg.
What is u? How do we get the mass in kg?
QuickCheck 18.3
A rigid container holds both hydrogen gas (H2) and nitrogen gas (N2) at 100°C. Which statement describes their rms speeds? A. vrms of H2 < vrms of N2. B. vrms of H2 = vrms of N2. C. vrms of H2 > vrms of N2.
© 2013 Pearson Education, Inc.
Slide 18-34
QuickCheck 18.3
A rigid container holds both hydrogen gas (H2) and nitrogen gas (N2) at 100°C. Which statement describes their rms speeds? A. vrms of H2 < vrms of N2. B. vrms of H2 = vrms of N2. C. vrms of H2 > vrms of N2.
© 2013 Pearson Education, Inc.
Slide 18-35
Kinetic Theory Problem Calculate the RMS speed of an oxygen molecule in the air if the temperature is 5.00 °C. The mass of an oxygen molecule is 32.00 u (k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)
vrms
3kT = m
What is m?
m is the mass of one oxygen molecule.
−23
3(1.38 x10 J / K )278 K = (32u )(1.66 x10−27 kg / u ) = 466m / s
Is this fast?
YES!
Speed of sound: 343m/s!
A cylinder contains a mixture of helium and argon gas in equilibrium at 150°C. (a) What is the average kinetic energy for each type of gas molecule? (b) What is the root-mean-square speed of each type of molecule?
More Kinetic Theory Problems A gas molecule with a molecular mass of 32.0 u has a speed of 325 m/s. What is the temperature of the gas molecule? A) 72.0 K
B) 136 K
C) 305 K
D) 459 K
E) A temperature cannot be assigned to a single molecule.
Temperature ~ Average KE of all particles
Equipartition of Energy •
•
• • • •
Each translational degree of freedom contributes an equal amount to the energy of the gas – In general, a degree of freedom refers to an independent means by which a molecule can possess energy Each degree of freedom contributes ½kBT to the energy of a system, where possible degrees of freedom are those associated with translation, rotation and vibration of molecules With complex molecules, other contributions to internal energy must be taken into account One possible energy is the translational motion of the center of mass Rotational motion about the various axes also contributes There is kinetic energy and potential energy associated with the vibrations
Monatomic and Diatomic Gases The thermal energy of a monatomic gas of N atoms is
A diatomic gas has more thermal energy than a monatomic gas at the same temperature because the molecules have rotational as well as translational kinetic energy.
Molar Specific Heat • We define specific heats for two processes that frequently occur: – Changes with constant pressure – Changes with constant volume
• Using the number of moles, n, we can define molar specific heats for these processes • Molar specific heats: – Q = nCV DT for constant-volume processes – Q = nCP DT for constant-pressure processes
Ideal Monatomic Gas • Therefore, ∆Eint = 3/2 nRT − ∆E is a function of T only
• In general, the internal energy of an ideal gas is a function of T only – The exact relationship depends on the type of gas
• At constant volume, Q = ∆Eint = nCV ∆T – This applies to all ideal gases, not just monatomic ones
Ratio of Molar Specific Heats • We can also define the ratio of molar specific heats CP 5R / 2 = γ = = 1.67 CV 3R / 2 • Theoretical values of CV , CP , and γ are in excellent agreement for monatomic gases • But they are in serious disagreement with the values for more complex molecules – Not surprising since the analysis was for monatomic gases
Agreement with Experiment • • •
•
Molar specific heat is a function of temperature At low temperatures, a diatomic gas acts like a monatomic gas CV = 3/2 R At about room temperature, the value increases to CV = 5/2 R – This is consistent with adding rotational energy but not vibrational energy At high temperatures, the value increases to CV = 7/2 R – This includes vibrational energy as well as rotational and translational
Sample Values of Molar Specific Heats
In a constant-volume process, 209 J of energy is transferred by heat to 1.00 mol of an ideal monatomic gas initially at 300 K. Find (a) the increase in internal energy of the gas, (b) the work done on it, and (c) its final temperature
Molar Specific Heats of Other Materials • The internal energy of more complex gases must include contributions from the rotational and vibrational motions of the molecules • In the cases of solids and liquids heated at constant pressure, very little work is done, since the thermal expansion is small, and CP and CV are approximately equal
Adiabatic Processes for an Ideal Gas • An adiabatic process is one in which no energy is transferred by heat between a system and its surroundings (think styrofoam cup) • Assume an ideal gas is in an equilibrium state and so PV = nRT is valid • The pressure and volume of an ideal gas at any time during an adiabatic process are related by PV γ = constant γ = CP / CV is assumed to be constant All three variables in the ideal gas law (P, V, T ) can change during an adiabatic process
Special Case: Adiabatic Free Expansion • This is an example of adiabatic free expansion • The process is adiabatic because it takes place in an insulated container • Because the gas expands into a vacuum, it does not apply a force on a piston and W = 0 • Since Q = 0 and W = 0, ∆Eint = 0 and the initial and final states are the same and no change in temperature is expected. – No change in temperature is expected
Reading Question 18.2
What additional kind of energy makes CV larger for a diatomic than for a monatomic gas?
A. B. C. D. E.
Charismatic energy. Translational energy. Heat energy. Rotational energy. Solar energy.
Slide 18-12
Reading Question 18.2
What additional kind of energy makes CV larger for a diatomic than for a monatomic gas?
A. B. C. D. E.
Charismatic energy. Translational energy. Heat energy. Rotational energy. Solar energy.
Slide 18-13
Reading Question 18.3
The second law of thermodynamics says that A. The entropy of an isolated system never decreases. B. Heat never flows spontaneously from cold to hot. C. The total thermal energy of an isolated system is constant. D. Both A and B. E. Both A and C. Slide 18-15
Reading Question 18.4
In general, A. Both microscopic and macroscopic processes are reversible. B. Both microscopic and macroscopic processes are irreversible. C. Microscopic processes are reversible and macroscopic processes are irreversible. D. Microscopic processes are irreversible and macroscopic processes are reversible.
Slide 18-16
QuickCheck 18.6 Systems A and B are both monatomic gases. At this instant,
A.
TA > TB.
B.
TA = TB.
C.
TA < TB.
D.
There’s not enough information to compare their temperatures. Slide 18-51
QuickCheck 18.6 Systems A and B are both monatomic gases. At this instant,
A.
TA > TB.
B.
TA = TB.
C.
TA < TB.
D.
There’s not enough information to compare their temperatures.
A has the larger average energy per atom.
Slide 18-52
Adiabatic Process • The PV diagram shows an adiabatic expansion of an ideal gas • The temperature of the gas decreases – Tf < Ti in this process
• For this process Pi Viγ = Pf Vfγ and Ti Viγ-1 = Tf Vfγ-1
A 2.00-mol sample of a diatomic ideal gas expands slowly and adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L. (a) What is the final pressure of the gas? (b) What are the initial and final temperatures? (c) Find Q, W, and ∆Eint.
Important Concepts
Thermo Processes • Adiabatic – No heat exchanged – Q = 0 and ∆Eint = W • Isobaric – Constant pressure – W = P (Vf – Vi) and ∆Eint = Q + W • Isovolumetric – Constant Volume – W = 0 and ∆Eint = Q • Isothermal – Constant temperature ∆Eint = 0 and Q = -W
∆Eint =Q + W
Vi W = nRT ln Vf
Cyclic Processes ∆Eint =Q + W = 0
• A cyclic process is one that starts and ends in the same state – On a PV diagram, a cyclic process appears as a closed curve
• ∆Eint = 0, Q = -W • In a cyclic process, the net work done on the system per cycle equals the area enclosed by the path representing the process on a PV diagram