Monatomic and Diatomic Gases the Thermal Energy of a Monatomic Gas of N Atoms Is

Home , KT (energy), Monatomic gas

• 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.

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.

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. λ depends only on N/V, not T. C. Decreases.

• 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 21N 2 increase the number of molecules per P  mo v unit volume 32V  • The pressure can also be increased by increasing the speed (kinetic energy) of the molecules 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

___ 21N2 nRT NkB T P mv   32VVV • Temperature is a direct measure of the average molecular kinetic energy 13___ m v2  k T 22o B The Kelvin Temperature of an ideal gas is a measure of the average translational kinetic energy per particle:

___ 12 3 3 3 Ktot trans N mv  Nk B T  nRT  PV 2 2 2 2 k =1.38 x 10-23 J/K Boltzmann’s Constant

2 3kT Root-mean-square speed: vv rms m Total Kinetic Energy

• The total kinetic energy is just N times the kinetic energy of each molecule ___ 12 3 3 Ktot trans N mv  Nk B T  nRT 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 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. 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)

3kT m is the mass of one v  What is m? oxygen molecule in kg. rms m

What is u? How do we get the mass in kg? 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)

3kT m is the mass of one v  What is m? oxygen molecule. rms m 3(1.38x 1023 J / K )278 K  (32u )(1.66 x 1027 kg / u )

Speed of  466ms / Is this fast? sound: YES! 343m/s! 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 3 / 2 m 2 o 2 mv/2 kB T NV  4 N v e 2kTB Ludwig Boltzmann 1844 – 1906 Austrian physicist Contributed to Kinetic Theory of Gases Electromagnetism Thermodynamics Pioneer in statistical mechanics 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.

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.

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 Molecular Interpretation of Temperature • Simplifying the equation relating temperature and kinetic energy gives 13___ m v2  k T 22o B • This can be applied to each direction, 11___ mv2  k T 22x B

– with similar expressions for vy and vz 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 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 Heats Isobaric requires MORE HEAT than Isochoric for the same change in Temperature!!!! .The total change in thermal energy for ANY PROCESS, due to work and heat, is:

This applies to all ideal gases, not just monatomic ones! WOW! C and C P V Note that for all ideal gases:

where R = 8.31 J/mol K is the universal gas constant.

Slide 17-80 Important Concepts Agreement with Experiment: Diatomic Hydrogen acts like a monatomic gas at low temperature! •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. Quantization of Energy.

•To explain the results of the various molar specific heats, we must use some quantum mechanics. – Classical mechanics is not sufficient •This energy level diagram shows the rotational and vibrational states of a diatomic molecule. •The lowest allowed state is the ground state. •The vibrational states are separated by larger energy gaps than are rotational states. •At low temperatures, the energy gained during collisions is generally not enough to raise it to the first excited state of either rotation or vibration.

Section 21.4 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 Reading Question 18.2

What additional kind of energy makes CV larger for a diatomic than for a monatomic gas?

A. Charismatic energy. B. Translational energy. C. Heat energy. D. Rotational energy. E. 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. Charismatic energy. B. Translational energy. C. Heat energy. D. Rotational energy. E. Solar energy.

Slide 18-13 AdiabaticAdiabatic Processes Processes: Q=0 . An adiabatic process is one for which:

g-1 g-1 Ti Vi = Tf Vf where:

. Adiabats are steeper than hyperbolic isotherms because only work is being done to change the Temperature. The temperature falls during an adiabatic expansion, and rises during an adiabatic compression. Slide 17-88 A 4.00-L sample of a nitrogen gas confined to a cylinder, is carried through a closed cycle. The gas is initially at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. (a) Draw a PV diagram of this cycle. (b) Find the number of moles of the gas. (c) Find the volumes and temperatures at the end of each process (d) Find the Work and heat for each process. (e) What was the net work done on the gas for this cycle? 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 g = constant

g = 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, DEint = 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 18.3

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. A has the larger average energy per atom.

B. TA = TB.

C. TA < TB. D. There’s not enough information to compare their temperatures.

Slide 18-52