Why the round trip instead collision time? 
From Newton's second law, the $\Delta t$ is defined as the collision time, but why in this case, it can be assumed to be the value of time between successive collisions on 1 wall? If I have an infinitely large vessel, doesn't that make the $\Delta t$ infinitely large, and thus exert approximately 0N of force on the wall? What are the assumptions being made here?
 A: The assumptions that are being made here are perfectly elastic collisions with the walls where the atom will rebound with the same speed as it came it with.  Its trying to calculate the average force based on the number of interactions per unit time with the wall, and is therefore fundamentally related to how long it takes for successive collisions between the walls.  As it states above, $F = \Delta p/\Delta t$, and the only time that $\Delta p$ changes for an atom interacting with the wall is during a collision with the wall.  To calculate the average force using the time in between collisions allows us to ignore the actual interaction time of a single collision.  The assumption here is that there are many particles interacting with the walls at all times, so that the average force on the walls stays relatively consistent.  If there are $N$ collisions in time $\Delta t$, the total momentum change is $2Nm\bar{v}_x$, for $N$ particles with an average velocity perpendicular to the wall $\bar{v}_x$.  If $\Delta t = 2L/\bar{v}_X$, the average round trip time, then this gives an average force for $N$ molecules of:
$$\bar{F} = mN\bar{v}_x^2/L$$
The point of all of this is that its taking a statistical average of many collisions over a long time period, even when its considering the single particle case.  Since for an ideal gas, we know that there will be a perfect rebound, we know precisely what $\Delta p$ should be for every collision even if we don't know how long each collision will take individually.  But, we can calculate the number of collisions in a time $\Delta t$ in order to calculate the average force over the entire time period, which is what they did in the above example.
Also, note that it states that the gas is an ideal gas, meaning that the relationship $PV = nRT$ can be invoked.  The pressure, $P$, is the force per unit area on the walls, and $V$ is the volume of the container.  Since temperature is just related to the energy spread of the gas, we can determine the following:  If the walls are infinitely far apart, then the volume, $V$ is infinite, therefore, $P = nRT/V \rightarrow 0$, and thus the force per unit area is zero.  This is consistent with the fact that for infinitely spaced walls, $\Delta t$ between collisions $\rightarrow \infty$.
A: Here is another derivation of the same formula:

Newton's third law tells us that the molecule exerts a force on the wall.  The greater the number of molecules hitting a wall, the greater is the force on the wall.  In a container with different size walls, the bigger walls will receive more hits than the smaller walls and therefore experience a greater force.  The pressure in the container is the magnitude of the normal force F on a wall divided by the surface area A of the wall. 
P = F/A
The faster the molecules move in the container, the greater is the change in momentum when they bounce off a wall, and the more often do they hit the walls.  Assume a molecule moves horizontally with speed |v_x| back and forth between two infinitely-massive walls, which are a distance L apart.  When it hits the right wall its momentum changes from p1 = +m|v_x| to p2 = -m|v_x|.  The change in the molecule's momentum is Δp_mol = p2 - p1 = -2m|v_x|.  The time interval between successive hits on the right wall is Δt = 2L/|v_x|.  So the average force the wall exerts on this molecule is F_mol = Δp_mol/Δt = -2m|v_x|/(2L/|v_x|) = -mv_x^2/L.  By Newton's third law, the average force that the molecule exerts on the wall is F_wall = mv_x^2/L, it is proportional to the square of the speed of the molecule or its kinetic energy.



Assume that there are N molecules in the volume V, moving horizontally with speed |v_x|.  Not all the molecules have the same kinetic energy.  The force exerted by the molecules on the walls of a container is therefore F = Nm/L, where  is the average value of vx2.

It looks at it with respect to  the force the wall exerts on the molecule and invokes the third law. It is considering the wall as the "actor" and the molecule "acted upon" . The interval the wall "sees"  in space traveled is twice the length because one hit on the same wall by the same molecule does not constitute an interval, the molecule has to come back. Then the wall can estimate a time from the velocity.
Seems to me also contrived, but at least consistent.
A: To answer your question concisely, yes, the derivation finds the average force of a single gas molecule in a gigantic container is zero because it hardly ever hits the wall.
But the next thing they do is multiply by the number of molecules.  If you have a known density $N/L$, the number of molecules $N$ is just as infinite as the length $L$.  So the two infinities cancel (L'Hôpital's rule) and you have an average force that is finite.
A: Review:
Assume we have a collection of gas molecules in a container with volume V at absolute temperature T. 
Then each molecule moves along with constant velocity in a straight line, until it hits another molecule, or a container wall.  A collision between two molecules is similar to a collision between two balls.  The molecules exchange momentum, but the total momentum of the two molecules is conserved.  When a molecule hits a wall, it bounces back.  Its momentum changes.  To change the molecule's momentum, the wall must exert a force on the molecule.  Newton's third law tells us that the molecule exerts a force on the wall.  The greater the number of molecules hitting a wall, the greater is the force on the wall.  In a container with different size walls, the bigger walls will receive more hits than the smaller walls and therefore experience a greater force.  The pressure in the container is the magnitude of the normal force F on a wall divided by the surface area A of the wall. 
P = F/A
The faster the molecules move in the container, the greater is the change in momentum when they bounce off a wall, and the more often do they hit the walls.  Assume a molecule moves horizontally with speed |vx| back and forth between two infinitely-massive walls, which are a distance L apart.  When it hits the right wall its momentum changes from p1 = +m|vx| to p2 = -m|vx|.  The change in the molecule's momentum is Δpmol = p2 - p1 = -2m|vx|.  The time interval between successive hits on the right wall is Δt = 2L/|vx|.  So the average force the wall exerts on this molecule is Fmol = Δpmol/Δt = -2m|vx|/(2L/|vx|) = -mvx2/L.  By Newton's third law, the average force that the molecule exerts on the wall is Fwall = mvx2/L, it is proportional to the square of the speed of the molecule or its kinetic energy.
Assume that there are N molecules in the volume V, moving horizontally with speed |vx|.  Not all the molecules have the same kinetic energy.  The force exerted by the molecules on the walls of a container is therefore F = Nm/L, where  is the average value of vx2.
The pressure is P = F/A = Nm/V, since L*A = V.  With ρparticle = N/V we have
P = F/A = ρparticlemvx2.
There is nothing special about the x-direction.  The atoms can move up and down, back and forth, in and out.  The average velocity components in all directions are all going to be equal to each other.
 =  = .
They are each equal to one-third of their sum, which is the square of the magnitude of the average velocity.
 =  +  + .
 = (1/3).
We may therefore write
P = (1/3)ρparticlem = (2/3)ρparticle(m/2)
m/2 is the kinetic energy of the center-of-mass or translational motion of an atom or molecule.  Using ρparticle = N/V we have
PV = (2/3)N(m/2)
This equation relates the pressure to the kinetic energy of the atoms or molecules.  The ideal gas law states PV = NkBT.  Comparing the two expressions we find
.
The temperature is a direct measure of the average translational molecular kinetic energy.
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The root-mean-square speed of the atoms or molecules is
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Links:  Please explore these interactive animations.
bullet  
Particle Speeds and Temperatures
bullet  
Welcome to the pressure chamber
Problem:
A cylinder contains a mixture of helium and argon gas in equilibrium at 150 oC.
(a) What is the average kinetic energy of each gas molecule?
(b) What is the root-mean-square speed of each type of molecule?
bullet  Solution:
bullet  (a) The average kinetic energy of each molecule is 
(3/2)kBT = (3/2)1.38*10-23 J/K(423 K) = 8.76*10-21 J.
The average kinetic energy is the same for both types of atoms,
bullet  (b) vrms2 = 2*8.76*10-21 J/m.
mHe = 4 u, mAr = 39.9 u.
1 u = 1 atomic mass unit = 1.66*10-27 kg.
vrms(He) = 1.62*103 m/s.
vrms(Ar) = 514 m/s.
The more massive molecules have a lower average speed.
Problem:
A 5-liter vessel contains 0.125 moles of an ideal gas at a pressure of 1.5 atm.  What is the average translational kinetic energy of a single molecule?
bullet  Solution:
PV = nRT yields T.
T = PV/(nR)
T = (1.5*1.01*105 Pa)(5000 cm3*1 m3/106 cm3)/(0.125*8.31 J/K)
T = 729 K.
The average translational kinetic energy of a single molecule is 1.51*10-20 J.
Heat flow
When you bring two objects of different temperature together, energy will always be transferred from the hotter to the cooler object.  The objects will exchange thermal energy, until thermal equilibrium is reached, i.e.  until their temperatures are equal.  We say that heat flows from the hotter to the cooler object.  Heat is energy on the move. 
Units of heat are units of energy.  The SI unit of energy is Joule.  Other often encountered units of energy are 1 Cal = 1 kcal = 4186 J, 1 cal = 4.186 J, 1 Btu = 1054 J. 
Without an external agent doing work, heat will always flow from a hotter to a cooler object.  Two objects of different temperature always interact.  There are three different ways for heat to flow from one object to another.  They are conduction, convection, and radiation.
Conduction
The atoms in a solid vibrate about their equilibrium positions.  As they vibrate, they bump into their neighbors.  In those collisions they exchange energy with their neighbors.  If the different regions of a solid object or of several solid objects placed in contact with each other have the same temperature, then all atoms are just as likely to gain energy as to loose energy in the collisions.  Their average random kinetic energy does not change.  If, however, one region has a higher temperature than another region, then the atoms in the high temperature region will, on average, loose energy in the collisions, and the atoms in the low temperature region will, on average, gain energy.  In this way heat flows through a solid by conduction.
The stiffness of the springs (strength of the chemical bonds) determines how easily the atoms can exchange energy and therefore determines if the material is a good or bad conductor of heat.  Each atom has a nucleus, surrounded by electrons.  In a solid metal all nuclei are bound to their equilibrium positions.  But some electrons are free to move throughout the solid.  They can easily pick up kinetic energy in collisions with hot cores and loose it again in collision with cooler cores.  Since their mean free path between collisions is larger than the distance between neighboring atoms, thermal energy can move quickly through the material.  Metals are, in general, much better conductors of heat than insulators. 
Convection
Convection transfers heat via the motion of a fluid which contains thermal energy.  In an environment where a constant gravitational force F = mg acts on every object of mass m, convection develops naturally because of changes in the fluids density with temperature.  When a fluid, such as air or water, is in contact with a hotter object, it picks up thermal energy by conduction.  Its density decreases.  For a given volume of the fluid, the upward buoyant force equals the weight of this volume of cool fluid.  The downward force is the weight of this volume of hot fluid.  The upward force has a larger magnitude than the downward force and the volume of hot fluid rises.  Similarly, when a fluid is in contact with a colder object, it cools and sinks.  When a volume of fluid such as air or water starts to move, the surrounding fluid has to rush in to fill the void.  Otherwise large pressure differences would develop.  This sets up a convection current and the looping path that follows is a convection cell.  Since fluid cannot pile up at some point in space without creating a high-pressure area, it will flow in a closed loop.  Convection can be increased if the fluid is forced to circulate.  A fan, for example, forces the air to circulate.
Video: Convection Current
Radiation
Nuclei and electrons are charged particles.  When charged particles accelerate, they emit electromagnetic radiation and loose energy.  Vibrating particles are always accelerating since their velocity is always changing.  They therefore always emit electromagnetic radiation.  Charged particles also absorb electromagnetic radiation.  When they absorb the radiation they accelerate.  Their random kinetic energy increases.  In thermal equilibrium, the amount of energy they lose to radiation equals the amount of energy they gain from radiation.  But hotter objects emit more radiation than they absorb from their cooler environment.  Radiation can therefore transport heat from a hotter to a cooler object.
Electromagnetic radiation refers to electromagnetic waves, which travel through space with the speed of light.  We classify electromagnetic waves according to their wavelength.  A graphical representation of the electromagnetic spectrum is shown in the figure on the right.   
The visible part of the spectrum may be further subdivided according to color, with red at the long wavelength end and violet at the short wavelength end, as illustrated in the next figure.   
Hot objects emit radiation with a distribution of wavelengths.  But the average wavelength of the radiation decreases as the temperature of the object increases.  Most thermal radiation lies in the infrared region of the spectrum.  We cannot see this radiation, but we can feel it warming our skin.  Different objects emit and absorb infrared radiation at different rates.  Dark surfaces are generally good emitters.    
Examples of all heat transfer processes:
When a wood stove is used to heat the air in a room, conduction, convection, and radiation play a role. 
When the wood burns, chemical energy stored in the wood is converted into thermal energy of the reaction products.  By conduction, these reaction products heat the surfaces and the air they are in contact with. 
Convection draws the hot smoke up a long black pipe and out of the room and draws fresh air into the stove.  When the smoke is in contact with inner the surface of the pipe, it heats the pipe by conduction.  Conduction also carries the thermal energy from the inner surfaces of the stove and the pipe to the outer surfaces, and heats the air close to the surfaces.  The hot air then begins to rise by convection.  Cooler air rushes in to replace the rising air, and a convection current begins to flow in a convection cell.
This distributes the warm air throughout the room.  The hot, black, outer surface of the stove is also a good emitter of infrared thermal radiation.  This thermal radiation is absorbed by the surfaces of different objects in the room.
Please complete assignment 4 now.  
