course Phy 202 100127Copy into Notepad or another text editor, type in your answers as instructed, and submit with your access code, according to instructions at the Submit Work Form (link: http://vhcc2.vhcc.edu/dsmith/submit_work.htm) The following conventions will allow your instructor to quickly locate your answers and separate them from the rest of any submitted document, which will significantly increase the quality of the instructor's feedback to you and to other students. When answering these questions, give your answer to a question starting in the line after the **** and before the #$&*. First measurement of atmospheric pressure `q001. What were the lengths of the air column, and the height of the water column, for your measurement of the system observed in the lab? **** Air column = 41 cm Water column = 67 cm #$&* By what percent did the length of the air column change when you squeezed the bottle? **** Percent change = 4/45 = 0.09%
100127
Copy into Notepad or another text editor, type in your answers as instructed, and submit with your access code, according to instructions at the Submit Work Form (link: http://vhcc2.vhcc.edu/dsmith/submit_work.htm)
The following conventions will allow your instructor to quickly locate your
answers and separate them from the rest of any submitted document, which will significantly increase the quality of the instructor's feedback to you and to other students.When answering these questions, give your answer to a question starting in the line after the **** and before the
#$&*.
First measurement of atmospheric pressure
`q001. What were the lengths of the air column, and the the height of the water column, for your measurement of the system observed in the lab?
****
For a different system I observed air column lengths 33 cm and 30 cm, and water column height 80 cm.
#$&*
By what percent did the length of the air column change when you squeezed the bottle?
****
The 3 cm difference was approximately 9% of the length of the original column, corresponding to about a 9% difference in pressure.
#$&*
By what percent did the volume of the air column change when you squeezed the bottle?
****
The volume of the air decreased by about 9%.
#$&*
By how much did the pressure increase, in Pa, when you squeezed the bottle, based on your measurement of the water column (you can for now use the rule of thumb that a 10-cm-high water column requires about 1000 Pa of pressure to support it)?
****
An 80 cm column corresponds to a pressure difference of about 8000 Pa (recall the rule of thumb that the pressure just below a stationary 10 cm water column is about 1000 Pa higher than the pressure just above the column).
More accurately the pressure difference is rho g h = 1000 kg / m^3 * 9.8 m/s^2 * .80 m = 7800 kg * m^2 / (m^3 * s^2).
kg * m^2 / (m^3 * s^2) = kg m / s^2 * m / m^3 = N / m^2, or Pa.
#$&*
By what percent did the pressure in the air column change when you squeezed the bottle?
****
P V = n R T.
The temperature of the air column remained the same, or nearly the same, as that of the tubing. The temperature of the system increased a bit when the air was compressed, but not by a significant amount. The gas was confined between the closed end of the tubing and the water. So n and T can be considered to constant.
Thus P V is constant, and the 9% decrease in V is matched by a 9% increase in P.
#$&*
Assuming the original pressure in the column to be 1 atmosphere, what do you therefore conclude to be the atmospheric pressure, in Pa (compare the percent change in pressure with the number of Pascals of pressure necessary to support the water column)?
****
The original pressure was 1 atmosphere, and the increase in pressure was therefore 9% of 1 atmosphere.
The pressure increase, according to the water column in the vertical tubing, was about 8000 Pa.
If 9% of atmospheric pressure is 8000 Pa, then atmospheric pressure is 8000 Pa / .09 = 90 000 Pa, very approximately.
More formally we can write
.09 P_atm = 8000 Pa,
which is easily solved to give us
P_atm = 8000 Pa / .09 = 90 000 Pa, approx..
#$&*
Impulse-momentum applied to pressure in a gas
The impulse of a force F acting during time interval `dt is F * `dt.
The impulse of the net force F_net acting on a system during time interval `dt is F_net * `dt, and this impulse is equal to the change in momentum m * v of the system. This is the impulse-momentum theorem:
`q002. A particle of mass m bounces back and forth between two parallel walls separated by distance L, moving between the walls at speed v in a direction perpendicular to the walls, and colliding elastically with the walls.
****
Assuming that the positive direction is the direction away from the wall, the velocity changes from -v to +v, so the momentum changes from - m v to m v. The change is therefore
`dp = p_f - p_0.
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****
The ball has to move distance 2 L at speed v, so it requires time `dt = 2 L / v to complete the round trip.
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****
The average net force, by the impulse-momentum theorem, is
F_ave = `dp / `dt = 2 m v / (2 L / v) = m v^2 / L.
Note that m v^2 is twice as great at 1/2 m v^2, which is the KE of the particle. Thus m v^2 / L can be written as 2 KE / L, and
F_ave = 2 KE / L.
#$&*
The result that follows from the preceding question is easily expressed as follows:
where KE stands for the kinetic energy 1/2 m v^2 of the particle.
If the direction of motion of the particle in the preceding problem is randomly distributed in 3 dimensions, then its average kinetic energy in the direction perpendicular to the walls is 1/3 of its kinetic energy.
So if the direction of the particle's velocity is random in space, our previous conclusion F_ave = 2 KE / L becomes
`q003. If the average force exerted by the particle is F_ave = 1/3 * (2 KE / L), and if the cross-sectional area of the wall is A, then what is the average pressure exerted by the particle on the wall?
****
P_ave = F_ave / A = 1/3 * (2 KE / L) / A = 1/3 * 2 KE / (A * L).
#$&*
If the particle is confined to a container with cross-sectional area A and length L, what is the volume of the container, and how is this volume related to the expression you got in the first part of this problem?
****
The volume is equal to cross-sectional area multiplied by length:
V = A * L.
This is just the denominator of the expression obtained in the first part of this question. So
PE = 1/3 * 2 KE / (A * L) becomes
PE = 1/3 * 2 KE / V, written more simply as
PE = 2/3 KE / V.
#$&*
Our conclusion is that
where V is the volume of the container and KE the kinetic energy of the particle.
Now suppose we have N particles. They might start out all moving in the same direction, but they're going to quickly start running into one another and changing direction, which will cause even more collisions. In short order the velocities of the particles will be randomly distributed in space.
In any collision between two particles, one of the particles will typically speed up and the other will slow down (strictly speaking, in fact, there is 0 probability that the speed of either particle will be unchanged; in a perfectly aligned collision this could happen, but the number of 'perfect alignments' is of a lesser cardinality that the number of possible alignments, but that's a topic for a more advanced course; you should just understand that the speeds of the particles change when the start colliding).
`q004. If we have N particles, each of mass m, with average rms velocity v_rms, and their total KE is 1/2 m v_rms^2, then:
What is their total KE?
****
Their total KE would be
- KE_total = N * (1/2 m v_rms^2).
#$&*
What is the expression for P_ave, in terms of the symbols V for the volume of the container, m and v_rms?
****
P_ave = 2/3 KE / V = 2/3 * 1/2 m v_rms^2 = 1/3 m v_rms^2.
#$&*
If the number N of particles in a given volume is very large, then (provided the height of the region is small enough that the average gravitational PE of the particles is much less than their average KE, which will pretty much be the case at room temperature as long as the container is less than, say, 100 meters high) the pressure throughout the container is very nearly uniform, and we can drop the _ave on the symbol P_ave. We can therefore say
Since 1/2 m v_rms^2 is the average KE of the particles, we can of course still make the equivalent statement that
Now we know that for an ideal gas
The previous equation P = 2/3 N * KE_ave / V, which we obtained above simply by applying impulse-momentum to N particles in a container of volume v, is easily rearranged to the form
We therefore have two expressions for P V. Setting them equal we obtain
Recall that N stands for the number of particles and n for the number of moles.
The number of particles and the number of moles have a simple relationship: The number of particles in a mole is Avagadro's Number (approximately 6.02 * 10^23 particles; this number is defined in terms of other quantities and must be determined experimentally; refined experiments keep measuring it more and more precisely, and its best-know 3-significant-figure value actually changes from time to time; since your instructor's high school years not far past the middle of the last century, its rounded-off value has in fact changed from about 6.03 * 10^23 to 6.02 * 10^23). That is,
We can rewrite the above equation in terms of N_A:
2/3 N * KE_ave = n R T so
2/3 KE_ave = n / N * R T so
2/3 KE_ave = (1 / N_A) * R T and
2/3 KE_ave = (R / N_A) * T. We use the symbol k for R / N_A and write
- 2/3 KE_ave = k T.
The symbol k as used in this equation has a distribution).
This equation can be used to determine the average velocity of any particle, as long as we know its mass.
`q005. R = 8.31 Joules / (mole * Kelvin) and N_A = 6.02 * 10^23 particles / mole. What therefore is the value of k = R / N_A, and what are its units?
****
k = R / N_A = (8.31 Joules / (mole K) ) / (6.02 * 10^23 particles / mole) = 1.38 * 10^-23 Joules / (particle Kelvin).
#$&*
`q006. At a temperature of 300 Kelvin, what is the average kinetic energy of the particles?
****
At 300 K we have
2/3 KE_ave = k T, so
KE_ave = 3/2 k T.
k = R / N_A = 1.38 * 10^-23 Joules / (particle Kelvin), as shown in the preceding problem.
Thus
KE_ave = 3/2 k T = 3/2 * 1.38 * 10^-23 Joules / (particle Kelvin) * 300 Kelvin = 6 * 10^-21 Joules, very approximately.
#$&*
The atomic number of oxygen is 16, so a mole of monatomic oxygen atoms would have a mass of 16 grams. Oxygen occurs in a diatomic molecule, so a mole of molecular oxygen has a mass of 32 grams. What therefore is the mass of a single oxygen molecule?
#$&*
The atomic number of oxygen is 16, so a mole of monatomic oxygen atoms would have a mass of 16 grams. Oxygen occurs in a diatomic molecule, so a mole of molecular oxygen has a mass of 32 grams. What therefore is the mass of a single oxygen molecule?
****
A mole consists of 6.02 * 10^23 particles, and a mole of oxygen has mass 32 grams. Therefore a single atom has mass
(32 grams / mole) / (6.02 * 10^23 particles / mole) = 5 * 10^-22 gram / particle, or 5 * 10^-25 kg / particle,
where the particle is an oxygen atom (which in another sense isn't a particle at all, but we'll worry about that when we come to it).
The answer to this question: an oxygen atom has mass approximately 5 * 10^-25 kg.
#$&*
You now know the average kinetic energy of diatomic oxygen models and you know the mass of a single molecule. So you can determine the associated velocity (KE is associated with v^2, so the associated velocity will be the rms velocity). What therefore is the rms speed of oxygen molecules at 300 Kelvin?
****
The average KE is about 6 * 10^-21 Joules. The mass of the particle is about 5 * 10^-25 kg.
Since KE = 1/2 m v^2, we easily conclude that
v = sqrt( 2 KE / m) = sqrt( 2 * 6 * 10^-21 Joules / (5 * 10^-25 kg) ) = 1.5 * 10^2 m/s, about 150 m/s.
#$&*
`q007. The bottles we have been using in lab each contain roughly .02 moles of air.
What therefore is the total KE of the molecules in a bottle, at a temperature of 300 K?
****
.02 moles consists of .02 moles * N_A = .02 moles * 6.02 * 10^-23 particles / mole = 1.2 * 10^-21 particles, approx..
As seen previously the average KE of particles at 300 K is 3/2 k T = 6 * 10^-21 Joules, approx..
Thus the total kinetic energy of the particles in the bottle at 300 K is about
(This idea will soon be modified; what we have just calculated is the translational KE of the particles. There is another form of KE present.)
#$&*
How much KE will they lose if the temperature drops to the freezing point? (qualifying note: the molecules in the air aren't particles, they are mostly diatomic atoms; however for the present we will treat them as particles; the difference is that particles can't spin and diatomic molecules can, so they actually contain more KE than will be indicated by your solution; it will later be fairly easy to correct our model to correctly accommodate diatomic molecules).
****
At the freezing point the average KE is
KE_ave = 3/2 k T = 3/2 * 1.38 * 10^-23 J / (particle Kelvin) * 273 K = 5 * 10^-21 Joules, approx..
We get
It appears that the energy loss would be from about 70 J to about 60 J, a loss of about 10 Joules.
Calculation of the total kinetic energies to three significant figures will give a result accurate to about 2 significant figures; this result will be closer to 7 Joules than to 10 Joules.
#$&*
You might have worked on these questions previously. If so, you can just answer the ones you think you can now answer better than before.
Four principles are the basis for the work-energy theorems:
1. For any system the work done on the system by the net force can be broken into the work done on the system by conservative forces, and by nonconservative forces:
2. Potential energy is defined in terms of work done by conservative forces:
3. The work done on a system by a force and the work done by the system against that force are related by
4. For any system the work done by the net force on the system is equal to the change in the system's kinetic energy:
`q008. Explain your understanding of each of these principles, and how each makes sense to you (or fails to do so). Examples are often appropriate in explaining why something makes sense.
****
Answers will vary. Your examples are very unlikely to match the examples given here.
1. The net force is the sum of the conservative forces and the nonconservative force, so the work done by the net force must equal the work done by the conservative forces and the work done by the nonconservative forces. For example the net force on a falling object is the sum of the conservative force of gravity and the nonconservative force of air resistance, the first doing positive and the second doing negative work on the object. Add the two and you get the work done by the net force.
2. If you raise an object the displacement is upward, the gravitational force acting on it is downward, so the work done by the gravitational force is negative. The object's potential energy increases (intuitively, if you release it the PE can convert to KE). The PE change in positive, work by gravitational force is negative, and the two are equal and opposite.
3. The force exerted on an object is equal and opposite to the force exerted by the object, so the work done on the object is equal and opposite to the work done by the object. For example the frictional force exerted on a sliding object by the surface over which is slides is equal and opposite to the force exerted by the object on the surface. Friction does negative work on the sliding object; the object does positive work against friction.
4. An object is accelerated by the net force acting on it, which does work equal to the change in KE. If I push an object across the floor, my force tends to speed it up and friction tends to slow it. My force is in the direction of motion, frictional force is opposite the direction of motion, so my force does positive work and the frictional force does negative work. If my force exceeds that of friction then the net work is positive and the object speeds up. If frictional force is greater the net work is negative and the object slows. If the forces are equal in magnitude the net work is zero and the object neither speeds up nor slows down.
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`q009. The four principles contain two expressions for `dW_net_on. If you set these two expressions equal to one another, what do you get and why does it make sense? Examples are often appropriate in explaining why something makes sense.
****
`dW_net_on = `dW_cons_on + `dW_noncons_on and
`dW_net_on = `dKE so
`dKE = `dW_cons_on + `dW_noncons_on.
For a falling object the gravitational force acts in the direction of the displacement so `dW_cons_on is positive, air resistance acts in the direction opposite motion so `dW_noncons_on is negative. If the former is greater in magnitude than the latter the object speeds up, if the latter is greater (as it can be if the object is falling very fast) the object slows. In either case `dKE is equal to the sum of the work done by conservative and the work done by nonconservative forces.
#$&*
`q010. The third principle can be applied to the work done by any force. The principle says that
For example, the principle could be applied to the net force, to conservative forces, or to nonconservative forces. We could write
How would the third principles be applied to the nonconservative forces associated with a system?
****
Applied to nononservative forces, the general principle that `dW_on = `dW_by becomes
`dW_noncons_on = - `dW_noncons_by.
For example the work done by a force I exert when pushing a car is equal and opposite to the work done by the car against my force.
#$&*
To what quantity in the second principle could the third principle be applied? What statement do you get when you do so, and why does it makes sense?
****
The second principle relates PE change to work done on a system by conservative forces (`dPE = - `dW_cons_on)
The third principle `dW_on = `dW_by tells us that `dW_cons_on = - `dW_cons_by so that
`dPE = - ( - `dW_cons_by ) = `dW_cons_by.
#$&*
`q011. Explain how the first principle can be rewritten in terms of `dPE by applying the second and third principles (see also the preceding problem).
****
As shown in the preceding,
`dPE = - ( - `dW_cons_by ) = `dW_cons_by.
This has a straightforward interpretation:
The change in PE is equal to the work done by the system against conservative forces.
#$&*
`q012. Rewrite the first principle from the perspective of the system, replacing the word 'on' with the word 'by' according to the third principle, and incorporating `dPE.
****
The first principle is `dW_net_on = `dW_cons_on + `dW_noncons_on.
Applying the second principle `dW_on = -`dW_by, this becomes
-`dW_net_by = -`dW_cons_by + (-`dW_noncons_by).
`dPE = `dW_cons_by so
-`dW_net_by = - `dPE - `dW_noncons_by. Multiplying through by -1 we get
`dW_net_by = `dPE + `dW_noncons_by.
#$&*
`q013. The total mechanical energy of a system is KE + PE, the sum of its kinetic and potential energies. The change in total mechanical energy is therefore `dKE + `dPE. From the expression you obtained in the preceding question, to what is `dKE + `dPE equal, and why does it make sense that it should be so?
****
`dW_noncons_on = `dKE + `dPE.
If no other conservative forces (like friction) intervene, the work I do on a system is all changed to KE and/or PE, and the total of these changes is equal to the work I do.
My work is in this case (to repeat, the case mine is the only nonconservative force) is therefore present in the system in the form of its increased mechanical energy.
#$&*
`q014. If the total mechanical energy of a
system decreases, then does the system do positive or negative work against nonconservative forces?****
`dW_noncons_on = `dKE + `dPE so a decrease in total mechanical energy implies that the work done on the system by nonconservative forces is negative. Since `dW_by = - `dW_on, it follows that the system does positive work against nonconservative forces.
In a common non-ideal system the nonconservative force is friction. Friction does negative work on the system, and the system does positive work against friction.
#$&*
`q015. If the total mechanical energy of a
system increases, then do nonconservative forces do positive or negative work on the system?****
`dW_noncons_on = `dKE + `dPE so a decrease in total mechanical energy implies that the work done on the system by nonconservative forces is negative.
#$&*
`q016. If the total kinetic energy of a
system remains constant while conservative forces do 5000 Joules of work on the system, then by how much does the total mechanical energy of the system change?****
If conservative forces do 5000 Joules of work on the system, then since `dPE is equal and opposite to the work done by conservative forces, the PE of the system decreases by 5000 Joules. The KE remains constant, so the total mechanical energy PE + KE changes by -5000 Joules.
For example if gravitational and frictional forces are balances, a box can slide down an incline with constant KE, which implies constant velocity. Gravity does positive work on the box as it descends, balancing the negative work done by friction. The box ends up with unchanged velocity but at a lower point, implying lower PE. Thus its total mechanical energy PE + KE decreases.
#$&*
`q017. If the total mechanical energy of a
system decreases by 5000 Joules, how much work is done on the system by nonconservative forces, and how much is done by the system against nonconservative forces?****
`dW_noncons_on = `dKE + `dPE, so `dW_noncons_on = change in total mechanical energy.
If total mechanical energy decreases by 5000 Joules, then `dW_noncons_on also decreases by 5000 Joules.
This means that nonconservative forces to -5000 Joules of work on the system, and the system does +5000 Joules of work against nonconservative forces.
#$&*