Query 8

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course Phy 122

Sir, the questions at the bottom, I spent way to much time trying to grasp. Is there a website that explains these a little betterVR

"Question: Query set 5 problems 16-20

explain how we calculate the average force exerted by a particle of mass m moving at speed v which collides at a time interval `dt elastically and at a right angle with a solid wall

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Your Solution:

I think I remembered this as MEFAT. J = impulse and (Je is me in French) so J=F(force) (Delta t) or change in time. This sounds like N2L to me. I understand the impulse momentum theorem put impulse = the change in momentum but don’t really get the elastically or right angle part. Unless these particles have a controlled vector.

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Given Solution:

** The impulse exerted on a particle in a collision is the change in the momentum of that particle during a collision.

The impulse-momentum theorem says that the change in momentum in a collision is equal to the impulse, the average force * the time interval between collisions.

The average force is thus change in momentum / time interval; the time interval is the round-trip distance divided by the velocity, or 2L / v so the average force is -2 m v / ( 2L / v) = m v^2 / L

If there were N such particles the total average force would be N * m v^2 / L

If the directions are random we distribute the force equally over the 3 dimensions of space and for one direction we get get 1/3 the force found above, or 1/3 N * m v^2 / L.

This 3-way distribution of force is related to the fact that for the average velocity vector we have v^2 = vx^2 + vy^2 + vz^2, where v is average magnitude of velocity and vx, vy and vz the x, y and z components of the velocity (more specifically the rms averages--the square root of the average of the squared components). **

STUDENT QUESTION

I'm not sure why you multiply the velocity by 2. I understand multiplying the distance by 2 to make the

round trip.

INSTRUCTOR RESPONSE

The given solution doesn't multiply the velocity by 2. The solution does, however, involve change in momentum and that results in a factor of 2.

The momentum changes from + m v to - m v when the particle bounces off the wall. The change in momentum is

change in momentum = final momentum - initial momentum = -mv - mv = - 2 mv.

The 2 in -2 m v results from a subtraction, not a doubling.

Your Self-Critique: Confusing. I am going to go back over the lesson a few more times concerning this one.

Your Self-Critique Rating: 2

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The impulse-momentum theorem is a standard first-semester topic that you should review. Impulse is F * `dt, change in momentum is m * `dv, and the two are equal.

Kinetic energy is retained in an elastic collision (also a first-semester topic worth reviewing). The wall doesn't change velocity, so the speed of the particle cannot change in an elastic collision.

Collision at a right angle to the wall means that the particle is moving straight at the wall. If the two opposite walls are parallel to one another, this ensures that the particle will bounce back and forth between the walls without bouncing off of anything else. Also that the momentum change will be maximized.

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Question: Summarize the relationship between the thermal energy that goes into the system during a cycle, the work done by the system during a cycle, and the thermal energy removed or dissipated during the cycle.

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Your Solution:

I know that the isolated system is an (in and out) kind of deal, anything that goes in must come out. The output, since we are talking about thermal energy, will either be thermal energy or work done by the system.

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Given Solution:

** Work-energy is conserved within an isolated system. So the thermal energy that goes into the system must equal the total of the work done by the system and the thermal energy removed from the system. What goes in must come out, either in the form of work or thermal energy. **

Your Self-Critique: I believe I am trying to state the same thing.

Your Self-Critique Rating: 2

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Question: If you know the work done by a thermodynamic system during a cycle and the thermal energy removed or dissipated during the cycle, how would you calculate the efficiency of the cycle?

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Your Solution:

Similar to above, the systems work is divided by the energy input.

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Given Solution:

** STUDENT SOLUTION: Efficiency is work done / energy input. Add the amount thermal energy removed to the amount of work done to get the input. Then, divide work by the energy input. **

Your Self-Critique: I had to look at the solution but was quickly reminded. Trying to do this from memory to see if I can actually figure it out. I believe I cheated a little on this one though.

Your Self-Critique Rating:3

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Perfectly OKas long as you acknowledge it, as you have. This is an important self-discipline, ensuring that you're not deluding and/or cheating yourself.

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Question: prin phy and gen phy problem 15.2, cylinder with light frictionless piston atm pressure, 1400 kcal added, volume increases slowly from 12.0 m^3 to 18.2 m^3. Find work and change in internal energy.

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Your Solution:

18.2m^3-12m^3 = 6.2m^3

101.3E3 N/m^2 * 6.2m^3 = 6.3E5 J

1400kcal = 5.9E6 J

So, 5.9E6 -.63E6 = 5.3E6

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Given Solution:

Work done at constant pressure is P `dV, so the work done in this situation is

`dW = P `dV = 1 atm * (18.2 m^3 - 12 m^3) = (101.3 * 10^3 N/m^2) * (6.2 m^3) = 630 * 10^3 N * m = 6.3 * 10^5 J.

A total of 1400 kcal = 1400 * 4200 J = 5.9 * 10^6 J of thermal energy is added to the system, the change in internal energy is

`dU = `dQ - `dW = 5.9*10^6 J - 6.3 * 10^5 J = 5.9 * 10^6 J - .63 * 10^6 J = 5.3 * 10^6 J.

It is worth thinking about the P vs. V graph of this process. The pressure P remains constant at 101.3 * 10^3 J as the volume changes from 12 m^3 to 18.2 m^3, so the graph will be a straight line segment from the point (12 m^3, 101.3 * 10^3 J) to the point (18.2 m^3, 101.3 * 10^3 J). This line segment is horizontaland the region above the horizontal axis and beneath the segment is a rectangle whose width is 6.2 * 10^3 m^3 and whose altitude is 101.3 * 10^3 N/m^2; its area is therefore the product of its altitude and width, which is 6.3 * 10^5 N m, or 6.3 * 10^5 J, the same as the word we calculated above.

STUDENT COMMENT:

My answer was way off, but I see where I made my mistake. I didn’t convert the m^3 into

Joules. Is (101.3 x 10^3 N/m^2) the conversion factor for m^3 into Joules??

INSTRUCTOR RESPONSE:

You calculated the right quantities, but you didn't use compatible units.

m^3 measures volume, Joules measure work\energy.

The units of your calculation 1atm * (18.2m^3 - 12.0 m^3) = 6.2 m^3 don't make sense. The units of this calculation would be atm * m^3, not m^3.

It's hard to make sense of the unit atm * m^3, but 1 atm = 101.3 kPa or 101 300 Pa, which is 101 300 N / m^2.

Your calculation should therefore have been

1atm * (18.2m^3 - 12.0 m^3) = 101 300 N/m^2 * (18.2 m^3 - 12.0 m^3).

The units will come out N * m, which is Joules.

Your Self-Critique:

Your Self-Critique Rating:

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Question: prin phy and gen phy problem 15.5, 1.0 L at 4.5 atm isothermally expanded until pressure is 1 atm then compressed at const pressure to init volume, final heated to return to original volume. Sketch and label graph.

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Your Solution:

I am lost on this one, I will head back to the book…

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Given Solution:

When a confined ideal gas is expanded isothermally its pressure and volume change, while the number of moled and the temperature remain constant. Since PV = n R T, it follows that P V remains constant.

In the initial state P = 4.5 atm and V = 1 liter, so P V = 4.5 atm * 1 liter = 4.5 atm * liter (this could be expressed in standard units since 1 atm = 101.3 kPa = 101.3 * 10^3 N/m^2 and 1 liters = .001 m^3, but it's more convenient to first sketch and label the graph in units of atm and liters). During the isothermal expansion, therefore, since P V remains constant we have

P V = 4.5 atm liters. At a pressure of 1 atm, therefore, the volume will be V = 4.5 atm liter / P = 4.5 atm liter / (1 atm) = 4.5 liters.

The graph follows a curved path from (1 liter, 4.5 atm) to (4.5 liters, 1 atm).

At the gas is compressed at constant pressure back to its initial 1 liter volume, the pressure remains constant so the graph follows a horizontal line from (4.5liters, 1 atm) to (1 liter, 1 atm). Note that this compression is accomplished by cooling the gas, or allowing it to cool.

Finally the gas is heated at constant volume until its pressure returns to 4.5 atm. The constant volume dictates that the graph follow a vertical line from (1 liter, 1 atm) back to (4.5 liters, 1 atm).

The graph could easily be relabeled to usestandard metric units.

1 atm = 101.3 kPa = 101.3 * 10^3 Pa = 101.3 * 10^3 N/m^2, so

4.5 atm = 4.5 * 101.3 * 10^3 Pa = 4.6 * 10^3 Pa = 4.6 * 10^3 N/m^2.

1 liter = .001 m^3 so 4.5 liters = 4.5 m^3.

Since P V = 4.5 atm liters, P = 4.5 atm liters / V. This is of the form P = c / V, with c a constant. For positive values of V, this curve descendsfrom a vertical asymptote with the vertical axis (the V axis) through the point (1, c) then approaches a horizontal asymptote with the horizontal axis. For c = 4.5 atm liters, the curve therefore passes through the point (1 liter, 4.5 atm). As we have seen it also passes through (4.5 liters, 1 atm).

Your Self-Critique:0

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With more information I could help clarify this for you.

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&#You need a detailed self-critique here.

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Your Self-Critique Rating:

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Question:

Openstax: A system does 1.80×10^8 J of work while 7.50×10^8 J of heat transfer occurs to the environment. What is the change in internal energy of the system assuming no other changes (such as in

temperature or by the addition of fuel)?

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Your Solution:

I will need your help on these sir. I am finding myself looking at the answers from this and trying to figure it out…

confidence rating #$&*:

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Given Solution:

If no energy goes into the system, then the work it does and the energy transferred to the environment are both at the expense of its internal energy.

So the system experiences an internal energy change of

-1.80 * 10^8 J - 7.50 * 10^8 J = -9.30 * 10^8 J.

Your Self-Critique:

Your Self-Critique Rating:

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A self-critique here would allow me to address what you don't understand, and reinforce what you do.

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Question:

Openstax: Steam to drive an old-fashioned steam locomotive is supplied at a constant gauge pressure of 1.75×10^6 N/m2 (about 250 psi) to a piston with a 0.200-m radius. (a) By calculating PΔV , find the work done by the steam when the piston moves 0.800 m. Note that this is the net work output, since gauge pressure is used. (b) Now find the amount of work by calculating the force exerted times the distance traveled. Is the answer the same as in part (a)?

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Your Solution:

I will need your help on these sir. I am finding myself looking at the answers from this and trying to figure it out…

confidence rating #$&*:

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Given Solution:

The cross-sectional area of the piston is

A = pi r^2 = pi * (0.20 meters)^2 = .13 m^2, approx..

It moves 0.60 meters, so the change in the volume of the cylinder is

`dV = 0.13 m^2 * 0.80 m = 0.10 m^3.

so, assuming no pressure loss as the piston recedes, the product P `dV is

P `dV = 1.75 * 10^6 Pa * 0.10 m^3 = 1.75 * 10^5 N/m^2 * m^3 = 1.75 * 10^5 N * m = 1.75 * 10^5 Joules.

The absolute pressure in the cylinder is equal to the gauge pressure plus atmospheric pressure, so P_abs = 1.75 * 10^6 Pa + 1.0 * 10^5 Pa = 1.85 * 10^6 Pa. The force exerted by this pressure on the piston is therefore

0.13 m^2 * 1.85 * 10^6 Pa = 2.4 * 10^5 Newtons.

Multiplying this force by the 0.60 meter change in the position of the piston we get

`dW = 2.4 * 10^5 N * 0.60 m = 1.4 * 10^5 Joules,

which is greater than the work calculated based on the gauge pressure.

Your Self-Critique:

Your Self-Critique Rating:

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i don't believe this problem was part of the assignment for your course. it goes a little beyond the scope of Phy 122.

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Question: Openstax: Calculate the net work output of a heat engine following path ABCDA in the figure below.

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Your Solution:

Sir, our these for our class???

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Given Solution:

The work done by a thermodynamic cycle is equal to the work enclosed by that cycle on a P vs. V diagram.

The path ABCDA is a parallogram with altitude 1.6 * 10^6 N/m^2 and base 3.0 * 10^-3 m^3, so it encloses area

area = `dW = 1.6 * 10^6 N/m^2 * 3.0 * 10^-3 m^3 = 4.8 * 10^3 N * m = 4800 Joules.

Your Self-Critique:

Your Self-Critique Rating:

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Check the text assignments for your course. I don't believe this was assigned for Phy 122. Let me know if you find differently.

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Question:

What is the net work output of a heat engine that follows path ABDA in the figure below, with a straight line from B to D?

Why is the work output less than for path ABCDA?

Explicitly show how you follow the steps in the Problem-Solving Strategies for Thermodynamics.

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Your Solution:

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Given Solution:

The path ABDA is a triangle with base 1.6 * 10^6 N/m^2 and altitude 3.0 * 10^-3 m^3. The area of the triangle is

area = `dW = 1/2 * 1.6 * 10^6 N/m^2 * 3.0 * 10^-3 m^3 = 2400 Joules.

Along the path BD negative work is done against the pressure, as is the case along path CD. However the area beneath BD is greater than that beneath CD, so the negative work done in this case is greater in magnitude than the negative work done along path ABCDA.

Your Self-Critique:

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&#Your work looks good. See my notes. Let me know if you have any questions. &#