Assignment 08

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course PHY 232

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

008. `query 7

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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:

Since the collision with the wall is elastic, the particle will reverse direction, keeping the same velocity. Therefore, the change in velocity is 2v. The change in momentum is the mass times the change in velocity, or 2*m*v. The average force of the particle, according to the Impulse-Momentum Theorem, is the change in momentum over time. The final calculation would then be:

(2mv)/’delta t

confidence rating #$&*:

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confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

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3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

• 2 indicates that you believe you addressed most of the discrepancies between the given solution and your solution but are unsure of some aspects of the situation; you would at this point consider including a question or a statement of what you're not sure you understand

• 1 indicates that you believe you understand the overall idea of the solution but have not been able to address the specifics of the discrepancies between your solution and the given solution; in this case you would normally include a question or a statement of what you're not sure you understand

• 0 indicates that you don't understand the given solution, and/or can't make a reasonable judgement about whether or not your solution is correct; in this case you would be expected to address the given solution phrase-by-phrase and state what you do and do not understand about each phrase)

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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:

A system does a certain amount of work. The net work of the system consists of the thermal energy that goes into the system and the energy removed or dissipated from the system:

Work done during cycle = thermal energy in - thermal energy out

confidence rating #$&*:

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confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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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:

First I would need to find the amount of thermal energy that was put into the system. This is the amount of work done plus the thermal energy removed or dissipated.

The efficiency would then be found by dividing the amount of work done by the amount of thermal energy put in the system. The calculation would look something like

this:

work/(work+out)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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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:

I’m in university physics.

confidence rating #$&*:

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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’m in university physics

confidence rating #$&*:

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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:

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:

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

confidence rating #$&*:

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confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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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:

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.

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.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

Your Self-Critique Rating:

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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:

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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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:

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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Question: gen phy problem 15.12, a-c curved path `dW = -35 J, `dQ = -63 J; a-b-c `dW = - 48 J

gen phy how much thermal energy goes into the system along path a-b-c and why?

Your Solution:

I’m in university physics

confidence rating #$&*:

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Given Solution:

** I'll need to look at the graph in the text to give a reliably correct answer to this question. However the gist of the argument goes something like this:

`dQ is the energy transferred to the system, `dW the work done by the system along the path. Along the curved path the system does -35 J of work and -63 J of thermal energy is added--meaning that 35 J of work are done on the system and the system loses 63 J of thermal energy.

If a system gains 35 J of energy by having work done on it while losing 63 J of thermal energy, its internal energy goes down by 28 J (losing thermal energy take internal energy from the system, doing work would take energy from the system so doing negative work adds energy to the system). So between a and c along the curved path the system loses 28 J of internal energy.

In terms of the equation, `dU = `dQ - `dW = -63 J -(-35 J) = -28 J.

It follows that at point c, the internal energy of the system is 28 J less than at point a, and this will be the case no matter what path is followed from a to c.

Along the path a-b-c we have -48 J of work done by the system, which means that the system tends to gain 48 J in the process, while as just observed the internal energy goes down by 28 Joules. The system therefore have `dQ = `dU + `dW = -28 J + (-48 J) = -76 J, and 76 J of internal energy must be removed from the system.**

STUDENT COMMENT:

I don’t understand the transition from the first step in which we found the work of the system and the

energy not used and then the second part where you combine this thermal energy with the work done in the second part

of the system. Why would you combine energy lost to work done?

INSTRUCTOR RESPONSE

During a complete cycle, energy is put into the system. The cycle ends in the same energy state as it began. So the energy put into the system has to go somewhere; it isn't retained by the system.

Some of this energy is converted to mechanical work. Whatever isn't converted to mechanical work has to be removed from the system (for example, as exhaust).

STUDENT COMMENT

I really just guessed on this problem. I figured that if the problem gave me the info to

find ‘dU, I might as well find it. But I didn’t really think that the ‘dU for a-c would also serve as the ‘dU for a-b-c.

INSTRUCTOR RESPONSE

The internal energy of a system is purely a function of its state. So when we go from state a to state c, it doesn't matter how we get there, the change in internal energy is the same.

The amount of thermal energy required to take the system from one state to another varies with the path, because different paths correspond to different amounts of work done on or by the system.

The amount of thermal energy required is equal to the change in the internal energy of the system, plus the work done by the system:

`dQ = `dW + `dU.

Your Self-Critique:

Your Self-Critique Rating:

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Question: gen phy How are the work done by the system, the thermal energy added to the system and the change in the internal energy of the system related, and what is this relationship have to do with conservation of energy?

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Your Solution:

I’m in university physics.

confidence rating #$&*:

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Given Solution:

** If a system does work it tends to reduce internal energy, so `dW tends to decrease `dU. If thermal energy is added to the system `dQ tends to increase `dU. This leads to the conclusion that `dU = `dQ - `dW. Thus for example if `dW = -48 J and `dU = -28 J, `dQ = `dU + `dW = -28 J + -48 J = -76 J. **

Your Self-Critique:

Your Self-Critique Rating:

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Question: gen phy How does the halving of pressure caused a halving of the magnitude of the work, and why is the work positive instead of negative as it was in the process a-b-c?

Your Solution:

I’m in university physics

confidence rating #$&*:

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Given Solution:

** Work is the area under the pressure vs. volume curve. If you have half the pressure between two volumes the graph has half the altitude, which leads to half the area.

The 'width' of a region is final volume - initial volume. If the direction of the process is such that final volume is less than initial volume (i.e., going 'backwards', in the negative x direction) then with 'width' is negative and the area is negative. **

Your Self-Critique:

Your Self-Critique Rating:

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Question: query univ phy problem (not in 12th edition; solve using the statement of the problem given here) 11th edition 19.56 (17.40 10th edition)

In a compressed air engine, input pressure is 1.6 * 10^6 Pa, output pressure is 2.8 * 10^5 Pa. Assume the process to be adiabatic.

If we are to avoid frost forming on the output valve, which occurs if the temperature of the exiting air is below freezing, what must be the temperature of the compressed air?

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Your Solution:

I’m in university physics

confidence rating #$&*:

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Given Solution:

** For an adiabatic process in an ideal gas you know that PV = nRT and PV^`gamma is constant.

You are given P1 and P2, and you want T2 > 273 K to prevent formation of frost.

Assume T2 = P2 V2 / (n R) = 273 K and n R = (P2 V2) / 273 K .

Then T1 = P1 V1 / (n R) = P1 V1 * 273 K / (P2 V2) = (P1 / P2) * (V1 / V2) * 273 K.

Since PV^`gamma = constant it follows that V1 / V2 = (P2 / P1)^(1/`gamma) = (P1 / P2)^(-1/`gamma).

Thus T1 = (P1 / P2) ( P1 / P2)^(-1/`gamma) * 273 K = (P1 / P2)^(1 - 1/`gamma) = (P1 / P2)^(1-1/1.4) * 273 K = (P1 / P2)^.29 * 273 K = 5.6^.29 * 273 K = 443 K, approx. **

STUDENT QUESTION:

i see how we substitute this expression for v1/v2. not why there is ^(1-1/gamma)

INSTRUCTOR RESPONSE

gamma = c_p / c_v, the ratio of specific heat at constant pressure to specific heat at constant volume.

Molar specific heat for an ideal monatomic or diatomic gas is 1/2 R per degree of freedom at constant volume, plus R if the expansion is at constant pressure.

PV^`gamma = constant. Doing the algebra:

P1 V1^gamma = P2 V2^gamma so

(V1 / V2)^gamma = P2 / P1. Taking the 1 / gamma power of both sides

V1 / V2 = (P2 / P1)^(1/`gamma)

1 / ((P2 / P1)^(1/`gamma) ) = (P1 / P2)^(-1/`gamma) since the reciprocal of a power is the negative power.

Then

(P1 / P2) ( P1 / P2)^(-1/`gamma)

= (P1 / P2)^1 * ( P1 / P2)^(-1/`gamma)

= ( P1 / P2)^(1 -1/`gamma) (just adding the exponents of the two like bases)

STUDENT COMMENT

If find the idea of 'gamma' to be difficult.

INSTRUCTOR COMMENT:

At one level, you simply need to know that an adiabatic expansion is characterized by P V^gamma = constant.

You should understand that during an adiabatic expansion, since some of the internal energy is used to do the work of

expansion, the temperature decreases. Thus P, V and T all change.

If only P and V changed, then P V would be constant.

Since T also changes, we can not say that PV remains constant.

The notes and your text explain the derivation of the formula and the reason for gamma. Basically gamma depends on what

fraction of the internal energy of the gas resides in its translational motion, and what fraction in rotational. It's the

changes in translational momentum that provide the force for the expansion.

Your Self-Critique:

Your Self-Critique Rating:

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Question: query univ 19.64 / 19.62 (17.46 10th edition) .25 mol oxygen 240 kPa 355 K. Isobaric to double vol, isothermal back, isochoric to original pressure.

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Your Solution:

First, we will use PV = nRT to compute the volume.

So V = (nRT)/P = [(0.25 mol)(8.31 J/(mol K))(355 K)]/ (2.4x10^5 Pa) = 0.0031 m^3

Isobaric has a constant pressure.

If the volume is doubled, the new Temperature = [(2.4 x 10^5 Pa)(0.0062 M^3)]/[(0.25 mol)(8.31 K/(mol K)] = 716 K

In this process, Work is equal to p(V2-V1), and V2 is double the volume as V1, so W = (2.4x10^5 Pa)(2*0.0031 - 0.0031) = 744 J

Isothermal compression has a constant temperature.

If temperature is constant, and n and R are constant, then PV must be constant. Since volume is decreased in half, pressure will have to double to remain constant. So Pressure = 2 x 2.4 x 10^5 Pa = 4.8 x 10^5 Pa

To find this work, I will have to take the integral of P dv. P = nRT/v. The antiderivative of 1/v = ln|v|. So integrating from V1 to V2, we get W = nRTln|v2| - nRTln|v1| = nRT(ln|v2| - ln|v1|) = nRT(ln|V2/V1|) = nRT(ln|V/2V|) = nRT(ln|1/2|) = (0.25 mol)(8.31 J/(mol K))(716 K)(ln|0.5|) = -1031 J

In an isochoric system, the volume is constant, so no work is done.

The net work = 744 - 1031 = -287 J

confidence rating #$&*:

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confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

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3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Given Solution:

** .25 mol oxygen at 240 kPa occupies about V = n R T / P = .25 mol * 8.31 J / (mol K) * 355 K / (2.4 * 10^5 N/m^2) = .003 m^3, very approximately.

Doubling volume, `dV = 2 * V - V = V = .003 m^2 and P = 2.4 * 10^5 Pa so P `dv = 700 J, very approximately.

During isothermal compression we have n = const and T = const so P = n R T / V. Compressing to half the volume, since PV = const, gets us to double the pressure, so max pressure is 2 * 240 kPA = 480 kPa.

To get work we integrate P dV. Integral of P dV is calculated from antiderivative n R T ln | V |; integrating between V1 and V2 we have n R T ln | V2 | - n R T ln | V1 | = n R T ln | V2 / V1 |. In this case V2 = V and V1 = 2 V so V2 / V1 = 1/2 and we have `dW = n R T ln(1/2) = .25 mol * 8.31 J/(mol K) * 710 K * (-.7) = -1000 J, approx.

So net work is about 700 J - 1000 J = -300 J **

Your Self-Critique:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Your Self-Critique Rating:

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Question: univ phy describe your graph of P vs. V

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Your Solution:

My graph starts off going from the first point to the second point, in a horizontal manner to the right. This signifies the doubling volume. Next, it goes from the second to the third point, which is located above the first point. The line is not quite straight as it goes across the graph. Finally, it goes from the third to the first point, singifying the volume and pressure going back to normal.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Given Solution:

** The graph proceeds horizontally to the right from original P and V to doubled V, then to the left along a curve that increases at an incr rate as we move to the left (equation P = 2 P0 V0 / V) until we're just above the starting point, then vertically down to the starting pt. **

STUDENT COMMENT

I still had no idea after the explanation.

INSTRUCTOR RESPONSE

To understand this graph, do the following:

• Sketch a graph of y = 1 / x, using, for example, x values .1, .5, 1, 2, 10. Having done so it should not be difficult to understand the shape of this graph.

• If you wanted to sketch the graph of y = 100 / x, you could use the same graph, and just relabel your vertical axis (for example you would replace 1 with 100, 2 with 200, etc). Do so.

• Now you can use the same graph for P = (2 P0 V0) / V. Instead of y, label the vertical axis P. Instead of x, label the horizontal axis V. Relabel the vertical axis in terms of your original x and y coordinates, multiplying each of your vertical coordinates by 2 P0. Relabel the horizontal axis in terms of V, replacing 1 with V0, 2 with 2 V0, etc.. [ The original coordinates of your five plotted points were (.1, 10), (.5, 2), (1, 1), (2, .5) and (10, .1). Relabeled they would be (.1 V0, 20 P0), (.5 V0, 4 P0), (V0, 2 P0), (2 V0, .5 P0) and (10 V0, .1 P0) ]

Your Self-Critique:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

Your Self-Critique Rating:

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Question: univ phy What is the temperature during the isothermal compression?

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Your Solution:

In this first step of the process, the volume doubled and was isobaric, so the pressure was constant. Because of this, the tempearture had to double as well to keep the equation constant. So 355 K x 2 = 710 K.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

.............................................

Given Solution:

** If vol doubles at const pressure then temp doubles to 710 K, from which isothermal compression commences. So the compression is at 710 K. **

Your Self-Critique:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Question: univ phy What is the max pressure?

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Your Solution:

WHen it is compressed isothermally, the temperature is constant, and the volume is cut in half, so in order to keep the equation constant the pressure had to double. So 240 kPa x 2 = 480 kPa.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

.............................................

Given Solution:

** It starts the isothermal at the original 240 kPa and its volume is halved at const temp. So the pressure doubles to 480 kPa. **

Your Self-Critique:

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Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

Self-critique (if necessary):

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Self-critique rating:

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

Self-critique (if necessary):

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Assignment 08

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