Query 8

course Phy 232

008. `query 7*********************************************

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 don’t understand the meaning behind the elastically part. I don’t know how to incorporate that.

confidence rating:

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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). **

Your Self-Critique:I don’t see where the elastic part was used, so maybe it was just there in an attempt to confuse me.

Your Self-Critique Rating:ok

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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:Since work energy is conserved within an isolated system so the thermal energy that goes in equals the total work wdone by the system and the thermal energy taken away.

confidence rating:

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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:ok

Your Self-Critique Rating:ok

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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:Efficiency=work done/energy input. You add the amount of thermal energy to the amount of work done to get the input. You then divide the work by the energy input.

confidence rating:

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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:ok

Your Self-Critique Rating:ok

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Question: query univ phy problem 19.56 (17.40 10th edition) compressed air engine, input pressure 1.6 * 10^6 Pa, output 2.8 * 10^5 Pa, assume adiabatic.

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Your Solution: You start using the equation PV=nRT. T2 must be greater than 273K to prevent frost so, T2=P2V2/nR=273K which would make nR=P2V2/273K. I wasn’t sure what to do next, so I looked down and the gamma part kind of confused me.

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

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)

Your Self-Critique:I think I get that, but I still am a little confused by the gamma part.

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.

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 Rating:1

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

Your Solution: Goes to the right from original P and V to doubled V. It then goes left on a curve increasing as it goes more to the left. When it is right above the starting point it goes vertically downward to it.

confidence rating:

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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. **

Your Self-Critique:ok

Your Self-Critique Rating:ok

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

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Your Solution: Temperature also doubles because the volume doubles at constant pressure. Which would mean the compression is 710 K.

confidence rating:

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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:ok

Your Self-Critique Rating:ok

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Question: univ phy What is the max pressure?

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Your Solution: It starts at 240 kPa and its volume is halved at constant temperature so pressure doubles to 480 kPa.

confidence rating:

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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:ok

Your Self-Critique Rating:ok

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&#Good responses. See my notes and let me know if you have questions. &#