course Phy 202 In response to your request of my address for replacement DVD shipment,until the end of this week my address is:
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10:26:30 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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RESPONSE --> Average force is equal to the change in total momentum over the time interval. The change in total momentum is equal to two times the total mass of the particles time the velocity. Thus the long equation that can be used is: Fave = (2*n*m*v)/time interval n in this case appears to be 1
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10:31:11 11-11-2007 10:31:11 ** 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). **
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NOTES -------> Saving as notes
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10:33:14 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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RESPONSE --> The work done by the system is equal to the difference between the thermal energy that goes into the system and the thermal energy that is removed or dissipated during the cycle.
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10:34:09 ** 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. **
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RESPONSE --> What goes in must come out, according to the Law of Conservation of energy
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10:36:57 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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RESPONSE --> efficiency is equal to the work done by the system over the thermal energy that goes into the system. Given the thermal energy dissipated during the cycle, we have to find the thermal energy that goes in. This number is equal to the sum of the work and the thermal energy dissipated. efficiency = Work / Work + dissipated thermal energy
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10:37:05 ** 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. **
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10:46:05 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 chagne in internal energy.
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RESPONSE --> W = P*dV = (1.013*10^5 N/m^2)(18.2 m^3 - 12.0 m^3) = 628060 J dU = (1400 kcal*4186 J/kcal) - 628060 J = 5232340 J
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10:48:14 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.
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RESPONSE --> Work is the area under the P vs V graph, a straight horizontal line in this problem.
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10:54:01 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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RESPONSE --> To sketch this graph I drew a vertical line connecting the two Pressure volumes, assuming an isovolumetric scenario. Then I drew a horizontal line connecting the two volumes, assuming isobaric. Then I connected the far ends of these two lines using an approximate curve. I don't have much of an understanding of these graphs.
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11:05:11 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).
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RESPONSE --> Oh I think I see how this one works. 1. The curved path from (1 liter, 4.5 atm) to (4.5 liters, 1 atm) representing the isothermic process described. 2. The horizontal line from (4.5 L, 1 atm) to (1 L, 1 atm) representing the compression at a constant pressure to lower the volume. 3. The vertical line from (1 L, 1 atm) to (1 L, 4.5 atm) representing the holding of the volume while the pressure returns to 4.5 atm.
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12:18:35 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?
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RESPONSE --> After attempting several attacks, I can't figure out how to approach this problem.
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12:29:31 11-11-2007 12:29:31 ** 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.**
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NOTES -------> dU on the curved path: dU = dQ - dW = - 63 J -(-35 J) = -28 J a-b-c dW = -48 J (system will gain 48 Joules in the process) The internal energy will decrease by 28 Joules between a and c, regardless of the path. dQ = dU + dW = -28 J + (-48 J) = -76 J. Therefore 76 J of energy will be removed from the system. This is not as complicated as I was making it in my attempts. I didn't think about the fact that the internal energy would be the same regardless of path.
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12:38:21 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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RESPONSE --> I feel like I've answered this already, but I'm not sure. The change in internal energy is equal to the difference between the thermal energy added to the system and work done by it. This relationship is related to the conservation of energy because the change in internal energy is directly related to the thermal energy and the work.
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12:39:43 ** 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. **
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RESPONSE --> Work leads to decreased internal energy. Addition of thermal energy leads to increased internal energy.
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12:53:36 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?
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RESPONSE --> The values are proportional, as indicated by the formula W = P'dV. If the work is positive, it means work was done by the system instead of performed ON the system.
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12:56:00 ** 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. **
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RESPONSE --> Half the pressure leads to half the altitude of the graph, which causes the area to halve. If the final volume < initial volume, the width of the area is negative, making the area negative.
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