assignment #007 Lÿ`weߴЇ Physics II 02-13-2006
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15:43:09 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 --> Speed = v time interval= delta t angle = 90 deg (right) FIND AVE FORCE We would use the velocity and the impulse momentum theorum. That is honestly as far as I can get today.
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15:45:32 ** 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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RESPONSE --> Delta momentum = Fave * delta t Fave = momentum / time Time is round trip ( i wouldnt have gotten that on my own) 2L/v Fave = -2*m*v / ( 2*L/v) = m*v^2*L 3 dimensions = 1/3 N * m * v^2 / L v^2 = vx^2 + vy^2 + vz^2 Okay. V average magnitude of vel.
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15:46:36 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 --> Thermal energy = Total work & Total energy taken out
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15:46:59 ** 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 (work or thermal energy). I understand this.
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15:48:22 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 of cycle Find input = work done + t energy removed Then divide work by input
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15:48:36 ** 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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RESPONSE --> efficiency = work done/energy input
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15:52:11 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 --> 1400 kcal added volume increase 12.0 m^3 to 18.2 m^3 Find work, change in internal energy Work done = Pressure * Delta Volume Work done = 1 atm * 6.2 m^3 Work done = 6.2 atm m^3 (This needs to be in Joules, but how?) Internal energy = delta q - delta w Internal energy =??
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15:54:47 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 --> Okay. P* delta v = work done ( I had that down) Here is where I went wrong Work done = (101.3 * 10^3 N/m^2) * (6.2 m^3) ***atm to N/m^2 work done = 6.3 * 10^5 J 1400 kcal * 4200 J = 5.9 * 10^6 J Internal energy = delta q * delta q = 5.9 * 10^6 J - .63 * 10^6 J = 5.3 * 10^6 J
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15:58:59 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 --> PV = n*R*T P= 4.5 atm v= 1 L Left side of the equation solved = 4.5 atm * L The graph here would go to 4.5 atm and 1 L. When compressed to a different volume, the two values would switch (they are equal). The graph would then move to 1 atm and 4.5 L.
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16:01:31 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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16:05:52 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 --> delta w= work done by system delta q = energy put into system -35 J of work done by system -63 J of energy put into system internal energy = delta q - delta w internal energy = -63 - -35 J = -28 J At the abc delta w we have -48 J of work done by system and we know delta u SOO delta q = delta u + delta w delta q = -28 J + -48 J = -76 J
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16:06:34 ** 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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RESPONSE --> A B C has -48 J of work done by the systems, gains 48 J in the process. We go from 76 to 48 losing 28 J. Okay.
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16:08:11 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 --> The internal energy is reduced as the work is done. This in turn reduces delta w. This would make delta u smaller because du = dq - dw.
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16:08:19 ** 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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16:09:32 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 quantities are all related. When you slice the pressure in half the graph has half the area or work.
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