course Phy 232 008. `query 7
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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 did not take the solution out as far as yours, I understand the concept very well now though. Your Self-Critique Rating: 2 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Thermal energy put into a system during a cycle can be converted to work by the expansion of the system. So when you put the thermal energy into the system, it then heats up and expands the gasses causing them to do work. This work done by the system removing the work put into it. Essential, what goes in must come out or vise versa. Confidence Rating: 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: 3 Your Self-Critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Efficiency is the work done by the system divided by the energy put into the system. So if you know how much thermal energy you put into you system and how much work the system did, all you need to do is divide the work by the energy input. Confidence Rating: 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: Okay Your Self-Critique Rating: 3 ********************************************* 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 chagne in internal energy. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I am not in these classes. Confidence Rating: 3
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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. Your Self-Critique: Okay. Your Self-Critique Rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I am not in these classes. Confidence Rating: 3
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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: Okay. Your Self-Critique Rating: 3 ********************************************* 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 am not in this class. Confidence Rating: 3
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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.** Your Self-Critique: Okay. Your Self-Critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I am not in this class. Confidence Rating: 3
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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: Okay. Your Self-Critique Rating: 3 ********************************************* 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 am not in this class. Confidence Rating: 3
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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: Okay. Your Self-Critique Rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: PV=nRT PV^gamma=constant. We want T2 = 273 K. V = T * n * R / P , plugging that into P1 * V1^gamma = P2* V2^gamma P1 * (T1 * n * R / P1 )^gamma = P2 * (T2 * n * R / P2 )^gamma, we know that gamma=1.4, so all we do is solve for T1 which is 449.2 Kelvin. Confidence Rating: 3
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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: Okay. Your Self-Critique Rating: 3 ********************************************* Question: query univ 19.62 (17.46 10th edition) .25 mol oxygen 240 kPa 355 K. Isobaric to double vol, isothermal back, isochoric to original pressure. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The volume is doubled while the pressure remains constant, that means that the temperature had to be doubled according to the ideal gas law. So T2 = 710 Kelvin. Since the volume is being restored its original size and the temperature has now been doubled, the pressure will have to be doubled according to the ideal gas law. So P3 = 4.8*10^5 Pa. From position 1 to 2, the pressure does not change so the work is just simply W12 = P * delV. So W12 = 737.9 J. From position 2 to 3, the pressure does change. This means that we have to first set up the differential form of the work before we can solve it. dW = P * dV = nRTdV/V. To get the work from 2 to 3, we just integrate that expression and get W23 = nRT*(lnV2-lnV3) = nRT*ln(1/2). W23 = -1022.96 J. The total work done was -285.1 J. Confidence Rating: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: Okay. Your Self-Critique Rating:3 ********************************************* Question: univ phy describe your graph of P vs. V YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: From the first position, the graph moves horizontally to a value of V that is twice the distance from the y-axis. From this second position, the graph makes a line that has a positive concavity moving up to a position twice that of the original pressure and back to the original volume. From there it moves vertically back to its original spot. Confidence Rating: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. ** Your Self-Critique: Okay. Your Self-Critique Rating: 3 ********************************************* Question: univ phy What is the temperature during the isothermal compression? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The temperature during the isothermal compression is 710 Kelvin. Confidence Rating:3
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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: Okay. Your Self-Critique Rating: 3 ********************************************* Question: univ phy What is the max pressure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 4.8*10^5 Pa. Confidence Rating:3
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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: Okay. Your Self-Critique Rating:3 "