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
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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: OK Your Self-Critique rating #$&* OK ********************************************* Question: Summarize the relationship between the thermal energy that goes into the system during a cycle, the work done by the system during a cycle, and the thermal energy removed or dissipated during the cycle. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Thermal energy that goes into the system during a cycle equals the total work done by the system and the thermal energy removed or dissipated during the cycle. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* OK ********************************************* Question: If you know the work done by a thermodynamic system during a cycle and the thermal energy removed or dissipated during the cycle, how would you calculate the efficiency of the cycle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Efficiency of the cycle = work done/energy input To calculate the energy input you need to add the thermal energy removed to the work done. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* OK ********************************************* Question: prin phy and gen phy problem 15.2, cylinder with light frictionless piston atm pressure, 1400 kcal added, volume increases slowly from 12.0 m^3 to 18.2 m^3. Find work and change in internal energy. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* ********************************************* Question: prin phy and gen phy problem 15.5, 1.0 L at 4.5 atm isothermally expanded until pressure is 1 atm then compressed at const pressure to init volume, final heated to return to original volume. Sketch and label graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* ********************************************* Question: gen phy problem 15.12, a-c curved path `dW = -35 J, `dQ = -63 J; a-b-c `dW = - 48 J gen phy how much thermal energy goes into the system along path a-b-c and why? Your Solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* ********************************************* Question: gen phy How are the work done by the system, the thermal energy added to the system and the change in the internal energy of the system related, and what is this relationship have to do with conservation of energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* ********************************************* Question: gen phy How does the halving of pressure caused a halving of the magnitude of the work, and why is the work positive instead of negative as it was in the process a-b-c? Your Solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* ********************************************* Question: query univ phy problem (not in 12th edition; solve using the statement of the problem given here) 11th edition 19.56 (17.40 10th edition) In a compressed air engine, input pressure is 1.6 * 10^6 Pa, output pressure is 2.8 * 10^5 Pa. Assume the process to be adiabatic. If we are to avoid frost forming on the output valve, which occurs if the temperature of the exiting air is below freezing, what must be the temperature of the compressed air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: PV = nRT The final temp (T2) needs to be higher than 273K to prevent frost So you could say T2 = P2 V2 / (n R) = 273 K, and therefore n R = (P2 V2) / 273 K T1 = P1 V1 / (n R) T1 = P1 V1 * 273 K / (P2 V2) T1 = (P1 / P2) * (V1 / V2) * 273 K Now using the PV^’ gamma constant V1 / V2 = (P2 / P1)^(-1/`gamma) So now … T1 = (P1 / P2) ( P1 / P2)^(-1/`gamma) * 273 K T1 = (P1 / P2)^(1-1/1.4) * 273 K = (P1 / P2)^.2857 * 273 K = 5.6^.2857* 273 K = 447 K confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK Your Self-Critique rating #$&*OK ********************************************* Question: query univ 19.62 (17.46 10th edition) .25 mol oxygen 240 kPa 355 K. Isobaric to double vol, isothermal back, isochoric to original pressure. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: V = n R T / P =.25 mol * 8.31 J / (mol K) * 355 K / (2.4 * 10^5 N/m^2) =.00307 m^3 `dV = 2 * V – V V = .00307m^3, P = 2.4 *10^5 Pa P `dv = 737 J Max pressure in this case is equal to double the pressure 2 * 240 kPA = 480 kPa Find work… n R T ln | V2 | - n R T ln | V1 | = n R T ln | V2 / V1 | V2 = V, V1 = 2 V, thus V2 / V1 = 1/2 `dW = n R T ln(1/2) = .25 mol * 8.31 J/(mol K) * 710 K * (-.7) = -1033 J Net work 737 – 1033 = -296 J confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK Your Self-Critique rating #$&*OK ********************************************* Question: univ phy describe your graph of P vs. V YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: It goes from the right to P (original) and V then to the doubled V, the left is increasing at an increasing rate until the original point then it goes straight down. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* OK ********************************************* Question: univ phy What is the temperature during the isothermal compression? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The temp double to 710 K, so the temperature during isothermal compression is at 710K confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* OK ********************************************* Question: univ phy What is the max pressure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The pressure (isothermal org. pressure = 240Kpa) doubles 240 *2 = 480kPa confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 "