#$&* course PHY 202 3/19 about 11 am 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). ** 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: The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system, or conservation of energy. According to conservation of energy, Q_in = W + Q_out. 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: The efficiency can be calculated by W / Q_in. This is give a percentage of how much of the thermal energy was converted into W and not lost as Q_out. 1 - efficiency would give you the percentage of the thermal energy not converted to work. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION: Efficiency is work done / energy input. Add the amount thermal energy removed to the amount of work done to get the input. Then, divide work by the energy input. ** Your Self-Critique:OK Your Self-Critique Rating: ********************************************* Question: prin phy and gen phy problem 15.2, cylinder with light frictionless piston atm pressure, 1400 kcal added, volume increases slowly from 12.0 m^3 to 18.2 m^3. Find work and change in internal energy. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Q_in = ‘dQ = 1400 kcal * (4184 J / 1 kcal) = 5857600 J P = 1 atm = 101,300 N/m^2 ‘dV = 18.2 m^3 - 12.0 m^3 = 6.2 m^3 ‘dW = P*’dV = 101,300 N/m^2 * 6.2 m^3 = 628060 N*m = 628060 J ‘dU = ‘dQ - ‘dW = 5857600 J - 628060 J = 5.3 * 10^6 J confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK Your Self-Critique Rating:OK ********************************************* 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: The initial P is 4.5 atm, y axis, and V is 1.0 L, x axis. This point on my graph is 1 unit R of origin and 4.5 units up from x axis. P1V1 = P2V2 so 4.5 atm * 1.0 L = 1.0 atm * V2, V2 = 4.5 L. The point after isothermal expansion is 4.5 L, 1.0 atm. I read the given solution to see that this would not be linear but rather curved from point 1 to point 2. Since the pressure remains constant as the container returns to the original volume, the horizontal line runs from point 2 to the point 1.0 L, 1.0 atm. After heating, the line runs vertical back to point 1. 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:I understand the reason the first line is not linear due to the P=c/V form. Your Self-Critique Rating:3 ********************************************* Question: Openstax: A system does 1.80×10^8 J of work while 7.50×10^8 J of heat transfer occurs to the environment. What is the change in internal energy of the system assuming no other changes (such as in temperature or by the addition of fuel)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ‘dU = ‘dQ - ‘dW. Since the heat transfers from the system to the environment, ‘dQ = -7.5*10^8 J and ‘dW = 1.80*10^8 J. ‘dU = 9.3*10^8 J when put into the able equation. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If no energy goes into the system, then the work it does and the energy transferred to the environment are both at the expense of its internal energy. So the system experiences an internal energy change of -1.80 * 10^8 J - 7.50 * 10^8 J = -9.30 * 10^8 J. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Openstax: Steam to drive an old-fashioned steam locomotive is supplied at a constant gauge pressure of 1.75×10^6 N/m2 (about 250 psi) to a piston with a 0.200-m radius. (a) By calculating PΔV , find the work done by the steam when the piston moves 0.800 m. Note that this is the net work output, since gauge pressure is used. (b) Now find the amount of work by calculating the force exerted times the distance traveled. Is the answer the same as in part (a)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ‘dV = ‘pi*r^2*’ds - 0 = ‘pi*(0.200 m)^2 * 0.800 m = 0.101 m^3 ‘dW_net = P*’dV = 1.75*10^6 N/m^2 * 0.101 m^3 = 176,750 J P_abs. = 1.75*10^6 N/m^2 + 101300 N/m^2 = 1.85*10^6 N/m^2 A = ‘pi*0.200m^2 = 0.126 m^2 F = P*A = 1.85*10^6 N/m^2 * 0.126 m^2 = 232,641 N ‘dW = F*d = 232,641 N * 0.800 m = 186,113 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The cross-sectional area of the piston is A = pi r^2 = pi * (0.20 meters)^2 = .13 m^2, approx.. It moves 0.60 meters, so the change in the volume of the cylinder is `dV = 0.13 m^2 * 0.80 m = 0.10 m^3. so, assuming no pressure loss as the piston recedes, the product P `dV is P `dV = 1.75 * 10^6 Pa * 0.10 m^3 = 1.75 * 10^5 N/m^2 * m^3 = 1.75 * 10^5 N * m = 1.75 * 10^5 Joules. The absolute pressure in the cylinder is equal to the gauge pressure plus atmospheric pressure, so P_abs = 1.75 * 10^6 Pa + 1.0 * 10^5 Pa = 1.85 * 10^6 Pa. The force exerted by this pressure on the piston is therefore 0.13 m^2 * 1.85 * 10^6 Pa = 2.4 * 10^5 Newtons. Multiplying this force by the 0.60 meter change in the position of the piston we get `dW = 2.4 * 10^5 N * 0.60 m = 1.4 * 10^5 Joules, which is greater than the work calculated based on the gauge pressure. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Openstax: Calculate the net work output of a heat engine following path ABCDA in the figure below. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The area on a P vs. V graph of a closed loop is equal to the work done. Area = 1.6*10^6 N/m^2 * 3.0*10^-3 m^3 = 4800 J confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The work done by a thermodynamic cycle is equal to the work enclosed by that cycle on a P vs. V diagram. The path ABCDA is a parallogram with altitude 1.6 * 10^6 N/m^2 and base 3.0 * 10^-3 m^3, so it encloses area area = `dW = 1.6 * 10^6 N/m^2 * 3.0 * 10^-3 m^3 = 4.8 * 10^3 N * m = 4800 Joules. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: What is the net work output of a heat engine that follows path ABDA in the figure below, with a straight line from B to D? Why is the work output less than for path ABCDA? Explicitly show how you follow the steps in the Problem-Solving Strategies for Thermodynamics. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The area of the triangle is equal to the work. Area = .5 * 1.6*10^6 N/m^2 * 3.0*10^-3 m^3 = 2400 J The work for each line is the area beneath the line to the x axis. Line BD has a greater area with the x axis than does line CD. These would represent negative work so a larger amount of work is deducted along BD than CD from line AB. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The path ABDA is a triangle with base 1.6 * 10^6 N/m^2 and altitude 3.0 * 10^-3 m^3. The area of the triangle is area = `dW = 1/2 * 1.6 * 10^6 N/m^2 * 3.0 * 10^-3 m^3 = 2400 Joules. Along the path BD negative work is done against the pressure, as is the case along path CD. However the area beneath BD is greater than that beneath CD, so the negative work done in this case is greater in magnitude than the negative work done along path ABCDA. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: gen phy problem 15.12, a-c curved path `dW = -35 J, `dQ = -63 J; a-b-c `dW = - 48 J gen phy how much thermal energy goes into the system along path a-b-c and why? Your Solution: ‘dU will be equal on both curves since start point and end point are the same. ‘dU_A-C = ‘dQ - ‘dW = -63 J - -35 J = -28 J ‘dU_A-C = ‘dU_A-B-C = -28 J = ‘dQ - ‘dW = ‘dQ - - 48J, ‘dQ = -76 J, the negative indicates that the system loses -76 J of thermal energy. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK Your Self-Critique Rating:OK ********************************************* 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: Change in internal energy = ‘dU Work done by the system = ‘dW Thermal energy added to system = ‘dQ The change in internal energy is equal to the thermal energy added to the system minus the work done by the system, or ‘dU = ‘dQ - ‘dW. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In other words, the energy put into the system is conserved and must either perform work by the system of be released in another form of energy, just like conservation of energy. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If a system does work it tends to reduce internal energy, so `dW tends to decrease `dU. If thermal energy is added to the system `dQ tends to increase `dU. This leads to the conclusion that `dU = `dQ - `dW. Thus for example if `dW = -48 J and `dU = -28 J, `dQ = `dU + `dW = -28 J + -48 J = -76 J. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: gen phy How does the halving of pressure caused a halving of the magnitude of the work, and why is the work positive instead of negative as it was in the process a-b-c? Your Solution: Since the work is the area on a P vs. V graph of the enclosed area or on a curve the area from the curve down to the x axis, if the pressure is half, it will result in half the area. When the volume increases, there is positive net work done by the system. When the volume decreases, the work done by the system is negative, or the work done on the system is positive. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK Your Self-Critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: