#$&* course phy 231 July 5th, 5:47 pm Copy the problem below into a text editor or word processor. This form accepts only text so a text editor such as Notepad is fine.
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Given Solution: ** STUDENT SOLUTION: Efficiency is work done / energy input. Add the amount thermal energy removed to the amount of work done to get the input. Then, divide work by the energy input. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: query univ phy problem (not in 12th edition; solve using the statement of the problem given here) 11th edition 19.56 (17.40 10th edition) In a compressed air engine, input pressure is 1.6 * 10^6 Pa, output pressure is 2.8 * 10^5 Pa. Assume the process to be adiabatic. If we are to avoid frost forming on the output valve, which occurs if the temperature of the exiting air is below freezing, what must be the temperature of the compressed air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: PV = nRT If V, n, R are all constant, then P/T = constant or (P1/T1) = (P2/T2) (1.6x10^6 Pa) / T1 = (2.8x10^5 Pa) / (273 K) T1 = 1560 K or 1287 C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: My first mistake was that I assumed volume was constant, when it is not. However, I was on the right track with using PV = nRT to begin solving the problem. Instead of P1/T1 = P2/T2, I should have started with (P1V1)/T1 = (P2V2)/T2. I know that T2 is 273 K, so I have (P1V1)/T1 = (P2V2)/273. Rearranging this, I get T1 = (P1/P2)*(V1/V2)*273 K Like others, I had a hard time with gamma at first. I got about this far before getting stuck again: P1V1^gamma = P2V2^gamma (P1/P2) = (V2/V1)^gamma (P1/P2)^(1/gamma) = (V2/V1) (P1/P2)^(-1/gamma) = (V1/V2) I know that I can plug this into my first equation to get T1 = (P1/P2)*(P1/P2)^(-1/gamma)*273 K This is where I got stuck again, because I wasn’t sure how to make this (P1/P2)^(1-1/gamma). However, I went back to the given solution and noticed at the bottom someone else had this problem. I see that I just need to add the exponents since they have the same base to get it. Thus, T1 = [(1.6x10^6 Pa)/(2.8x10^5 Pa)]^[1-(1/1.4)]*273 K = 449 K or 176 C 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 temperature of the gas during the isothermal process is found using the ideal gas equation: PV = nRT V/T = constant V1/T1 = V2/T2 T2 = (V2/V1)T1 T2 = (2/1)*355 K = 710 K The maximum pressure is at point 3. It can also be found using the ideal gas equation: PV = nRT PV = constant P2V2 = P3V3 P3 = P2(V2/V3) P3 = (2.4x10^5 Pa)(2/1) = 4.8x10^5 Pa The total amount of work has to be calculated separately by each step. The first step (from point 1 to point 2) is found first. W = P(’delta V) P(‘delta V) = nR (‘delta T), so W = nR(‘delta T) W = (0.25 mol)(8.314 J/mol*K)(710 K – 355 K) = 738 J From point 2 to point 3 I had a harder time since both pressure and volume are changing. I ended up looking up the formula to use for an isothermal expansion/compression. W = nRT*ln(V3/V2) W = (0.25 mol)(8.314 J/mol*K)(710 K)ln(1/2) = -1023 J From point 3 back to the original point 1, there is no change in volume, which means that no work was done. To find the total amount of work, I simply add the two that I found. Total work = 738 J – 1023 J = -285 J. However, this total is the amount of work done by the gas. The work done by the piston is then + 285 J. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** .25 mol oxygen at 240 kPa occupies about V = n R T / P = .25 mol * 8.31 J / (mol K) * 355 K / (2.4 * 10^5 N/m^2) = .003 m^3, very approximately. Doubling volume, `dV = 2 * V - V = V = .003 m^2 and P = 2.4 * 10^5 Pa so P `dv = 700 J, very approximately. During isothermal compression we have n = const and T = const so P = n R T / V. Compressing to half the volume, since PV = const, gets us to double the pressure, so max pressure is 2 * 240 kPA = 480 kPa. To get work we integrate P dV. Integral of P dV is calculated from antiderivative n R T ln | V |; integrating between V1 and V2 we have n R T ln | V2 | - n R T ln | V1 | = n R T ln | V2 / V1 |. In this case V2 = V and V1 = 2 V so V2 / V1 = 1/2 and we have `dW = n R T ln(1/2) = .25 mol * 8.31 J/(mol K) * 710 K * (-.7) = -1000 J, approx. So net work is about 700 J - 1000 J = -300 J ** Your Self-Critique: I was glad to see the derivation of the work formula for the isothermal compression. I found it, I just wasn’t sure how to get it. Your Self-Critique Rating: 3 ********************************************* Question: univ phy describe your graph of P vs. V YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: V is on the x-axis and P is on the y-axis. Point 1 is towards the bottom left, which was the original position of pressure and volume. Point 2 is in the same y-position as Point 1, but farther down the x-axis. This creates a horizontal line segment. Point 3 is in the same x-position as point 1, but further up on the y-axis. I made a straight, diagonal line with a negative slope between points 2 and 3. The system then returns back to the original position, which means that I simply drew a vertical line down from point 3 to point 1. When finished, my graph looked like a right triangle. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT COMMENT I still had no idea after the explanation. INSTRUCTOR RESPONSE To understand this graph, do the following: • Sketch a graph of y = 1 / x, using, for example, x values .1, .5, 1, 2, 10. Having done so it should not be difficult to understand the shape of this graph. • If you wanted to sketch the graph of y = 100 / x, you could use the same graph, and just relabel your vertical axis (for example you would replace 1 with 100, 2 with 200, etc). Do so. • Now you can use the same graph for P = (2 P0 V0) / V. Instead of y, label the vertical axis P. Instead of x, label the horizontal axis V. Relabel the vertical axis in terms of your original x and y coordinates, multiplying each of your vertical coordinates by 2 P0. Relabel the horizontal axis in terms of V, replacing 1 with V0, 2 with 2 V0, etc.. [ The original coordinates of your five plotted points were (.1, 10), (.5, 2), (1, 1), (2, .5) and (10, .1). Relabeled they would be (.1 V0, 20 P0), (.5 V0, 4 P0), (V0, 2 P0), (2 V0, .5 P0) and (10 V0, .1 P0) ] Your Self-Critique: I drew the second line straight instead of a curve. I wasn’t sure what the behavior of the graph would be, but I just wanted a brief sketch to give me an idea. I understand why the line is curved, though, especially after reading the explanation. Your Self-Critique Rating: 3 "
#$&* course phy 231 July 5th, 5:47 pm Copy the problem below into a text editor or word processor. This form accepts only text so a text editor such as Notepad is fine.
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Given Solution: ** STUDENT SOLUTION: Efficiency is work done / energy input. Add the amount thermal energy removed to the amount of work done to get the input. Then, divide work by the energy input. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: query univ phy problem (not in 12th edition; solve using the statement of the problem given here) 11th edition 19.56 (17.40 10th edition) In a compressed air engine, input pressure is 1.6 * 10^6 Pa, output pressure is 2.8 * 10^5 Pa. Assume the process to be adiabatic. If we are to avoid frost forming on the output valve, which occurs if the temperature of the exiting air is below freezing, what must be the temperature of the compressed air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: PV = nRT If V, n, R are all constant, then P/T = constant or (P1/T1) = (P2/T2) (1.6x10^6 Pa) / T1 = (2.8x10^5 Pa) / (273 K) T1 = 1560 K or 1287 C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: My first mistake was that I assumed volume was constant, when it is not. However, I was on the right track with using PV = nRT to begin solving the problem. Instead of P1/T1 = P2/T2, I should have started with (P1V1)/T1 = (P2V2)/T2. I know that T2 is 273 K, so I have (P1V1)/T1 = (P2V2)/273. Rearranging this, I get T1 = (P1/P2)*(V1/V2)*273 K Like others, I had a hard time with gamma at first. I got about this far before getting stuck again: P1V1^gamma = P2V2^gamma (P1/P2) = (V2/V1)^gamma (P1/P2)^(1/gamma) = (V2/V1) (P1/P2)^(-1/gamma) = (V1/V2) I know that I can plug this into my first equation to get T1 = (P1/P2)*(P1/P2)^(-1/gamma)*273 K This is where I got stuck again, because I wasn’t sure how to make this (P1/P2)^(1-1/gamma). However, I went back to the given solution and noticed at the bottom someone else had this problem. I see that I just need to add the exponents since they have the same base to get it. Thus, T1 = [(1.6x10^6 Pa)/(2.8x10^5 Pa)]^[1-(1/1.4)]*273 K = 449 K or 176 C 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 temperature of the gas during the isothermal process is found using the ideal gas equation: PV = nRT V/T = constant V1/T1 = V2/T2 T2 = (V2/V1)T1 T2 = (2/1)*355 K = 710 K The maximum pressure is at point 3. It can also be found using the ideal gas equation: PV = nRT PV = constant P2V2 = P3V3 P3 = P2(V2/V3) P3 = (2.4x10^5 Pa)(2/1) = 4.8x10^5 Pa The total amount of work has to be calculated separately by each step. The first step (from point 1 to point 2) is found first. W = P(’delta V) P(‘delta V) = nR (‘delta T), so W = nR(‘delta T) W = (0.25 mol)(8.314 J/mol*K)(710 K – 355 K) = 738 J From point 2 to point 3 I had a harder time since both pressure and volume are changing. I ended up looking up the formula to use for an isothermal expansion/compression. W = nRT*ln(V3/V2) W = (0.25 mol)(8.314 J/mol*K)(710 K)ln(1/2) = -1023 J From point 3 back to the original point 1, there is no change in volume, which means that no work was done. To find the total amount of work, I simply add the two that I found. Total work = 738 J – 1023 J = -285 J. However, this total is the amount of work done by the gas. The work done by the piston is then + 285 J. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** .25 mol oxygen at 240 kPa occupies about V = n R T / P = .25 mol * 8.31 J / (mol K) * 355 K / (2.4 * 10^5 N/m^2) = .003 m^3, very approximately. Doubling volume, `dV = 2 * V - V = V = .003 m^2 and P = 2.4 * 10^5 Pa so P `dv = 700 J, very approximately. During isothermal compression we have n = const and T = const so P = n R T / V. Compressing to half the volume, since PV = const, gets us to double the pressure, so max pressure is 2 * 240 kPA = 480 kPa. To get work we integrate P dV. Integral of P dV is calculated from antiderivative n R T ln | V |; integrating between V1 and V2 we have n R T ln | V2 | - n R T ln | V1 | = n R T ln | V2 / V1 |. In this case V2 = V and V1 = 2 V so V2 / V1 = 1/2 and we have `dW = n R T ln(1/2) = .25 mol * 8.31 J/(mol K) * 710 K * (-.7) = -1000 J, approx. So net work is about 700 J - 1000 J = -300 J ** Your Self-Critique: I was glad to see the derivation of the work formula for the isothermal compression. I found it, I just wasn’t sure how to get it. Your Self-Critique Rating: 3 ********************************************* Question: univ phy describe your graph of P vs. V YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: V is on the x-axis and P is on the y-axis. Point 1 is towards the bottom left, which was the original position of pressure and volume. Point 2 is in the same y-position as Point 1, but farther down the x-axis. This creates a horizontal line segment. Point 3 is in the same x-position as point 1, but further up on the y-axis. I made a straight, diagonal line with a negative slope between points 2 and 3. The system then returns back to the original position, which means that I simply drew a vertical line down from point 3 to point 1. When finished, my graph looked like a right triangle. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT COMMENT I still had no idea after the explanation. INSTRUCTOR RESPONSE To understand this graph, do the following: • Sketch a graph of y = 1 / x, using, for example, x values .1, .5, 1, 2, 10. Having done so it should not be difficult to understand the shape of this graph. • If you wanted to sketch the graph of y = 100 / x, you could use the same graph, and just relabel your vertical axis (for example you would replace 1 with 100, 2 with 200, etc). Do so. • Now you can use the same graph for P = (2 P0 V0) / V. Instead of y, label the vertical axis P. Instead of x, label the horizontal axis V. Relabel the vertical axis in terms of your original x and y coordinates, multiplying each of your vertical coordinates by 2 P0. Relabel the horizontal axis in terms of V, replacing 1 with V0, 2 with 2 V0, etc.. [ The original coordinates of your five plotted points were (.1, 10), (.5, 2), (1, 1), (2, .5) and (10, .1). Relabeled they would be (.1 V0, 20 P0), (.5 V0, 4 P0), (V0, 2 P0), (2 V0, .5 P0) and (10 V0, .1 P0) ] Your Self-Critique: I drew the second line straight instead of a curve. I wasn’t sure what the behavior of the graph would be, but I just wanted a brief sketch to give me an idea. I understand why the line is curved, though, especially after reading the explanation. Your Self-Critique Rating: 3 "
#$&* course phy 231 July 5th, 5:47 pm Copy the problem below into a text editor or word processor. This form accepts only text so a text editor such as Notepad is fine.