#$&* course Phy 232 7/31/138:00 Question: Query set 5 problems 16-20
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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 net work of the system consists of the thermal energy that enters the system and the energy that leaves the system. This means that work done during the cycle is equal to the difference between the thermal energy that enters and exits the system. 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: Same thing. 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: To find efficiency, you need to have the amount of thermal energy put into a system, the amount or work done, and the thermal energy that left the system. Then all you have to do is: work/(energy input) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION: Efficiency is work done / energy input. Add the amount thermal energy removed to the amount of work done to get the input. Then, divide work by the energy input. ** Your Self-Critique: OK Your Self-Critique Rating:OK ********************************************* Question: query univ phy problem (not in 12th edition; solve using the statement of the problem given here) 11th edition 19.56 (17.40 10th edition) In a compressed air engine, input pressure is 1.6 * 10^6 Pa, output pressure is 2.8 * 10^5 Pa. Assume the process to be adiabatic. If we are to avoid frost forming on the output valve, which occurs if the temperature of the exiting air is below freezing, what must be the temperature of the compressed air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: PV=nRT so P/T is constant since V, n, and R are constant. This means P1/T1=P2/T2 1.6x10^6 Pa/T1=2.8x10^5 Pa/273 K T1=1560 K confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: T1=(P1/P2)*(V1/V2)* 273K P1V1^gamma=PsV2^gamma (P1/P2)^(-1/gamma)=V1/V2 T1=(P1/P2)*(P1/P2)^(-1/gamma)* 273K T1 = (1.6x10^6/2.8x10^5)^[1-(1/1.4)]*273 K = 449 K Your Self-Critique Rating:OK ********************************************* Question: query univ 19.64 / 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: PV=nRT V/T is constnat V1/T1=V2/T2 T2=(V2/V1)*T1 T2=(2/1)*355K=710 K Maximum pressure is at point three, so we find it with the same equation: PV=nRT P2V2=P3V3 P3=2.4x10^5pa*(2/1)=4.8x10^5 Work done: each point has to be calculated to find total work: from 1 to 2 W=P*'deltaV P*'deltaV=nR*'deltaT W=.25 mol*8.314 J/molK * (710-355K)=738 J from 2 to 3 Take the derivative to get W=nRT*ln(v3/v2) W=.25 mol * 8.314 J/mol K * 719 K * ln(1/2)=-1023 J from 3 to 1 there is no work done because there was no change in volume total work: W1+W2+W3 738-1023+0=-285 J of total work done this means the gas had 285 J of work done on it confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** .25 mol oxygen at 240 kPa occupies about V = n R T / P = .25 mol * 8.31 J / (mol K) * 355 K / (2.4 * 10^5 N/m^2) = .003 m^3, very approximately. Doubling volume, `dV = 2 * V - V = V = .003 m^2 and P = 2.4 * 10^5 Pa so P `dv = 700 J, very approximately. During isothermal compression we have n = const and T = const so P = n R T / V. Compressing to half the volume, since PV = const, gets us to double the pressure, so max pressure is 2 * 240 kPA = 480 kPa. To get work we integrate P dV. Integral of P dV is calculated from antiderivative n R T ln | V |; integrating between V1 and V2 we have n R T ln | V2 | - n R T ln | V1 | = n R T ln | V2 / V1 |. In this case V2 = V and V1 = 2 V so V2 / V1 = 1/2 and we have `dW = n R T ln(1/2) = .25 mol * 8.31 J/(mol K) * 710 K * (-.7) = -1000 J, approx. So net work is about 700 J - 1000 J = -300 J ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: univ phy describe your graph of P vs. V YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: V is the x axis and P is on the y axis. Point 1 is on the bottom left of the graph. This is the original pressure and volume. Point two is the same y coordinate as point 1, but the x coordinate shifted to the right. Point three is in the same x coordinate as point 2 but the y coordinate is higher. Then, the system resturns to normal so the graph ends up looking like a right triangle. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK Your Self-Critique Rating:OK ********************************************* Question: univ phy What is the temperature during the isothermal compression? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The temperature is 710 K because when volume doubles and pressure is held constant, that means temperature has to double. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If vol doubles at const pressure then temp doubles to 710 K, from which isothermal compression commences. So the compression is at 710 K. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: univ phy What is the max pressure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: It has to be 480 kPa because it originally starts at 240 kPa and since volume is reduced by 2 and temp is constnat, pressure is increased by 2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!