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: 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 says that the change in energy in a system is equal to the Qty. of Heat – Work done. If there is no work done by the system, the thermal energy removed from the system will be the same as that which was added to it. If the amount of work done is equal to the energy that was put into the system, then no thermal energy will leave the system. 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: OK 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: You could calculate the efficiency by dividing the work done by the maximum amount of work that could be done at 100 percent efficiency. Efficiency = work done/max work = W/(Q – deltaU) 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: I was assuming that you know the original amount of work. I figured if you know how much work you did, you could compare it to the maximum amount of work that could possibly be done. Your Self-Critique Rating: 3 of the system. Why would you combine energy lost to work done? Your Self-Critique Rating: ********************************************* 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: In this problem the pressure entering and leaving the system is known. We also know that the final temperature needs to be at least 273 K. - I tried working this problem but I cant get anywhere because I don’t know the volume of air. 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 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: This makes sense algebraically. However, I am still confused as to how PV^gamma is the same thing as (P1/P2)*(V1/V2)^gamma, this is a unit-less ratio, so I don’t understand how we can make that substitution. The rest makes sense though.
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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: Your Self-Critique Rating: ********************************************* Question: univ phy describe your graph of P vs. V YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: From P0 and V0 on the graph, when the volume is doubled the pressure stays the same so there is a line horizontally on the graph to V1,P1. Then when the original volume returns, the temperature stays constant so the pressure increases. So the next point is on a negative slope from V1,P1, directly above V0,P0. The last point ends up exactly were the first point was, so the graph makes a triangular shape. 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: OK Your Self-Critique Rating:OK ********************************************* Question: univ phy What is the temperature during the isothermal compression? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We know that the number of moles stays constant so: (PV/T)0 = (PV/T)1, since the expansion in isobaric, the pressure stays constant also, so: (V/T)0 = (V/T)1, we can solve for T1 = V1/V0*T0 = 2*355 = 710 K Since the compression that follows is isothermic, we know that the pressure stays the same. So the temperature during isothermic compression is 710 K 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: OK Your Self-Critique Rating: OK ********************************************* Question: univ phy What is the max pressure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The pressure doubles when the volume returns to its original amount. When the oxygen is cooled to its original temperature, the pressure is cut in half to its original value. So, the max pressure is 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: OK Your Self-Critique Rating: OK "