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- Physics 221
- Prell
- hw14_solutions.pdf

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Physics 221 Name: Homework 14 Spring 2012 14.1 (a) The temperature near the top of Jupiter’s multicolored cloud layer is about 140K. The temperature at the top of the earth’s troposphere, at an altitude of about 20km, is about 220K. Calculate the rms speed of hydrogen molecules in both thee environments. Give your answers in m/s and as a fraction of the escape speed from the respective planet. (b) Hydrogen gas (H2) is a rare element in the earth’s atmosphere. In the atmosphere of Jupiter, by contrast, 89% of all molecules are H2. Explain why. (c) Suppose an astronomer claims to have discovered an oxygen (O2) atmosphere on the asteroid Ceres. How likely is this to be true? Ceres has a mass equal to 0.014 times the mass of the moon, a density of 2400 kg/m3, and a surface temperature of about 200K. Solution: IDENTIFY: v rms = 3RT M . SET UP: M H 2 = 2.02 ×10 −3 kg/mol . M O 2 = 32.0 ×10 −3 kg/mol . For Earth, M = 5.97 ×10 24 kg and R = 6.38 ×10 6 m . For Jupiter, M = 1.90 ×10 27 kg and R = 6.91×10 7 m . For a sphere, M = ρV = ρ 4 3 πr3 . The escape speed is v escape = 2GM R . EXECUTE: (a) Jupiter: v rms = 3(8.3145J mol ⋅ K)(140K) (2.02 ×10 −3 kg mol) = 1.31×10 3 m s . v escape = 6.06 ×10 4 m/s . v rms = 0.022v escape . Earth: v rms = 3(8.3145J mol ⋅ K)(220K) (2.02 ×10 −3 kg mol) = 1.65×10 3 m s . v escape = 1.12 ×10 4 m/s . v rms = 0.15v escape . (b) Escape from Jupiter is not likely for any molecule, while escape from earth is much more probable. (c) v rms = 3(8.3145J mol ⋅ K)(200K) (32.0 ×10 −3 kg mol) = 395m s. The radius of the asteroid is R = (3M 4πρ)1/3 = 4.68 ×105 m, and the escape speed is v escape = 2GM R = 542m s . Over time the O 2 molecules would essentially all escape and there can be no such atmosphere. EVALUATE: As Figure 18.23 in the textbook shows, there are some molecules in the velocity distribution that have speeds greater than v rms . But as the speed increases above v rms the number with speeds in that range decreases. Physics 221 Name: Homework 14 Spring 2012 14.2 A single cylinder of the engine of a Ferrari F355 F1 sports car takes in 0.440L of 20.0°C air at 101 kPa and compresses it adiabatically to a volume of 9.00% of the original volume. The air may be treated as an ideal gas with . (a) Calculate the final temperature and pressure. (b) Draw a pV-diagram for this process. On this diagram also draw the isotherms for the initial and final temperatures. Solution: (a) For adiabatic compression, so . Thus p final = p initial V final V initial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −γ = (101 kPa)(.0900) −1.4 = 2940 kPa The initial absolute temperature is 293.2 K=20°C for adiabatic processes so thus T final = T initial V final V initial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1−γ = (293.2 K)(.0900) −0.4 = 768.1 K (= 495.0 o C) (b) The pV diagram for the entire process is shown below: Physics 221 Name: Homework 14 Spring 2012 14.3 0.250 mol of an ideal gas at 2.40 x 105 Pa and 355 K are contained in a cylinder with a piston. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume and finally cooled isochorically to its original pressure. a) Show the series of processes on a pV diagram. b) Calculate the behavior of the temperature of the system during the whole process, be quantitative. Sketch your answer. c) Compute the maximum pressure in the system during the whole process. d) Compute the total work done by the piston on the gas during each leg of the cycle. Solution: IDENTIFY: For the isobaric process, For the isothermal process, SET UP: EXECUTE: (a) The pV diagram for these processes is sketched in Figure on right.. (b) Find For process n, R, and p are constant so and (c) The maximum pressure is for state 3. For process n, R, and T are constant. and (d) process process process and The total work done is This is the work done by the gas. The work done on the gas is 285 J. EVALUATE: The final pressure and volume are the same as the initial pressure and volume, so the final state is the same as the initial state. For the cycle, and During the cycle, 285 J of heat energy must leave the gas. Physics 221 Name: Homework 14 Spring 2012 14.4 Consider the hypothetical thermodynamic cycle acting on 1 mole of helium (an ideal mono-atomic gas) depicted by the solid line on the pV-diagram below. The process moves from points and the curve on the graph is an ellipse. (a) Find the work done by the gas when it moves from A to C (via B) on the diagram. How much heat flows into the gas? Be sure to make the sign of these quantities clear. (b) Find the work done by the gas through the entire cycle (i. e. ). How much heat flows into the gas over the course of the cycle? (c) If, instead, the system cycle between the same four points following the diamond path indicated by the dashed line. What is the total work done by the gas during one complete cycle? Explain why it is less than the cycle in part (b) and what feature of the graph indicates this relationship. Physics 221 Name: Homework 14 Spring 2012 Solution (a) The work done by the gas is given by: Since the limits are backwards, the work is clearly negative (i.e. work must be done on the gas to compress it from A to C). Geometrically, referring to the diagram below, the magnitude of the integral is the area under the curve ABC. This is the area in the rectangle minus the area of the half ellipse ABC. The work is therefore Thus We now use the first law of thermodynamics to determine the heat flow in this process. To do this, we need to know the change in internal energy. Since this is an ideal mono-atomic gas, the internal energy is Thus for the initial and final states of this process the internal energy is: The first law of thermodynamics tells us that Physics 221 Name: Homework 14 Spring 2012 The negative sign indicates that the net heat flow is out of the system during this process. (b) The total work done by the gas through the full process is which is just the area of the elliptical curve. This total is positive because, while the first term is negative, the second term is positive and greater in magnitude since it measures the area under ADC. This latter area is clearly is larger than the area under ABC (the magnitude of the first term). The work, in terms of the area of this ellipse, is therefore: This is positive, indicating that the gas does a net positive amount of work. Since this is a cycle and therefore there is no change in internal energy between the initial and final states (because they are the same), the first law of thermodynamics implies that the heat flow into the gas must be the same as the work done by the gas: . Because this is positive, the net flow of heat is into the system. (c) For the diamond curve, the area is smaller than that of the ellipse (since it is enclosed in the ellipse) and therefore the work done by the gas will be smaller. The amount of that work is the area of this diamond: Soeren Prell hw14_solutions

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