#$&* course Phy 122 12:50 am June 11 Your solution, attempt at solution:If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: ** PV = n R T so n R / P = V / T Since T and V remain constant, V / T remains constant. · Therefore n R / P remain constant. · Since R is constant it follows that n / P remains constant. ** STUDENT QUESTION: I don’t understand why P is in the denominator when nR was moved to the left side of the equation INSTRUCTOR RESPONSE: The given equation was obtained by dividing both sides by P and by T, then reversing the sides. We could equally well have divided both sides by v and by n R to obtain P / (n R) = T / V, and would have concluded that P / n is constant. To say that P / n is constant is equivalent to saying the n / P is constant. Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: why, in terms of the ideal gas law, is T / V constant when only temperature and volume change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: They have to change to keep that proportion confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT ANSWER AND INSTRUCTOR RESPONSE: They are inversely proportional. They must change together to maintain that proportion. INSTRUCTOR RESPONSE: You haven't justified your answer in terms of the ideal gas law: PV = n R T so V / T = n R / P. If only T and V change, n and P don't change so n R / P is constant. Therefore V / T is constant, and so therefore is T / V. You need to be able to justify any of the less general relationships (Boyle's Law, Charles' Law, etc.) in terms of the general gas law, using a similar strategy. ** Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 86 C = 86 + 273.15 = 359.15 degrees K -100 C = -100 + 273.15 = 173.15 degrees K 5500 C = 5500 + 273.15 = 5773.15 degrees K 78 F is (78 F - 32 F) = 46 F above the freezing (5/9 C / F ) * 46 F = 26 C (26 + 273) K = 299 K. -459 F to Celsius -459 F is (459 + 32) F = 491 F below freezing (5/9 C / F) * ( -491 F) = 273 C below freezing. (-273 + 273) K = 0 K confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The Kelvin temperature is 273 K higher than the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree). · 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -100 + 273 K = 173 K, (5500 + 273) K = 5773 K. The freezing point of water is 0 C or 32 F, and a Fahrenheit degree is 5/9 the size of a Celsius degree. Therefore · 78 F is (78 F - 32 F) = 46 F above the freezing point of water. · 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing. · Since freezing is at 0 C, this means that the temperature is 26 C. · The Kelvin temperature is therefore (26 + 273) K = 299 K. Similar reasoning can be used to convert -459 F to Celsius · -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing. · This is -273 C or (-273 + 273) K = 0 K. · This is absolute zero, to the nearest degree. Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P V = n R T T2 = (P2 / P1) * (V2 / V1) * T1 T2 = 40 * 1/ 9 * 293 K confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: First we reason this out intuitively: If the air was compressed to 1/9 its original volume and the temperature didn’t change, it would end up with 9 times its original pressure. However the pressure changes from 1 atm to 40 atm, which is a 40-fold increase. The only way the pressure could end up at 40 times the original pressure, as opposed to 9 times the original, would be to heat up. Its absolute temperature would therefore have to rise by a factor of 40 / 9. Its original temperature was 20 C = 293 K, so the final temperature would be 293 K * 40/9, or over 1300 K. Now we reason in terms of the ideal gas law. P V = n R T. In this situation the number of moles n of the gas remains constant. Thus P V / T = n R, which is constant, and thus P1 V1 / T1 = P2 V2 /T2. The final temperature T2 is therefore · T2 = (P2 / P1) * (V2 / V1) * T1. From the given information P2 / P1 = 40 and V2 / V1 = 1/9 so · T2 = 40 * 1/9 * T1. The original temperature is 20 C = 293 K so that T1 = 293 K, and we get · T2 = 40 * 1/9 * 293 K, the same result as before. Your Self-Critique:ok Your Self-Critique Rating: ok ********************************************* Question: `q001. The temperature of a certain object increases from 50 Celsius to 150 Celsius. What is its change in temperature in Celsius? Convert 50 Celsius and 150 Celsius to Kelvin. What is the change in temperature in Kelvin? What is the change in temperature in Fahrenheit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 50 degrees C + 273.15 = 323.15 degrees Kelvin 150 degrees C + 273.15 = 423.15 degrees Kelvin Change in temperature in Kelvin is 100 degree increase F= (C * 9/ 5) + 32 F= (50 * 9/5) + 32 F= 90 + 32 F = 122 degrees F F = (C * 9/ 5) + 32 F = (150 * 9/ 5) + 32 F= 180 + 32 F= 212 degrees F Fahrenheit increased by 100 degrees confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok ********************************************* Question: `q002. A sample of gas originally at 0 Celsius and pressure 1 atmosphere is compressed from volume 450 milliliters to volume 50 milliliters, in which state its pressure is 16 atmospheres. What is its temperature in the new state? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P V= n R T = (1 atm) (.450 L) = n (0.08206) (273 K) n = 0.02 = (16 atm) (0.050 L) = (0.02) (0.08200) T new temperature= 487.4 K (214.4 degrees Celsius) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok ********************************************* Question: `q003. What product or ratio involving P, V, n and T would remain constant if V and T were held constant? Why does this make sense? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: V/ T would remain constant...P/ T would remain constant also because V/ T did
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Given Solution: Simple reasoning: the volume increases by a factor of 20, which by itself would imply that the pressure decreases by a factor of 20. But absolute temperature also decreases by a factor of 223 / 283, which would also imply a pressure decrease. Bottom line: The pressure changes by factor (1/20) * (223/283) = .039, ending up at .039 of its original value. More reliable analysis: If V_0 and V_f are the initial and final volumes, then V_f / V_0 = 20 The initial and final temperatures are 10 C = 283 K and -50 C = 223 K, so the ratio of temperatures is T_f / T_0 = 223 / 283. Since no helium escapes, n is constant to the PV = n R T implies that P V / T is constant. Thus P_0 V_0 / T_0 = P_f V_f / T_f. Solving this for the pressure P_f we obtain P_f = P_0 * V_0 / V_f * T_f / T_0 = P_0 * (1/20) * (223/283) = P_0 * .039. That is, the pressure at the new altitude is about .039 that at the surface. Assuming the surface pressure to have been 1 atmosphere, the gauge pressure at altitude would be (.039 atm - 1 atm) = -.961 atm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q004. A body with a 1/2 m^2 surface area, at temperature 25 Celsius, has an emissivity of approximately 1. It exchanges energy by radiation with a large surface at temperature -20 Celsius. At what net rate does it lose energy? By what percent would the rate of energy loss change if the large surface was at temperature -270 Celsius? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Hnet = Ae (stefan-boltzmann constant) (T ^4 - Ts ^4) Hnet = (1/2) (1) (5.67051 x 10^-8 W/m^2*K) ((298 K )^4 - (253 K)^4) = (1/2) (1) (5.67051 * 10^-8 W/m ^2 *K) (3788998335 K) = 107.43 W If the large surface was at -270 celsius, then the new rate would be: Hnet = (1/ 2) (1) (5.67051 * 10^-8 W/m ^2 * K)((298 K) ^4 - (3 K) ^4) = (1/ 2) (1)(5.67051 x 10 ^-8 W/ m ^2 *K)(7886150335 K) = 223.6 W change (223.6 W / 107.43 W) x 100 = 208.12 % confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!