course Phy 242 Tonight was not my night. I am not sure if I just did not understand this chapter well or if it is because I have a Cell Biology Exam in the morning. Either way I will make sure to review all of the problems and the answers so that I can better my understanding k??w???r??????assignment #003??????????
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20:39:53 query problem 15 introductory problem sets temperature and volume information find final temperature. When temperature and volume remain constant what ratio remains constant?
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RESPONSE --> pV=nRT V/RT = n/p If temperature and volume remain constant you must take and rearrange the ideal-gas equation so that all of the constants are on one side (ie. V,T, and R). Then you should have: constant = n/p. This means that when Temperature and Volume are constant then the ratio between the nuber of moles (n) and the pressure (p) is constant and directly proportional.
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20:40:00 ** PV = n R T so n R / P = T / V; since T and V remain constant T / V and therefore n R / P remain constant; since R is constant it follows that n / P remains constant. **
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RESPONSE --> ok
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20:42:24 why, in terms of the ideal gas law, is T / V constant when only temperature and volume change?
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RESPONSE --> pV=nRT When only temperature and Volume change the ratio of T/V is also constant because: T/V = p/(nR). Since one side of the equation is always constant the ratio on the other side of the equation must always remain the same for the equation to be mathematically equal.
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20:42:46 ** STUDENT ANSWER AND INSTRUCTOR RESPONSE: They are inverselt 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. **
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RESPONSE --> ok
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20:42:54 prin and gen problem 14.3: 2500 Cal per day is how many Joules? At a dime per kilowatt hour, how much would this cost?
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20:42:56 One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 13,000,000 Joules. A watt is a Joule per second, a kilowatt is 1000 Joules / second and a kiliowatt-hour is 1000 Joules / second * 3600 seconds = 3,600,000 Joules. 13,000,000 Joules / (3,600,000 Joules / kwh) = 4 kwh, rounded to the nearest whole kwh. This is about 40 cents worth of electricity. It's worth noting that you use 85% of this energy just keeping yourself warm, so the total amount of physical work you can produce in a day is worth less than a dime.
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20:42:59 prin phy and gen phy problem 14.07 how many Kcal of thermal energy would be generated in the process of stopping a 1200 kg car from a speec of 100 km/hr?
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20:43:01 **STUDENT SOLUTION for 1000 kg car at 100 km/hr WITH ERROR IN CONVERSION OF km/hr TO m/s: The book tells that according to energy conservation initial KE = final KE + heat or (Q) 100km/hr *3600*1/1000 = 360 m/s ** 100 km / hr is 100 * 1000 meters / (3600 sec) = 28 m/s or so. With units your conversion would be 100km/hr *3600 sec / hr *1/ (1000 m / km) = 360 km^2 sec / ( m hr^2). Correct conversion with conversion factors would be 100 km / hr * (1000 m / km) * (1 hr / (3600 sec) = 28 m/s, approx. Otherwise your solution is correct. STUDENT SOLUTION WITH DIFFERENT ERROR IN UNITS Ke=0.5(1000Kg)(100Km)^2 = 5MJ 1Kcal=4186J 5MJ/4186J==1194Kcal INSTRUCTOR COMMENT: Right idea but 100 km is not a velocity and kg * km^2 does not give you Joules. 100 km / hr = 100,000 m / (3600 sec) = 28 m/s, approx. so KE = .5(1000 kg)(28 m/s)^2 = 400,000 J (approx.). or 100 Kcal (approx). **
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20:43:04 query gen phy problem 14.13 .4 kg horseshoe into 1.35 L of water in .3 kg iron pot at 20 C, final temp 25 C, spec ht of hossshoe
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20:43:07 ** STUDENT SOLUTION BY EQUATION (using previous version in which the amount of water was 1.6 liters; adjust for the present 1.35 liters): M1*C1*dT1 + M2*C2*dT2 = 0 is the equation that should be used. 0.4kg * 450J/Kg*Celcius *(25celcius -T1) + 0.3kg * 450J/kg*celcius * (25 celcius - 20 celcius) + 1.6kg *4186J/kg*celcius * (25 celcius - 20 celcius) = 0 } Solve for T1, T1 = 214.8 Celsius Solution below is 189.8 C. GOOD STUDENT SOLUTION: This problem is very similar to the first video experiment. You can determine in the same way the thermal energy gained by the water. The pot gains a little thermal energy too,and you have to know how many Joules are gained by a kg of the iron per degree per kg, which is the specific heat of iron (give in text). You can therefore find the energy lost by the horseshoe, from which you can find the energy lost per degree per kg. For this problem I think we are assuming that none of the water is boiled off ass the hot horse shoe is dropped into the water. First I will find the initial temperature of the shoe. Since water's density is 1g/ml, 1 milliliter of water will weigh 1 gram. So 1.35 Liters of water will have a mass of 1.35 kg. 1.35kg of water is heated by 5 degrees The specific heat of water is 4186 J/kg/degrees celsius so 4186 J / kg / C *1.35 kg * 5 C = 28255 J is necessary to heat the water but since it is in equilibrium with the bucket it must be heated too. mass of bucket = 0.30 kg specific heat of iron = 450 J/kg/degrees 450 J / kg / C * 0.30 kg * 5 C = 675 J to heat bucket So it takes 675 J to heat bucket to 25 degrees celsius 28255 J to heat water to 25 degrees celsius so the horse shoe transferred 675+28255 = 28930 J of energy. Mass of horse shoe = 0.40 kg horse shoe is also iron specific heat of iron = 450 J/kg/degree 28930 J / 0.40kg =72,326 J / kg 72,326 J/kg / (450 J / kg / C) = 160.7 C, the initial temperature of the horseshoe. STUDENT ERROR: MISSING COMMON KNOWLEDGE: Estimate 1.60L of water = 1KG. Could not find a conversion for this anywhere. 1 liter = 1000 mL or 1000 cm^3. Density of water is 1 gram/cm^3 so 1 liter contains 1000 g = 1 kg. Alternatively, density of water is 1000 kg / m^3 and 1 liter = .001 m^3, leading to same conclusion. **
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21:32:40 query univ problem 18.60 (16.48 10th edition) 1.5 L flask, stopcock, ethane C2H6 at 300 K, atm pressure. Warm to 380 K, open system, then close and cool. What is the final pressure of the ethane and how many grams remain? Explain the process you used to solve this problem.
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RESPONSE --> Using the ideal-gas equation pV=nRT you can plug in what you are given to find out how many moles of ethan gas there are in the container. Step 1) Solving for total moles of ethane gas pV=nRT n = (pV)/(RT) n = [(1.013*10^5 Pa)(1.50 L)]/ [(8.314 J/mol*K)(300 K)] n = (151950 Pa*L)/(2494.2 J/mol) n = 60.291 moles, this is the total number of moles in the system Step 2)
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21:34:59 ** use pV = nRT and solve for n n = p V / (r T) = (1.283 *10^5 Pa )(1.50 * 10^-3 m^3 ) / [ (8.36 J / (mol K) )(380 K) ] = .062 mol, approx.. The atomic masses of 2 C and 6 H add up to 30.1, meaning 30.1 grams / mol. So total mass of the gas is initially m(tot) = (.062 mol)(30.1 g/mol) m(tot) = 1.86 g Now if the system is heated to 380 K while open to the atmosphere, pressure will remain constant so volume will be proportional to temperature. Therefore the volume of the gas will increase to V2 = 1.5 liters * 380 K / (300 K) = 1.9 liters. Only 1.5 liters, with mass 1.5 liters / (1.9 liters) * 1.86 grams = 1.47 grams, will stay in the flask. The pressure of the 1.47 grams of ethane is 1 atmosphere when the system is closed, and is at 380 K. As the temperature returns to 300 K volume and quantity of gas will remain constant so pressure will be proportional to temperature. Thus the pressure will drop to P3 = 1 atm * 300 K / (380 K) = .79 atm, approx.. **
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RESPONSE --> My answer differs from yours but I believe it is because we had different ""given"" information. I do believe that I did do the proper procedures despite my answers being different.
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22:33:37 univ phy query problem 18.62 (16.48 10th edition) unif cylinder .9 m high with tight piston depressed by pouring Hg on it. How high when Hg spills over? How high is the piston when mercury spills over the edges?
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RESPONSE --> I can take the ideal-gas equation and change a factor and have it as: pV = NkT Since T is constant you can plug in any nuber as long as you keep it constant in all the equations. (1.00 atm)*V = (6.022*10^23)(1.381*10^-23)(273) V = 2270.372 L = 2.27 m^3 Using this information we can determine the radius of the cylinder and this can be helpful. V = pi*r^2h r = 0.896 m Now you can plug the information known into the ideal-gas equation and solve for the n. pV = nRT (1.00)(2270.372) = n(0.08206)(273) n = 101.3 moles = 20269 grams Now, using F = m*a we can solve for the force required. F = (20269)(9.81) F = 198636.2 N At this point I am unable to determine the answer.
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22:37:07 ** Let y be the height of the mercury column. Since T and n for the gas in the cylinder remain constant we have P V = constant, and since cross-sectional area remains constant V = A * h, where h is the height of the air column, we have P * h = constant. Thus P1 h1 = P2 h2, with P1 = atmospheric pressure = Patm and h1 = .9 m, P2 = Patm + rho g y. Mercury spills over when the depth of the mercury plus that of the air column is .9 m, at which point h2 = h1 - y. So the equation becomes Patm * h1 = (Patm + rho g y) * (h1 - y). We can solve this equation for y (the equation is quadratic). We obtain two solutions; one y = 0 tells us what when there is no mercury (y = 0) there is no deflection below the .9 m level. The other solution is y = (g?1?ho - Pa)/(g?ho) = .140 m, which tells us that .140 m of mercury will again bring us to .9 m level. We might assume that this level must correspond to the level at which mercury begins spilling over. To completely validate this assumption we need to show that the level of the top of the column will be increasing at this point: The level of the top of the mercury column above the bottom of the cylinder can also be regarded as a function f (y) of the depth of the mercury. If mercury depth is y then the pressure in the cylinder is Patm + rho g y and the height of the gas in the cylinder is Patm / (Patm + rho g y ) * h1. The level of the mercury is therefore f(y) = Patm / (Patm + rho g y) * h1 + y The derivative of this function is f ' ( y ) = 1 - Patm??1?ho/(g?ho? + Patm)^2, which is a quadratic function of y. Multiplying both sides by (rho g y + Patm)^2 we solve for y to find that y = sqrt(Patm)?sqrt(g* h1 * rho) - sqrt(Patm) )/ (g?ho) = .067 m approx., is a critical point of f(y). The second derivative f '' (y) is 2 Patm?^2?1?ho^2/(g?ho? + Patm)^3, which is positive for y > 0. This tells us that any critical point of f(y) for which y > 0 will be a relative minimum. So for y = .0635 m we have the minimum possible total altitude of the air and mercury columns, and for any y > .0635 m the total altitude is increasing with increasing y. This proves that at y = .140 m the total height of the column is increasing and additional mercury will spill over. To check that y = .140 m results in a total level of .9 m we note that the air column would then be .9 m - .140 m = .760 m, resulting in air pressure .9 / .760 * 101300 Pa = 120,000 Pa. The pressure due to the .140 m mercury column is 19,000 Pa, which when added to the 101,300 Pa of atmospheric pressure gives us 120,000 Pa, accurate to 3 significant figures. The gauge pressure will be 19,000 Pa. A more direct but less rigorous solution: The cylinder is originally at STP. The volume of the air in the tube is inversely proportional to the pressure and the altitude of the air column is proportional to the volume, so the altitude of the air column is inversely proportional to the pressure. If you pour mercury to depth y then the mercury will exert pressure rho g y = 13,600 Kg/m^3 * 9.8 m/s^2 * y = 133,000 N / m^3 * y. Thus the pressure in the tube will thus be atmospheric pressure + mercury pressure = 101,000 N/m^2 + 133,000 N/m^3 * y. As a result the altitude of the air column will be altitude of air column when y cm of mercury are supported: altitude of air column = atmospheric pressure / (atmospheric pressure + mercury pressure) * .9 m =101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m. At the point where mercury spills over the altitude of the air column will be .9 m - y. Thus at this point 101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m = .9 m - y. This equation can be solved for y. The result is y = .14 m, approx. The pressure will be 101,000 N/m^2 + 133,000 N/m^3 * .14 m = 120,000 N/m^2. The gauge pressure will therefore be 120,000 N/m^2 - 101,000 N/m^2 = 19,000 N/m^2. **
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RESPONSE --> I now understand that I made things much harder than they needed to be. I have also printed out a copy of your answer and will make sure that I am able to understand this kind of problem in the future.
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23:06:38 query univ phy 18.75 (16.61 10th edition) univ phy problem 16.61 for what mass is rms vel .1 m/s; if ice how many molecules; if ice sphere what is diameter; is it visible? Give your solution to the problem.
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RESPONSE --> What is the mass of a molecule or particle at v(rms) equal to 1.00mm/s at 300 K? Vrms = 'sqrt((3kT)/m) = (3(1.38*10^-23)(300))/((1.00)^2) = 1.2429*10^-20 kg
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23:08:56 ** We can solve this problem knowing that ave KE per particle is 3/2 k T so the .5 m v^2 = 3/2 k T, where v is RMS velocity. Thus m = 3 k T / v^2. Then from the density of water and the mass of the particle we can determine its volume, which is equal to 4/3 pi r^3. From this we find r. We obtain volume 3 k T / (v^2 rho), where rho is the density of water. Setting this equal to 4/3 pi r^3 we get r = [ 9 k T / ( 4 v^2 rho) ] ^(1/3). From the mass, Avogadro's number and the mass of a mole of water we determine the number of molecules. The following analysis shows the intermediate quantities we obtain in the process. Some of the calculations, which were done mentally, might be in error so you should redo them using precise values of the constants. At 273 Kelvin we have ave KE = 3/2 k T = 5.5 * 10^-21 Joules. mass is found by solving .5 m v^2 = 3/2 k T for m, obtaining m = 3/2 k T / (.5 v^2) = 5.5 * 10^-21 J / (.5 * (.001 m/s)^2 ) = 1.2 * 10^-14 kg. The volume of the sphere is therefore 1.2 * 10^-14 kg / (1000 kg / m^2) = 1.2 * 10^-17 m^3. Setting this equal to 4/3 pi r^3 we obtain radius r = ( 1.2 * 10^-17 m^3 / 4.2)^(1/3) = ( 2.8 * 10^-18 m^3)^(1/3) = 1.4 * 10^-6 m. Diameter is double this, about 2.8 * 10^-6 m. This is only 3 microns, and is not visible. A water molecule contains 2 hydrogen and 1 oxygen molecule with total molar mass 18 grams = .018 kg. The 1.2 * 10^-14 kg mass of particle therefore consists of 1.2 * 10^-14 / (.018 kg / mole) = 6.7 * 10^-13 moles. With about 6 * 10^23 particles in a mole this consists of 6.7 * 10^-13 moles * 6 * 10^23 particles / mole = 4* 10^11 particles (about 400 billion water molecules). **
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RESPONSE --> I understand that I started out using the wrong equation and that is a key step, to start out correctly.