course PHY 232 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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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 #$&* OK ********************************************* Question: why, in terms of the ideal gas law, is T / V constant when only temperature and volume change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: In terms of the ideal gas law (PV = nRT), it only temperature (T) and volume (V) change and nothing else then that makes n(R/P) is constant. And n(R/P) is equal to V/T so it too is a constant which makes T/V constant as well. confidence #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: I felt like I wasn’t explaining it that well but after looking at the given solution I think I touched on all the bases. Your Self-Critique #$&*3 ********************************************* Question: prin and gen problem 14.3: 2500 Cal per day is how many Joules? At a dime per kilowatt hour, how much would this cost? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 10,500,000 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. 10,500,000 Joules / (3,600,000 Joules / kwh) = 3 kwh, rounded to the nearest whole kwh. This is about 30 cents worth of electricity, and a dime per kilowatt-hour. Relating this to your physiology: • You require daily food energy equivalent to 30 cents’ worth of electricity. • It's worth noting that you use 85% of the energy your metabolism produces just keeping yourself warm. • It follows that the total amount of physical work you can produce in a day is worth less than a dime. Your Self-Critique: Your Self-Critique #$&* ********************************************* Question: prin phy and gen phy problem 14.07 how many Kcal of thermal energy would be generated in the process of stopping a 1200 kg car from a speed of 100 km/hr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: NOTE: The given solution is based on student solutions for a previous edition of the text, in which the mass of the car was 1000 kg and its initial velocity 100 km/hr. The adjustments for the current mass and velocity (1200 kg and 95 km/hr as of the current edition) are easily made (see the student question below for a solution using these quantities). **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 INSTRUCTOR COMMENT: 100km/hr *3600 sec / hr *1/ (1000 m / km) = 360 km^2 sec / ( m hr^2), not 360 m/s. The correct conversion would be 100 km / hr is 100 * 1000 meters / (3600 sec) = 28 m/s or so. 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). ** STUDENT QUESTION: The book and this problem originally states *1200kg* as the mass. The solution uses 1000kg, giving the answer as about 100kcal. The book uses 95km/hr, 1200kg, and gets an answer of 100kcal. This problem shows 100km/hr, with a mass of 1000kg in the solution and still an answer of 100kcal. I just want to make sure I am doing it correctly. Where 26.39m/s should be used, I am using the conversion for this specific problem with 100km/hr = 1000m/1hr = 1hr/3600s = 27.78 ~28m/s. KE = 1/2mv^2 = ½(1200kg)(28 m/s)^2 = 470,400kg*m/s^2 = 470,400 J 470,400J = 1 cal/4.186J = 1 kcal/1000cal = 112.37 kcal = 112kcal INSTRUCTOR RESPONSE: I apparently missed the change in the latest edition of the text; or perhaps I chose not to edit the student solutions posted here. In any case your solution is good. Your Self-Critique: Your Self-Critique #$&* ********************************************* Question: query gen phy problem 14.13 .4 kg horseshoe into 1.35 L of water in .3 kg iron pot at 20 C, final temp 25 C, final temp. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 of energy 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 • energy transferred / mass = 28930 J / 0.40kg =72,326 J / kg • 72 330 J / kg, at 450 (J / kg) / C, implies `dT = 72,330 J/kg / (450 J / kg / C) = 160.7 C, the initial temperature of the horseshoe. A symbolic solution: m1 c1 `dT1 + m2 c2 `dT2 + m3 c3 `dT3 = 0. Let object 1 be the water, object 2 the pot and object 3 the horseshoe. Then `dT1 = `dT2 = + 5 C, and `dT3 = 25 C - T_03, where T_03 is the initial temperature of the horseshoe. We easily solve for `dT3: `dT3 = - (m1 c1 `dT1 + m2 c2 `dT2) / (m3 c3) so `dT3 = - (m1 c1 `dT1 + m2 c2 `dT2) / (m3 c3) = - (1.35 kg * 4200 J / (kg C) * 5 C + .3 kg * 450 J / (kg C) ) / ((.4 kg * 450 J / (kg C) ) = -160 C, approx. so 25 C - T_03 = -160 C and • T_03 = 160 C + 25 C = 185 C, approx.. STUDENT ERROR: MISSING COMMON KNOWLEDGE: Estimate 1.60L of water = 1KG. Could not find a conversion for this anywhere. INSTRUCTOR RESPONSE: Each of the following should be common knowledge: • 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. ** Your Self-Critique: Your Self-Critique #$&* ********************************************* Question: query univ problem 18.60 (16.48 10th edition) 1.5 L flask, stopcock, ethane C2H6 at 300 K, atm pressure. Warm to 380 K, open system, then close and cool. What is the final pressure of the ethane and how many grams remain? Explain the process you used to solve this problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Find the total mass of the gas… pV = nRT (1.03 * 10^5 Pa)(1.5*10^-3 m^3) = n(8.31J/molK)(380K) N = (1.03 *10^5 Pa )(1.5 * 10^-3 m^3 ) / ( (8.31 J /mol K )(380 K) ) N = 154.5Pam^3 / 3157.8 J/molK^2 N= .0489 mol Atomic mass of C = 12.0107 *2 = 24.0214 Atomic mass of H = 1.00794 *6 = 6.04764 24.0214 + 6.04764 = 30.06904 g/mol Total mass of gas M = (30.06904 g/mol) * (.0489 mol) = 1.47g The increased volume of the gas after being heated to 380K V = 1.5 liters * 380 K / (300 K) = 1.9L How much will stay in the container… (1.5 L/ 1.9L) * 1.4g = 1.11g Final pressure (after temp goes down) P = 1 atm * 300 K / (380 K) = .789 atm confidence #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** use pV = nRT and solve for n. • n = p V / (R T) = (1.03 *10^5 Pa )(1.5 * 10^-3 m^3 ) / [ (8.31 J / (mol K) )(380 K) ] = .048 mol, approx.. If the given quantities are accurate to 2 significant figures, then calculations may be done to 2 significant figures and more accurate values of the constants are not required. 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) = (.048 mol)(30.1 g/mol) • m(tot) = 1.4 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) * 1.4 grams = 1.1 grams, will stay in the flask. • The pressure of the 1.1 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.. ** Your Self-Critique: Ok Your Self-Critique #$&* OK ********************************************* Question: univ phy query problem 18.62 (16.48 10th edition) unif cylinder .9 m high with tight piston depressed by pouring Hg on it. How high when Hg spills over? How high is the piston when mercury spills over the edges? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P1 h1 = P2 h2 Where P1 = Patm (atmospheric pressure), h1 = .9m, P2 = Patm + rho g y, and h2 = h1 – y Patm (.9m) = (Patm + rho g y) (h1 – y) Y = 0 and y = .140m So the answer would be .140m. confidence #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 • 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 solution is y = 0; this 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•h1•rho - Pa)/(g•rho) = .140 m, which tells us that .140 m of mercury will again bring us to .9 m level. We might assume that this level corresponds 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 (if the height is not increasing the mercury will reach this level but won’t spill over). • The level of the top of the mercury column above the bottom of the cylinder can 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•g•h1•rho/(g•rho•y + 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•rho) = .067 m approx., is a critical point of f(y). The second derivative f '' (y) is 2 Patm•g^2•h1•rho^2/(g•rho•y + 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 the altitude of the 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. ** Your Self-Critique: I didn’t validate my assumption which I should have. I didn’t even think to prove that .14m is the correct answer. But now I see how to further conclude that the answer I got was the correct answer. Your Self-Critique #$&*3 ********************************************* Question: query univ phy 18.75 (16.61 10th edition) univ phy problem 16.61 for what mass is rms vel .1 m/s; if ice how many molecules; if ice sphere what is diameter; is it visible? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: First we need to find r 4/3 pi r^3 = 3 k T / (v^2 rho) r = ( 9 k T / ( 4 v^2 rho) ) ^(1/3) At 273 K avg ke= 3/2 k T = 5.5 * 10^-21 J 3/2 k T / (.5 v^2) = 5.5 * 10^-21 J / (.5 * (.001 m/s)^2 ) = 1.1 * 10^-14 kg 1.1* 10^-14 kg / (1000 kg / m^2) = 1.1 * 10^-17 m^3. 4/3 pi r^3 = 1.1 * 10^-17 m^3 R = (1.1 * 10^-17 m^3 / 4.2)^(1/3) =1.38 * 10^-6 m R^2 = (1.3 * 10^-6)^2 = 2.76 * 10^-6 is the diameter Mass of the water particle 1.2 * 10^-14 kg/ (.018 kg / mole) = 6.67 * 10^-13moles Now… (6.67 *10^-13moles) ( 6 * 10^23 particles/moles) = 4.002 * 10^11 particles confidence #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. 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 m / rho = 3 k T / (v^2 rho), where rho is the density of water. • Setting this equal to 4/3 pi r^3 we get the equation 4/3 pi r^3 = 3 k T / (v^2 rho). The solution is 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 to the naked eye, though it could easily be viewed using a miscroscope. 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). ** STUDENT COMMENT: I'm still not sure about the 'visible' thing. INSTRUCTOR COMMENT: In any case, visible light has a wavelength between about .4 microns and .7 microns. Nothing smaller than this is visible even in principle, in the sense that its image can't be resolved by visible light. If we mean 'visible to the naked eye', that limit occurs between 10 and 100 microns. So this object is in principle visible (wouldn't be hard to resolve with a microscope), but not to the naked eye. Your Self-Critique: OK Your Self-Critique #$&* OK "