#$&* course PHY 242 11:00; 4/11/2011 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: ** PV = n R T so n R / P = V / T Since T and V remain constant, V / T remains constant. • Therefore n R / P remain constant. • Since R is constant it follows that n / P remains constant. ** STUDENT QUESTION: I don’t understand why P is in the denominator when nR was moved to the left side of the equation INSTRUCTOR RESPONSE: The given equation was obtained by dividing both sides by P and by T, then reversing the sides. We could equally well have divided both sides by v and by n R to obtain P / (n R) = T / V, and would have concluded that P / n is constant. To say that P / n is constant is equivalent to saying the n / P is constant. Your Self-Critique:OK, totally understand so far =} Your Self-Critique Rating:3; OK ********************************************* Question: why, in terms of the ideal gas law, is T / V constant when only temperature and volume change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: PV = nRT Putting variables on one side and constants on the other side of the equation: P/T = nR/T Because nRT never chances => given any P and V, if P varies T varies, that is if one goes up the other must as well. They are inversely proportional. This is said that if P/T = 5N/m^2 / K => that if T is over P [T/P] then the constant is 1/5 K*m^2/N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ********************************************* 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: PV = nRT T0 = 300K V0=1.5 L R=8.3145 J/(mol*K) P0 = 1.013 E 5 Pa n = PV/(RT) = 1.013 E 5 N/m^2* 1.5 L / 300K / 8.3145 N*m /(mol*K) =1.013 E 5 N/m^2* .0015m^3 / 300K / 8.3145 N*m /(mol*K) =.06092 mol Mass = 0.06092 mol * 30.1 g/mol = 1.834 g = initial mass The pressure as T increases: Heating while V can change and T, then change => V/T = `V/`T V * `T/ T = 1.5N/m^2 * 380K/300K = 1.9L This is how much vol the gas occupies => the flask is only 1.5 L so we can set up a proportionality evaluation to get the grams left in the flask. 1.834 grams * 1.5L/1.9L =1.448 g The cock is closed so the Pressure changes, the moles remain constant and the Temp falls=> P0 = 1.013 E 5 Pa 1.013 E 5 Pa*300 K/380 K = about 80 kN/m^2 The number of etheline = 1.448g, Finial pressure = 80 000 N/m^2 confidence rating #$&*:I cheated some but it makes perfect sense! ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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; I did keep Pa through the equation but either way the same idea is complete Your Self-Critique Rating:3+ ********************************************* Question: univ phy query problem (publisher has omitted this problem from the 12th edition) 18.62 (16.48 10th edition) A uniform cylinder is .9 meters high, and contains air at atmospheric pressure. It is fitted at the top with a tightly sealed piston. A little bit of mercury (density 13600 kg / m^3) is poured on top of the piston, which increases the force exerted by the piston. The piston therefore descends, compressing the confined air until the pressures equalize. Mercury continues to be added, further lowering the piston and compressing the air. If this continues long enough, mercury will spill over the top of the cylinder. How high is the piston above the bottom of the cylinder when this occurs? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Knowing that the pressure created by the mercury = distance x *density rho * g created by the mercury where x = the height of the mercury. This must be added to the atm pressure = the pressure in the cylinder. The pressure was originally 101300N/m^2 inside and outside of the container! The pressure inside the cylinder must be equal and opposite => P_inside must equal P_outside so… P_outside is already defined, and P inside is related to the change in height of the volume of the cylinder = P_0(y0 * CSA)/ (y`*CSA) = P_0*y0/y`, where y0 = initial height under the piston, and y` = new height under the piston. Assuming the piston is very thin => y0 = y`+x => y` = .9m - xm Knowing this we can say: P_0*y0 = (P0 + rho*g*x) * (y0 - x) = P0*y0 - P0x + rho*g*x*y0 - rho*g*x^2 => 0 = - P0x + rho*g*x*y0 - rho*g*x^2 = -x*( rho*g*x + P0 - rho*g*y0) One soln is x = 0 and the other is (P0 - rho*g*y0)/(rho*g) = -P0/(rho*g) + y0 = -101300N/m^2 / (9.81 m/s^2 * 13600 kg/m^3) + .9m = .141 m confidence rating #$&*:w, I did it with the solutions help ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Makes sense, once again the initial stages were easy for me but after finding the mass of the mercury I wanted to use force equivalency as apposed to pressure. Now I understand that either is possible given a single cross sectional area! Your Self-Critique Rating:3 ********************************************* Question: query univ phy 18.79 was 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: If v_rms = .1 m/s => .1 m/s = sqrt(3*R*T/M), where M = particles / mass, R = gas constant, T = current Temp. = 273K. (.01 m^2/s^2) / 3 = 8.31451 J/mol/K * 273 K / M => M = 3* 8.31451 J/mol/K * 273 K / [(.01 m^2/s^2)] =748305.9 J*s^2/m^2 [kg] The number of g / mole = about 18 => 748305.9/18 mol = 41572 mol Knowing the mass = about 748305.9 kg and the density of ice = about 995 kg/m^3 We determine the vol of water = 748305.9 kg / (995 kg/m^3) = 41.781457286432165 m^3 Of course the vol of a sphere is 4/3 Pi r^3 So setting these values = and solving we find r to equal cubic root(41.781457286432165*3/4/Pi) = 2.15m => the diameter = 4.3m, this is of course visible to the human eye. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: The only thing I do not see is how k was obtained which is how I believe I wound up with a much bigger diameter! Anyway, just let me know where I went wrong and I will fix it. Your Self-Critique Rating:3