course Phy 232 004. `query 4vvvv
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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? Your Solution: If T and V change then n * R/P is constant. Since V and T are inversely proportional if V/T is constant then so is T/V. 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: 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: We would first solve for n using n = p*V/ (R*T) = 1.03*10^5*1.5*10^3/8.31*380 = approx 0.048 mol. You would then take 0.048/30.1 = 1.4 grams. The volume of the gas could then be found by taking V = 1.5 * 380/300 = 1.9 liters. You could then find the pressure drop by 1/ 380 *300 = approx. 0.79 atm. Confidence Rating:3
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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: I forgot to mention how much would stay in the flask. Your Self-Critique Rating: 3 ********************************************* 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? Your Solution: I’m not real sure how to solve this problem. I don’t really understand what formula I should use. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: That’s a lot to take in, so I don’t quite get it all yet. I’ll have to look over it several times.
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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: After looking through it several times, it seems pretty difficult to grasp. I think if I would have just written these formulas down I could have gotten to the answer, so I’ll write them in now. Your Self-Critique Rating: 1