course Phy 232 004. `query 4
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Given Solution: ** PV = n R T so n R / P = T / V Since T and V remain constant, T / V remains constant. • Therefore n R / P remain constant. • Since R is constant it follows that n / P remains 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: If Temperature and volume are the only two variables to change, then we are presuming that the mass and pressure stays the same. So as the temperature of an ideal gas increases, the volume will increase with it, and as the volume decreases the temperature will decrease along with it. The two variables are proportional, so if no other change is involved, the ratio of the two will stay the same. Confidence Rating:3
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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: I assumed that no ethane left the system when the flask was opened, so (PV/T)0 = (PV/T)1 So the pressure we are looking for can be found by manipulating the above equation to: P1 = (P0*V0*T2/T1*V2) = (110.3 kPa*1.5L*380K)/(1.5 L*300K) = 139.7 kPa = 1.27 atm To find the grams of ethane remaining in the flask, we use the final pressure shown above: m= (MPV)/RT = (30.1 g/mol)*(1.27 atm)*(1.5 L)/(0.0821 L*atm/mol*K * 380K) = 1.83 grams 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.. ** ********************************************* 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: The piston will stop moving when the force from the trapped air equals the force from the mercury. h = depth that piston will sink F(air) = P*A = (nRT)*A/V = nRT*(0.9 – h) F(hg) = (rho(hg)*h*A F(hg) = F(air), nRT/(0.9-h) = rho(hg)*h*A*g We don’t know T, only that its constant We can find n when the piston is at the top of the cylinder: n= PV/(RT) = (1 atm)*(0.9*A m^3)/(0.0*T) plugging this into the equation we get: (1 atm)*(0.9*A m^3)/(0.9 – h) = rho(hg)*h*A*g When we solve for h, we get a quadratic equation: h^2 – 0.9*h+(P(atm)*0.9m)/(rho(hg)*g) = = Unfortunately the roots of this equation are complex, so Im making a mistake somewhere. Confidence Rating:1
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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 also got a quadratic equation, but I should have just let the temperature and mass of the air remain constant and it would have simplified the problem a lot. Your Self-Critique Rating:1 ********************************************* 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: Mass = 3*k*T/(v(ms)^2) = 3*1.38*10^-23 J/K*300 K/(0.001 m/s)^2 = 1.24*10^-14 V = m/rho V = 4/3*pi*r^3 So the radius = (3*1.24*10^-14/(4*pi*1000))^(1/3) = 1.44*10^-6 Confidence Rating: 3
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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. 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). ** Your Self-Critique: I didn’t understand how the diameter couldn’t be viable. Im still sort of unclear about it.