ractice Exercises. psi = 70 mm Hg in. Hg = 70 mm Hg 4.7 psi 760 mm Hg 9.9 in. Hg 760 mm Hg = 4. psi = 8.7 in. Hg. ascals = 888 mbar torr = 888 mbar bar 0, 5 ascal 000 mbar.0 bar bar 000 mbar.0 bar = 666 torr = 88,800 ascal. mm Hg = 5 cm Hg 0 mm Hg cm Hg = 50 mm Hg 760 mm Hg mm Hg = 770 torr = 770 mm Hg The maximum pressure = 770 mm Hg + [(50 mm Hg) ] = 70 mm Hg The minimum pressure = 770 mm Hg [(50 mm Hg) ] = 70 mmhg.4 The pressure of the gas in the manometer is the pressure of the atmosphere less the pressure of the mercury,.7 cm Hg. Using the pressure in the atmosphere from the previous example: mm Hg =.7 cm Hg 0 mm Hg cm Hg 770 mm Hg 7 mm Hg = 65 mm Hg V V.5 T T V = V and T = T V V T T = / The pressure must change by /. = 7 mm Hg.6 Since volume is to decrease, pressure must increase, and we multiply the starting pressure by a volume ratio that is larger than one. Also, since V = V, we can solve for : = V V 740 torr 880 ml 870 ml = 750 torr.7 In general the combined gas law equation is: V T TV 50 m 98. K V T 745 torr 950 m. K V T 688 torr, and in particular, for this problem, we have: 7
.8 When gases are held at the same temperature and pressure, and dispensed in this fashion during chemical reactions, then they react in a ratio of volumes that is equal to the ratio of the coefficients (moles) in the balanced chemical equation for the given reaction. We can, therefore, directly use the stoichiometry of the balanced chemical equation to determine the combining ratio of the gas volumes: L O = (4.50 L CH 4 ).9 L O = (6.75 L CH 4 ) volume O volume CH volume O volume CH.50 L O 0.9% O = 00% x L air 00% L air = (.50 L O ) = 64.6 L air 0.9%.0 ml O ml V(NO) 4 4 = 9.00 L O =.50 L O V O = volume O volume NO x 84 ml NO = 9 ml O V = 7 torr 9.0 ml 95 K 755 torr 8 K = 8.7 ml O. First determine the number of moles of CO in the tank: n = V RT = 000 psig 4.696 psig = 40 atm 0.48 cm ml L V = 6.0 ft ft cm 000 ml R = 0.08 T = C + 7 = 95 K = 70 L 40 atm 70 L mol CO in the tank = 0.08 95 K = 980 mol CO Then find the total number of grams of CO in the tank, MW CO = 44.0 g/mol 44.0 g CO g CO in the tank = 980 mol CO = 4,000 g CO mol CO Amount of solid CO = 4,000 g CO 0.5 = 5,000 g solid CO V 57.8 atm.0 L. n = = 8. moles gas RT 0.08 98 K 8. mol Ar (9.95 g Ar/mol) =,0 g Ar 8
. Find the number of moles of argon V.0000 atm 0.544 L n = RT 0.08057 7.5 K Mass of the argon = (0.048 mol argon) = 0.048 mol argon 9.948 g arg on mol arg on = 0.96998 g argon The mass of the flask = 75.6898 g 0.96998 g air = 74.798 g Mass of the organic compound = 76.06 g 74.798 g =.4 g The number of moles of the organic compound equals the number of moles of air:.4 g organic compound MW = = 58. g/mol 0.048 mole organic compound The unknown could be butane, MW = 58. g/mol.4 Since V = nrt, then n = V/RT 685 torr 0.00 L V n RT 0.08 00. K.45 g molar mass g mol 0.00 mol The gas must be xenon. 0.00 moles gas.5 Density of air = mol air Density of radon at ST = mol Rn 8.8 g air mol air mol air.4 L =.9 g/l.0 g Rn mol Rn mol Rn.4 L = 9.9 g/l Since radon is almost eight times denser than air, the sensor should be in the lowest point in the house: the basement..6 d = m/v Taking.00 mol SO : m = 64. g nrt V =.00 mol 0.08 68. K =. L =,00 ml 96.5 ka 0.5 ka density = 64. g. L =.77 g/l.7 In general V = nrt, where n = mass formula mass. Thus V mass RT formula mass We can rearrange this equation to get; formula mass (mass/v)rt drt 9
5.60 g L 0.08 96. K formula mass = = 8 g mol 750 torr The empirical mass is 69 g mol. The ratio of the molecular mass to the empirical mass is - 8 g mol - 69 g mol = Therefore, the molecular formula is times the empirical formula, i.e., F 4..8 formula mass formula mass = (mass/v)rt drt 5.55 g L 0.08. K.5 atm = 4 g mol Since the compound contains C and H it could be an alkane, C n H n+, an alkene, CV n H n, or an alkyne, C n H n-. 9 C and 6 H 8 C and 8 H 7 C and 0 H 6 C and 4 H 5 C and 54 H 4 C and 66 H C and 78 H C and 90 H C and 0 H The most probable compound is C 8 H 8 also known as octane..9 CS (g) + O (g) SO (g) + CO (g) mol CS mol of CS =.0 g CS 76. g CS mol CO mol CO = 0.445 mol CS mol CS mol SO mol SO = 0.445 mol CS mol CS L CO = 0.445 mol CO 0.08 0 K 88 torr 0.890 mol SO 0.08 0 K L SO = 88 torr The total number of liters is = 9. L. = 0.445 mol CS = 0.445 mol CO = 0.890 mol SO =.07 L CO = 6.4 L SO 0
.0 CaCO (s) CaO(s) + CO (s) mol CO = 78 torr 0.57 L V RT 0.08 96 K mol CaCO mol CaCO = 0.007 mol CO mol CO g CaCO = 0.007 mol CaCO 00.09 g CaCO mol CaCO. mol Ar =.0 g Ar Ar = mol Ar 9.95 g Ar = 0.75 mol Ar 0.75 mol Ar 0.08 0 K.00 L mol N mol N = 0.6 g N 8.0 g N N = = 0.78 mol N 0.78 mol N 0.08 0 K.00 L mol O mol O = 4. O.00 g O = 0.447 mol O = 0.007 mol CO = 0.007 mol CaCO = 6.80 atm = 9.4 atm =.0 g CaCO 0.447 mol O 0.08 0 K O = =.04 atm.00 L total = Ar + N + O = 6.80 atm + 9.4 atm +.04 atm = 7.8 atm. We can determine the pressure due to the oxygen since total = N + O. O = total N = 7.0 atm 5.0 atm =.0 atm. We can now use the ideal gas law to determine the number of moles of O : V (.0 atm)(7.00 L) n = = 84.8 mol O RT 0.08 (98 K) g O = (84.8 mol O ).0 g O mol O = 7 g O. The total pressure is the pressure of the methane and the pressure of the water. We can determine the pressure of the methane by subtracting the pressure of the water from the total pressure. The pressure of the water is determined by the temperature of the sample. At 8 C, the partial pressure of water is.8 torr. CH4 = T total water = 775 torr 8. torr = 747 torr The pressure in the flask is 74.8 torr. mol CH 4 = V RT (747 torr) (.50 L) 0.08 (0 K) = 0.0994 mol CH 4
.4 First we find the partial pressure of nitrogen, using the vapor pressure of water at 5 C: N = total water = 745 torr.79 torr = 7 torr. To calculate the volume of the nitrogen we can use the combined gas law V V T T For this problem, V V T (7. torr)(0.7 L)(7 K) T ()(88 K) 89 ml.5 Since the stoichiometric ratio of the SO and SO are the same, the pressure in the flask after the reaction when the only substance in the flask is SO will be the same as the pressure in flask when there is just SO, 0.750 atm. After the reaction additional oxygen gas is added to the flask. (O ) added to flask after reaction = 0.750 atm SO x O atm SO The total pressure of SO and O = 0.750 atm + 0.75 atm =.5 atm = 0.75 atm.6 Find the number of moles of both the H and NO then find the mol fractions. mol H = (.5 g H ) mol NO = (4.0 g NO) H = mol H.06 g H mol NO 0.0 g NO.07 mol H. mol NO +.07 mol H. mol NO =.07 mol H =. mol NO = 0.486 NO =. mol NO +.07 mol H = 0.54 H = ( total )( H ) = (.05 atm)(0.486) = 0.996 atm NO = ( total )( NO ) = (.05 atm)(0.54) =.05 atm.7 The mole fraction is defined in Equation.5: O 6 torr X O 0.47 or 4.7% total 788 torr.8 effusion rate (Br 8) MBr 79 78.9 effusion rate (Br 79) M 80.9 = 0.988 Br 8.9 Use Equation.7; effusion rate (HX) effusion rate (HCl) M M HCl HX effusion rate (HX) M HX M HCl 6.46 g mol (.88) 8.9 g mol effusion rate (HCl) The unknown gas must be HI.
Review Questions. The reason it hurts more to be jabbed by a point of a pencil rather than the eraser, even though the force is the same, is because the area of the point is smaller than the area of the eraser, and therefore, the pressure is higher.. (a) (b) (c) (d) (e) (f) 0.5 ka torr mm Hg.0 bar 0,5 a and 0.5 ka and mm Hg torr and 0,5 a.0 bar and 0,5 a and 0,5 a bar 0.9868 atm and 0.9868 atm bar. Since the density of water is approximately times smaller than that of mercury, a barometer constructed with water as the moveable liquid would have to be some times longer than one constructed using mercury. Also, the vapor pressure of water is large enough that the closed end of the barometer may fill with sufficient water vapor so as to affect atmospheric pressure readings. In fact, the measurement of atmospheric pressure at normal temperatures would be about 8 torr too low, due to the presence of water vapor in the closed end of the barometer..4 A closed end manometer reads pressure without the need to correct for atmospheric pressure..5 (a) Temperature Volume Law: The volume of a given mass of a gas is directly proportional to the Kelvin temperature, provided the pressure is held constant: V T or V /T = V /T, at constant. This is Charles' Law. (b) Temperature ressure Law: The pressure of a gas is directly proportional to the Kelvin temperature, provided the volume is held constant: T or /T = /T, at constant V. This is Gay-Lussac s Law. (c) ressure Volume Law: The volume of a given mass of a gas is inversely proportional to the pressure, provided the temperature is held constant: V / or V = V, at constant T. This is Boyle's Law. (d) Combined Gas Law: The pressure and volume of a gas are directly proportional to the Kelvin temperature, provided the number of particles is held constant. V T or V /T = V /T, at constant n..6 (a) number of moles and temperature (b) number of moles and pressure (c) number of moles and volume (d) number of moles.7 An ideal gas obeys the gas laws over all pressures and temperatures. A real gas behaves most like an ideal gas at low pressures and high temperatures.
.8 V = nrt (a) lot temperature in Kelvin versus Volume. (b) lot temperature in Kelvin versus ressure. (c) lot ressure versus the inverse Volume..9 total = a + b + c +.0 Mole fraction is the ratio of the number of moles of one component of a mixture to the total number of moles of all components.. Middle drawing Left drawing: A: 0.500 atm B: 0.500 atm Middle drawing: A: 0.600 atm B: 0.400 atm Right drawing A: 0.667 atm B: 0. atm. Diffusion is the spontaneous intermingling of one substance with another while effusion is the movement of a gas through a very tiny opening into a region of lower pressure. effusion rate (A) db M effusion rate (B) d M A B A. A gas consists of hard, super small or volumeless particles in random motion, and the particles neither attract nor repel one another..4 The temperature and pressure will decrease..5 The increase in temperature requires an increase in kinetic energy. This can happen only if the gas velocities increase. Higher velocities cause the gas particles to strike the walls of the container with more force, and this in turn causes the container to expand if a constant pressure is to be maintained..6 The increase in temperature causes an increase in the force with which the gas particles strike the container walls. If the container cannot expand, an increase in pressure must result..7 The minimum temperature corresponds to zero kinetic energy, which is accomplished only when velocity is zero. In other words, the molecules have ceased all movement..8 The answer (c) NH will have the largest rms since it has the lowest molecular mass..9 (a) As the pressure of a gas increases, the rate of effusion should increase since the molecules will hit the walls of the container more frequently and with greater force. If the molecules hit more frequently, they are more likely to go through the small openings in the walls of the container. (b) As the temperature of the gas increases, the rate of effusion will increase since temperature is proportional to kinetic energy which is dependent on the velocity of the particles. The faster the particles move, the more likely they are to hit the walls and pass through the small openings..0 It is not true that the gas particles occupy no volume themselves, apart from the volume between the gas particles. Also, it is not true that the gas particles exert no force on one another. In other words, real molecules occupy space and attract or repel one another. Because of short-range interactions, it is also not true that particles travel always in straight paths.. A small value for the constant a suggests that the gas molecules have weak forces of attraction among themselves. 4
. (b) has a larger value of the van der Waals constant b, since it is a larger molecule.. Under the same conditions of T and V, the pressure of a real gas is less than the pressure of an ideal gas because real gases do not have perfectly elastic collisions and may clump together and stick to the walls of the container, thus decreasing the number of collisions the gas makes. The volume of a real gas is greater than the volume of an ideal gas because the atoms and molecules take up space..4 The helium atoms are moving faster than the argon atoms because they have less mass. Review roblems.5 (a) torr = (.6 atm) (b) (c) (d) atm = (740 torr) mm Hg = 78 torr torr = (.45 0 a).6 (a) torr = (0.65 atm) (b) (c) (d) atm = (85 torr) torr = 6 mm Hg bar =. ka 760 mm Hg = 958 torr = 0.974 atm 5.05 0 a 760 mm Hg.0 bar 0. ka = 78 mm Hg = 475 torr =.09 atm = 6 torr = 0.0 bar = 0.9 torr.7 (a) torr = (0.9 atm) (b) torr = (0.460 atm) = 50 torr = 50 torr.8 (a) atm = (595 torr) (b) (c) atm = (60 torr) atm = (0.00 torr) = 0.78 atm = 0. =.95 0 4 atm.9 765 torr 70 torr = 45 torr 45 torr cm Hg = (45 mm Hg) cm 0 mm = 4.5 cm Hg 760 mm Hg = 45 mm Hg 5
gas 4.5 cm.0 80 torr 750 torr = 70 torr 70 torr cm Hg = (70 mm Hg) cm 0 mm = 7.0 cm Hg 760 mm Hg = 70 mm Hg gas 7.0 cm. 65 mm Hg 760 mm Hg = 65 torr 748 torr + 65 torr = 8 torr. 8 mm Hg 760 mm Hg = 8 torr 75 torr 8 torr = 670 torr. In a closed-end manometer the difference in height of the mercury levels in the two arms corresponds to the pressure of the gas. Therefore, the pressure of the gas is 5 mm Hg. 5 mm Hg 760 mm Hg = 5 torr 6
.4 The closed-end manometer data indicates that the pressure inside the flask is 6 mm Hg. The open-end manometer data indicate that atm = 5 mm Hg + 6 mm Hg = 748 mm Hg..5 Use Boyle s Law to solve for the second volume: V V 55 ml 75 torr 65 torr 507 ml.6 V = V V V = since the pump has a fixed diameter, the length of the tube is proportional to its volume = = (75.0 cm) 5.50 atm =.6 cm.7 Use Charles s Law to solve the second volume: V VT.86 L (60 K) T 8 K 4.7 L.8 Use Charles s Law to solve for the second volume: V VT.50 L (58 K) T 95 K.9 L.9 Compare pressure change to temperature to solve for temperature change: T 708 torr 558 K T = = 6 K = 0 K 6 K 7 K = 84 C 854 torr.40 In general the combined gas law equation is: V V T T volume does not change, we have: T T 45 lb in (6 K) 85. K 50 lb in.4 In general the combined gas law equation is: V V T T V T (745 torr)(.58 L)(48. K) 80 torr TV (97. K)(.8 L), and in particular, for this problem since the, and in particular, for this problem, we have:.4 In general the combined gas law equation is: V T V T (0.985 atm)(648 ml)(6. K) T V (89. K)(689 ml) V T.08 atm, and in particular, for this problem, we have: 7
.4 In general the combined gas law equation is V V T T V T (745 torr)(.68 L)(648. K) V 5.69 L T (97. K)(765 torr), and in particular, for this problem, we have:.44 In general the combined gas law equation is: V V, and in particular, for this problem, we have: T T V T (74 torr)(86 ml)(06. K) V 9 ml T (9. K)(765 torr).45 In general the combined gas law equation is: V V, and in particular, for this problem, we have: T T V T (7 torr)(9.45 L)(9. K) T 9.8 K 5 C V (76 torr)(6.8 L).46 In general the combined gas law equation is: V V, and in particular, for this problem, we have: T T V T (.00 atm)( ml)(98. K) T 9 K 80. C V (.5)(455 ml).47 000 ml 4 ml torr R 0.08 6.4 0 L.48 If V = nrt, then R = V/nT. Let = = 0,5 a, T = 7 K, and n =. Next, express the volume of the standard mole using the units m, instead of L, remembering that.4 L =,400 cm : m m, 400 cm 0.04 m 00 cm 0, 5 a 0.04 m R 8. m a mole 7 K.49 V nrt mol 0.6 g 0.08 9. K.0 g 748 torr 0.04 L.50 V nrt mol.67 g 0.08 95. K 8.0 g 756 torr.45 L 8
.5 mol 0.0 g 0.08 00. K nrt.0 g.08 atm 40 torr V.50 L.5.5.54 nrt V mol.0 g 0.08 8 K 8.0 g.60 L 5.80 atm atm 64 torr 0.065 L V 4 44.0 g n 9.04 0 mol 0.098 g RT mol 0.08 9. K V 758 torr 0.55 L n.0 x 0 mol RT 0.08 00. K 6.0 g grams of methane =.0 0 mol 0.65 g mol CH 4.55 (a) density C H 6 = (b) density N = (c) density Cl = (d) density Ar =.56 (a) density Ne = (b) density O = (c) density CH 4 = (d) density CF 4 = 0. g CH6 mol mol C H.4 L 6 8.0 g N mol mol N.4 L 70.9 g Cl mol mol Cl.4 L 9.9 g Ar mol mol Ar.4 L 0. g Ne mol mol Ne.4 L.0 g O mol mol O.4 L 6.0 g CH4 mol mol CH.4 L 88.0 g CF4 mol mol CF.4 L 4 4 =.4 g L =.5 g L =.7 g L =.78 g L = 0.90 g L =.4 g L = 0.74 g L =.9 g L.57 In general V = nrt, where n = mass formula mass. Thus mass V = RT (formula mass) and we arrive at the formula for the density (mass divided by volume) of a gas: 9
d = (formula mass) RT (74 torr) (.0 g/mol) d = 0.08 (97. K) d =.8 g/l for O.58 In general V = nrt, where n = mass formula mass. Thus mass V = RT (formula mass) and we arrive at the formula for the density (mass divided by volume) of a gas: d = (formula mass) RT (748.0 torr) (9.95 g/mol) d = 0.08 (9.80 K) d =.6 g/l for Ar.59 First determine the number of moles from the ideal gas law: L V (0.0 torr) 55 ml 000 ml 4 n =.7 0 mol RT 0.08 (98. K) Now calculate the molecular mass: molecular mass g 000 mg mass. mg # of moles 4.7 0 mol 88. g/mol.60 molecular mass.6 The balanced equation is C 4 H 0 + O 8CO + 0H O. g/l 0.08 95 K drt mass RT V (755 torr) ml O ml O (75 ml C4H 0).4 0 ml ml C H 4 0 O 7.6 g/mol.6 The balanced equation is C 6 H 4 + 9O CO + 4H O 9 ml O ml O (855 ml CO ).5 0 ml ml CO O.6 mol C H 6 = (8.0 g C H 6 ) mol H = (0.48 mol C H 6 ) mol C H 6 4.08 g C H 6 mol H mol CH6 = 0.48 mol C H 6 = 0.48 mol H 0
V = nrt = (0.48 mol H ) 0.08 (97. K) (740 torr) = 0.7 L H.64 mol HNO = (.0 g HNO ) mol NO = (0.90 mol HNO ) V = mole HNO 6.0 g HNO moles NO moles HNO = 0.90 mol HNO = 0.86 mol NO 0.86 moles NO 0.08 98 K nrt 75 torr.65 CH 4 + O CO + H O n CH 4 75 torr 6.8 0 L V RT 0.08 08. K mol O = (6.4 0 4 mol CH 4 ) mol O mol CH 4 = 7.07 L or 7.07 0 ml = 6.4 0 4 mol CH 4 =.7 0 mol O.7 0 moles 0.08 00. K nrt V O.6 0 L 6. ml O 654 torr.66 n NH 85 torr.6 0 L V RT 0.08 400 K mol H O =. x 0-6 mol H mol NH O 4 mol NH =. 0 mol NH =.67 0 mol H O.67 0 moles 0.08 59 K nrt V HO 8.6 0 L 8.6 ml 75 torr.67 CO + O CO moles CO 68 torr 0.00 L.0 0 moles 0.08 98. K moles O 75 torr 0.55 L = 4.46 0 moles 0.08 98. K O is the limiting reactant
mol CO moles CO (4.46 0 moles O ) 8.9 0 moles CO mol O 8.9 0 mol 0.08 00. K V =.4 0 L 4 ml 745 torr.68 0.750 atm 0.00 L moles NH 9.0 0 moles 0.08 0 K 0.780 atm 0.0 L moles O 6.47 0 moles 0.08 K Assume NH is the limiting reagent. mol N moles N (9.0 0 moles NH ) 4.55 0 moles N 4 moles NH Assume O is the limiting reagent: mol N moles N (6.47 0 moles O ) 4. 0 mol N O is limiting reactant ml N mol O 4. 0 mol 0.08 7 K 0.740 atm 000 ml L 78 ml.69 Tot = N + O + He Tot = (0 cm)(0 mm/cm)( torr/mm) + 55 torr + (0.450 atm)(/atm) Tot = 00 torr + 55 torr + 4 torr = 697 torr.70 Tot = N + O + CO CO = Tot N O CO = 740 torr 0 torr 40 torr = 580 torr.7 Assume all gases behave ideally and recall that mole of an ideal gas at 0 C and occupies a volume of.4 L. Therefore, the moles of gas equals the pressure of gas in atm: (RT/V =.000 atm mol - ) N = 0.0 atm O = 0.0 atm He = 0.40 atm CO = 0.0 atm N = 0.0 atm = 8 torr
O = 0.0 atm He = 0.40 atm CO = 0.0 atm = 5 torr = 04 torr = 76 torr N = 0.0 atm O = 0.0 atm He = 0.40 atm CO = 0.0 atm bar 0.9868 atm bar 0.9868 atm bar 0.9868 atm bar 0.9868 atm = 0.04 bar = 0.0 bar = 0.405 bar = 0.0 bar.7 CO = 845 torr torr = 5 torr 5 torr n CO (0.00 mol) 0.4 moles 845 torr.7 total = ( CO + H O) H O = 9.8 torr at C, from Table.. CO = 754 9.8 torr = 74 torr V /T = V /T, and V T (74. torr)(0.68 L)(98. K) V 0.6 L 6 ml T (95. K)().74 total = H + H O H O =.76 torr at 5 C, from Table.. H = total H O = 74.76 = 78 torr The temperature stays constant so, V = V, and V (78 torr)(88 ml) V = = = 7 ml ().75 From Table., the vapor pressure of water at 0 C is 7.54 torr. Thus only (74 7.54) = 74 torr is due to "dry" methane. In other words, the fraction of the wet methane sample that is pure methane is 74/74 = 0.976. The question can now be phrased: What volume of wet methane, when multiplied by 0.976, equals 44 ml? Volume "wet" methane 0.976 = 44 ml Volume "wet" methane = 44 ml/0.976 = 50 ml In other words, one must collect 50 total ml of "wet methane" gas in order to have collected the equivalent of 44 ml of pure methane.
.76 First convert the needed amount of oxygen at to the volume that would correspond to the laboratory conditions of 746 torr: V = V or V = V / V = 75 ml /746 torr = 80 ml of dry oxygen gas The wet sample of oxygen gas will also be collected at atmospheric pressure in the lab of 746 torr. The vapor pressure of water at 5 C is equal to.8 torr (from Table.), and the wet sample will have the following partial pressure of oxygen, once it is collected: O = total H O = 746.8 = 7 torr of oxygen in the wet sample. Thus the wet sample of oxygen is composed of the following % oxygen: % oxygen in the wet sample = 7/746 00 = 98. % The question now becomes what amount of a wet sample of oxygen will contain the equivalent of 80 ml of pure oxygen, if the wet sample is only 98. % oxygen (and.7 % water). 0.98 V wet = 80 ml, hence V wet = 85 ml. This means that 85 ml of a wet sample of oxygen must be collected in order to obtain as much oxygen as would be present in 80 ml of a pure sample of oxygen..77 Effusion rates for gases are inversely proportional to the square root of the gas density, and the gas with the lower density ought to effuse more rapidly. Nitrogen in this problem has the higher effusion rate because it has the lower density: rate(n ) rate(co ).96 g L = =.5.5 g L.78 Ethylene, C H 4, the lightest of these three, diffuses the most rapidly, and Cl, the heaviest, will diffuse the slowest. Cl < SO < C H 4.79 The relative rates are inversely proportional to the square roots of their molecular masses: 5 6 8 6 8 6 5 6 rate( UF ) molar mass ( UF ) 5 g mol = = =.004 rate( UF ) molar mass ( UF ) 49 g mol Meaning that the rate of effusion of the 5 UF 6 is only.004 times faster than the 8 UF 6 isotope..80 Use equation.7 effusion rate x effusion rate C H M M x C H Additional Exercises 8 CH 8 x M M 8 effusion rate C H effusion rate x 8 44. g/mol 6. g/mol.65.8 We found that =.9 ft of water. This is equivalent to.9 ft in./ft = 407 in. of water, which in this problem is equal to the height of a water column that is uniformly.00 in. in diameter. Next, we convert the given density of water from the units g/ml to the units lb/in. : 4
lb.00 g lb ml.54 cm lb 0.06.00 ml 454 g in. in. cm in. The area of the total column of water is now calculated:.00 in. 407 in. = 407 in., along with the mass of the total column of water: 407 in. 0.06 lb/in. = 4.7 lb. Finally, we can determine the pressure (force/unit area) that corresponds to one atm: = 4.7 lb.00 in. = 4.7 lb/in..8 Total footprint = (4 tires) 6.0 in. in tire = 76.8 in Total pressure 500 lb 45.6 lb/in 76.8 in Gauge pressure 45.6 lb/in - 4.7 lb/in 0.9 lb/in.8 000 lb 5 Total weight (45.6 tons 8. tons).08 0 lbs ton Total pressure = 85 psi + 4.7 psi = 99.7 psi/tire 5.08 0 lbs number of tires 0.8 tires (99.7 lbs in /tire)(00 in ) The minimum number of wheels is since tires are mounted in multiples of..84 Assume a sq in. cylinder of water V = (,468 ft) Mass = (.458 0 6 ml) in..54 cm ml in. (4966 in. ) ft in. cm =.458 06 ml.05 g 6 lb (.506 0 g) 5.5406 0 lb ml 45.6 g ressure (5.5406 0 lb in ) 76.89 atm 4.7 lb in.85 From the data we know that the pressure in flask is greater than atmospheric pressure, and greater than the pressure in flask. The pressure in flask can be determined from the manometer data. The pressure in flask is: 760 mm Hg cm = (0.87 atm) +.6 cm = 75. cm Hg 0 mm The pressure in flask is lower than flask 0.86 g ml = 75. cm Hg (6.4 cm oil) = 74. cm Hg = 74. torr.6 g ml.86 To calculate the pressure at 00 ft assume a cylinder of water 00 ft long and in. in.54 cm ml.05 g lb mass (00 ft) in 44.4 lb ft in cm ml 45.6 g (44.4 lb in ).0 atm 4.7 lb in Since the pressure decreases by a factor of, the volume must increase by a factor of. Divers exhale to decrease the amount of gas in their lungs, so it does not expand to a volume larger than the divers lungs. 5
.87 First calculate the initial volume (V ) and the final volume (V ) of the cylinder, using the given geometrical data, noting that the radius is half the diameter (0.7/ = 5.5 cm): V = (5.5 cm).4 cm =.0 0 cm V = (5.5 cm) (.4 cm.7 cm) = 6.9 cm In general the combined gas law equation is: V V =, and in particular, for this problem, we have: T T V T (4.0 atm)(6.9 cm )(64 K) T = = = 649 K = 76 C V (.00 atm)(.0 0 cm ).88 First convert the temperature data to the Kelvin scale: 7 + 5/9(60.0.0) = 89 K and 7 + 5/9(04 ) = K. Next, calculate the final pressure at the gauge, taking into account the temperature change only: T (64.7 lb in. )( K) = = = 70. lb in. T (89 K) This represents the actual pressure inside the tire. The pressure gauge measures only the difference between the pressure inside the tire and the pressure outside the tire (atmospheric pressure). Hence the gauge reading is equal to the internal pressure of the tire less atmospheric pressure: (70. 4.7) lb/in = 55.4 lb/in.89 The temperatures must first be converted to Kelvin: 5 5 C = ( F ) = ( 50 ) = 46 C or 7 K 9 9 5 5 C = ( F ) = (0 ) = 49 C or K 9 9 Next, the pressure calculation is done using the following equation: T (5 lb in. )( K) = = = 50 lb in. T (7 K).90 Using the ideal gas law, determine the number of moles of H and O gas initially present: For hydrogen: V 50 torr 0.400 L n = = =.5 0 mol H RT 0.08 8 K for oxygen: V 740 torr 0.00 L n = = =.9 0 mol O RT 0.08 98 K This problem is an example of a limiting reactant problem in that we know the amounts of H and O initially present. Since mol of O reacts completely with mol of H, we can see, by inspection, that there is excess H present. Using the amounts calculated above, we can make.8 0 mol of H O and have an excess of.4 0 mol of H. Thus, the total amount of gas present after complete reaction is.5 0 mol. Using this value for n, we can calculate the final pressure in the reaction vessel: 6
nrt.5 0 mol 0.08 95 K = = =.6 atm =.4 0 torr V 0.500 L.9 We first need to determine the pressure inside the apparatus. Since the water level is 8.5 cm higher inside than outside, the pressure inside the container is lower than the pressure outside. To determine the inside pressure, we first need to convert 8.5 cm of water to an equivalent dimension for mercury. This is done using the density of mercury: Hg = 85 mm/.6 = 6.5 mm (where the density of mercury,.6 g/ml, has been used.) inside = outside Hg = 746 torr 6 torr = 740 torr. In order to determine the H, we need to subtract the vapor pressure of water at 4 C. This value may be found in Table. and Appendix C.5 and is equal to.4 torr. The H = inside H O = 740 torr.4 torr = 77 torr. Now, we can use the ideal gas law in order to determine the number of moles of H present; L V 77 torr 8.45 ml 000 ml 4 n = = = 7.4 0 mol H RT 0.08 97 K The balanced equation described in this problem is: Zn(s) +HCl(aq) ZnCl (aq) + H (g) By inspection we can see that mole of Zn(s) reacts to form mole of H (g) and we must have reacted 7.4 0 4 mol Zn in this reaction. 4 65.9 g Zn g Zn = 7.4 0 mol Zn = 4.67 0 g Zn mol Zn.9 H + O H O mol H = (.7 g H ) mol O = (87.5 g O ) mol H.0 g H mol O.0 g O O is the limiting reactant mol H O = (.7 mol O ) mol H needed = (.7 mol O mol HO mol O = 6.9 mol H =.7 mol O mol H mol O = 5.46 mol H = 5.46 mol H Remaining mol H = 6.9 mol H 5.46 mol H = 0.8 mol H 5.46 mol 0.08 45 K HO 6. atm.0 L 0.8 mol 0.08 45 K H.5 atm.0 L Tot = 6. atm +.5 atm = 8.7 atm 7
.9 Note: 0.08747 mg/ml = 0.08747 g/l 0.08747 g/l 0.08 90. K drt molecular mass ( ) This gas must be H..08 g/mol.94 (a) The equation can be rearranged to give: V( ) HO 0.04489 = %N W 7 + t C This means that the left side of the above equation should be obtainable simply from the ideal gas law, applied to the nitrogen case. If V = nrt, then for nitrogen: V = (mass nitrogen)/(8.0 g/mol) RT, and the mass of nitrogen that is collected is given by: (mass nitrogen) = V(8.0)/RT, where R = 8. m/k mol /atm = 6.4 0 4 ml torr/k mol. Using this value for R in the above equation, we have the following result for the mass of nitrogen, remembering that the pressure of nitrogen is less than the total pressure, by an amount equal to the vapor pressure of water: (mass nitrogen) = 8.0 V (total H O) 4 6.4 0 ml torr 7 + C Finally, it is only necessary to realize that the value 8.0 6.4 0 4 00 = 0.04489 is exactly the value given in the problem. (b) (8.90 ml)(746 torr. torr) % N = 0.04489 = 8.69 % 0.94 g 7.5 +.80 Multi-Concept roblems.95 (a) Zn(s) + HCl(aq) H (g) + ZnCl (aq) Calculate the number of moles of hydrogen: V 765 torr.0 L n = = = 0.50 mol H RT 0.08 9. K and the number of moles of zinc: mol Zn mol Zn = 0.50 mol H = 0.50 mol Zn mol H The number of grams of zinc needed is, therefore: 8
65.9 g Zn g Zn = (0.50 mol Zn) =.8 g Zn mol Zn (b) mol HCl mol HCl = (0.50 mol Zn) =.004 mol HCl mol Zn 000 ml HCl ml HCl = (.004 mol HCl) = 6 ml HCl 8.00 mol HCl.96 First determine the % by mass S and O in the sample: % S =.448 g/.60 g 00 = 40.00 % S % O =.7 g/.60 g 00 = 60.00 % O Before we determine the molecular formula of the compound we need to find the empirical formula. Determine the number of moles of S and O in a sample of the material weighing 00 g exactly, in order to make the conversion from % by mass to grams straightforward: In 00 g of the material, there are 40.00 g S and 60.00 g O: 40.00 g S.07 g/mol =.47 mol S 60.00 g O 6.00 g/mol =.750 mol O Dividing each of these mole amounts by the smaller of the two gives the relative mole amounts of S and O in the material: for S,.47 mol.47 mol =.000 relative moles, for O,.750 mol.47 mol=.007 relative moles, and the empirical formula is, therefore, SO. We determine the formula mass of the material by use of the ideal gas law: V 750 torr.0 L n = = = 0.045 mol RT 0.08 98. K The formula mass is given by the mass in grams (given in the problem) divided by the moles determined here: formula mass =.60 g 0.045 mol = 80. g mol. Since this is equal to the formula mass of the empirical unit determined in step (b) above, namely SO, then the molecular formula is also SO..97 (a) total = 746.0 torr = H O + N N = 746.0 torr. torr = 7.9 torr Now, use the ideal gas equation to determine the moles of N that have been collected: L V 7.9 torr 8.90 ml 000 ml 4 n = = = 7.84 0 mol N RT 0.08 96.95 K Then the mass of nitrogen that has been collected is determined: 7.84 0 4 mol N 8.0 g/mol =.068 0 g N. Next, the % by mass nitrogen in the material is calculated: % N = (0.0068 g)/(0.94 g) 00 = 8.68 % N (b) mass of C in the sample: g C = (7.57 0 g CO ) = 4.795 0 g C mole CO mol C.0 g C 44.0 g CO mol CO mol C 9
mass of H in the sample: mole H O mol.008 g H g H = (4.9 0 g H O) 8.0 g H O mol H O mol H 4 = 4.8 0 g H mass of N in the sample: 8.68 g N 4 g N = (6.478 0 g sample) = 5.596 0 g N 00 g sample mass of O in the sample = total mass (mass C + H + N) mg O = 6.478 mg sample 4.795 mg C + 0.48 mg H + 0.5596 mg N = 0.640 mg O Next we convert each of these mass amounts into the corresponding mole values: for C, 4.795 0 g.0 g/mol =.99 0 4 mol C for H, 4.8 0 4 g.008 g/mol = 4.794 0 4 mol H for N, 5.596 0 4 g 4.0 g/mol =.994 0 5 mol N for O, 6.40 0 4 g 6.00 g/mol = 4.00 0 5 mol O Last, we convert these mole amounts into relative mole amounts by dividing each by the smallest of the four: for C,.99 0 4 mol/.994 0 5 mol = 9.998 for H, 4.794 0 4 mol/.994 0 5 mol =.00 for N,.994 0 5 mol/.994 0 5 mol =.000 for O, 4.00 0 5 mol/.994 0 5 mol =.00 The empirical formula is therefore C 0 H NO % C = (4.795 x 0 - g C/6.478 x 0 - g sample) x 00 = 74.0 % % H = (4.8 x 0-4 g H/6.478 x 0 - g sample) x 00 = 7.46 % The formula mass of the empirical unit is 6. Since this is half the value of the known molecular mass, the molecular formula must be twice the empirical formula, C 0 H 4 N O..98 (a) We begin by converting the dimensions of the room into cm: 4 ft 0.48 cm/ft =. 0 cm, 4 ft 0.48 cm/ft = 7. 0 cm, 8.6 ft 0.48 cm/ft =.6 0 cm. Next, the volume of the room is determined: V = (. 0 cm)(7. 0 cm)(.6 0 cm) =.4 0 8 cm. Since there are 000 cm in a liter, volume is: V =.4 0 5 L The calculation of the amount of H S goes as follows: 5 0.5 L HS 5 L HS =.4 0 L space =.6 0 L H 9 S 0 L space (b) Convert volume (in liters) to moles at ST: mol H S mol H S =.6 0 L H S =.6 0 mol H S 5 6.4 L HS Since the stoichiometry is :, we require the same number of moles of Na S: 40
000 ml Na S ml Na S =.6 0 mol Na S 0.00 mol Na S 6 =.6 0 ml Na S.99 Cl + SO + H O Cl + SO 4 + H + 0.00 moles Na SO mole SO mole Cl moles Cl 50.0 ml Na SO 000 ml Na SO mole Na SO mole SO.00 0 moles Cl.00 0 moles 0.08 98 K V Cl 0.5 L 5 ml 74 torr.00 total = 740 torr = H + water The vapor pressure of water at 5 C is available in Table.:.76 torr. Hence: H = (740 4) torr = 76 torr Next, we calculate the number of moles of hydrogen gas that this represents: V 76 torr 0.5 L n = = = 0.09 mol H RT 0.08 98. K The balanced chemical equation is: Zn(s) + HCl(aq) H (g) + ZnCl (aq) and the quantities of the reagents that are needed are: mol Zn 65.9 g Zn g Zn = (0.09 mol H ) = 0.844 g Zn mol H mol Zn mol HCl 000 ml HCl ml HCl = (0.09 mol H ) = 4.0 ml HCl mol H 6.00 mol HCl.0 This is a limiting reactant problem. First we need to calculate the moles of dry CO that can be produced from the given quantities of CaCO and HCl: mol CaCO mol CaCO = (. g CaCO ) = 0. mol CaCO 00.09 g CaCO 0.50 mol HCl mol HCl = (85 ml HCl) = 0.046 mol HCl 000 ml HCl Thus, HCl is limiting and we use this to determine the moles of CO that can be produced: mol CO mol CO = (0.046 mol HCl) = 0.0 mol CO mol HCl The pressure of the dry CO is 745 torr. Finally, the volume of this dry CO is calculated using the ideal gas equation: nrt 0.0 mol 0.08 9. K V = = = 0.567 L CO or 567 ml 745 torr 4
drt (mass)rt.0 formula mass = = V (6. 0 g) 0.08 (98. K) formula mass = ( torr) (85 ml) L 000 ml formula mass = 8 g mol The formula weights of the boron hydrides are: BH,.8 B H 6, 7.7 B 4 H 0, 5. And we conclude that the sample must have been B H 6. 4