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(a)23181000 g1 mol air6.02210molecules air(5.2510kg air)1 kg29.0 g air1 mol air×××××=441.0910molecules×(b)First, calculate the moles of air exhaled in every breath. (500 mL =0.500 L) 2(1 atm)(0.500 L)1.9610mol air/breathL atm0.0821(37273)Kmol K−===×⋅+⋅PVnRTNext, convert to molecules of air per breath. 2326.02210molecules air1.9610mol air/breath1 mol air−×××=221.1810molecules/breath×(c)221.1810molecules12 breaths60 min24 h365 days35 yr1 breath1 min1 h1 day1 yr××××××=302.6010molecules×(d)Fraction of molecules in the atmosphere exhaled by Mozart is: 30442.6010molecules1.0910molecules×=×142.3910−×Or, 131414.18102.3910−=××Thus, about 1 molecule of air in every 4 ×1013molecules was exhaled by Mozart. In a single breath containing 1.18 ×1022molecules, we would breathe in on average: 22131 Mozart air molecule(1.1810molecules)410air molecules××=×8310molecules that Mozart exhaled×(e)We made the following assumptions: 1. Complete mixing of air in the atmosphere. 2. That no molecules escaped to the outer atmosphere. 3. That no molecules were used up during metabolism, nitrogen fixation, and so on. 5.147 First, let’s calculate the root-mean-square speed of N2at 25°C. rms233(8.314 J/mol K)(298 K)(N )515 m/s0.02802 kg/mol⋅===RTuCHAPTER 5: GASES 174Now, we can calculate the temperature at which He atoms will have this same root-mean-square speed. rms3(He)=RTu3(8.314 J/mol K)515 m/s0.004003 kg/mol⋅=TT=42.6 K5.148The ideal gas law can be used to calculate the moles of water vapor per liter. 1.0 atmmol0.033L atmL(0.0821)(100273)Kmol K===⋅+⋅nPVRTWe eventually want to find the distance between molecules. Therefore, let's convert moles to molecules, and convert liters to a volume unit that will allow us to get to distance (m3). 2325330.033 mol6.02210molecules1000 Lmolecules2.0101 L1 mol1 mm×=×This is the number of ideal gas molecules in a cube that is 1 meter on each side. Assuming an equal distribution of molecules along the three mutually perpendicular directions defined by the cube, a linear density in one direction may be found: 1253832.010moleculesmolecules2.710m1 m×=×This is the number of molecules on a line onemeter in length. The distance between each molecule is given by: 981 m3.710m2.7010−=×=×3.7 nmAssuming a water molecule to be a sphere with a diameter of 0.3 nm, the water molecules are separated by over 12 times their diameter: 3.7 nm12 times.0.3 nm≈A similar calculation is done for liquid water. Starting with density, we convert to molecules per cubic meter. 32328233221 mol H O0.96 g6.02210molecules100 cmmolecules3.21018.02 g H O1 mol H O1 m1 cmm××××=×This is the number of liquid water molecules in onecubic meter. From this point, the calculation is the same as that for water vapor, and the space between molecules is found using the same assumptions.