a 23 18 1000 g 1 mol air 6022 10 molecules air 525 10 kg air 1 kg 290 g air 1

A 23 18 1000 g 1 mol air 6022 10 molecules air 525 10

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(a) 23 18 1000 g 1 mol air 6.022 10 molecules air (5.25 10 kg air) 1 kg 29.0 g air 1 mol air × × × × × = 44 1.09 10 molecules × (b) First, calculate the moles of air exhaled in every breath. (500 mL = 0.500 L) 2 (1 atm)(0.500 L) 1.96 10 mol air/breath L atm 0.0821 (37 273)K mol K = = = × + PV n RT Next, convert to molecules of air per breath. 23 2 6.022 10 molecules air 1.96 10 mol air/breath 1 mol air × × × = 22 1.18 10 molecules/breath × (c) 22 1.18 10 molecules 12 breaths 60 min 24 h 365 days 35 yr 1 breath 1 min 1 h 1 day 1 yr × × × × × × = 30 2.60 10 molecules × (d) Fraction of molecules in the atmosphere exhaled by Mozart is: 30 44 2.60 10 molecules 1.09 10 molecules × = × 14 2.39 10 × Or, 13 14 1 4.18 10 2.39 10 = × × Thus, about 1 molecule of air in every 4 × 10 13 molecules was exhaled by Mozart. In a single breath containing 1.18 × 10 22 molecules, we would breathe in on average: 22 13 1 Mozart air molecule (1.18 10 molecules) 4 10 air molecules × × = × 8 3 10 molecules 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 N 2 at 25 ° C. rms 2 3 3(8.314 J/mol K)(298 K) (N ) 515 m/s 0.02802 kg/mol = = = RT u CHAPTER 5: GASES 174 Now, we can calculate the temperature at which He atoms will have this same root-mean-square speed. rms 3 (He) = RT u 3(8.314 J/mol K) 515 m/s 0.004003 kg/mol = T T = 42.6 K 5.148 The ideal gas law can be used to calculate the moles of water vapor per liter. 1.0 atm mol 0.033 L atm L (0.0821 )(100 273)K mol K = = = + n P V RT We 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 (m 3 ). 23 25 3 3 0.033 mol 6.022 10 molecules 1000 L molecules 2.0 10 1 L 1 mol 1 m m × = × 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: 1 25 3 8 3 2.0 10 molecules molecules 2.7 10 m 1 m × = × This is the number of molecules on a line one meter in length. The distance between each molecule is given by: 9 8 1 m 3.7 10 m 2.70 10 = × = × 3.7 nm Assuming 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 nm 12 times. 0.3 nm A similar calculation is done for liquid water. Starting with density, we convert to molecules per cubic meter. 3 23 28 2 3 3 2 2 1 mol H O 0.96 g 6.022 10 molecules 100 cm molecules 3.2 10 18.02 g H O 1 mol H O 1 m 1 cm m × × × × = × This is the number of liquid water molecules in one cubic 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.
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