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Unformatted text preview: 797 Chapter 19 1. Each atom has a mass of m = M / N A , where M is the molar mass and N A is the Avogadro constant. The molar mass of arsenic is 74.9 g/mol or 74.9 10 –3 kg/mol. Therefore, 7.50 10 24 arsenic atoms have a total mass of (7.50 10 24 ) (74.9 10 –3 kg/mol)/(6.02 10 23 mol –1 ) = 0.933 kg. 2. (a) Equation 193 yields n = M sam / M = 1.5/197 = 0.00761 mol. (b) The number of atoms is found from Eq. 192: N = nN A = (0.00761)(6.02 10 23 ) = 4.58 10 21 . 3. In solving the idealgas law equation pV = nRT for n , we first convert the temperature to the Kelvin scale: (40.0 273.15) K 313.15 K i T , and the volume to SI units: 3 3 3 1000 cm 10 m i V . (a) The number of moles of oxygen present is 5 3 3 2 1.01 10 Pa 1.000 10 m 3.88 10 mol. 8.31J/mol K 313.15K i i pV n RT (b) Similarly, the ideal gas law pV = nRT leads to 5 3 3 2 1.06 10 Pa 1.500 10 m 493K 3.88 10 mol 8.31J/mol K f f pV T nR , which may be expressed in degrees Celsius as 220°C. Note that the final temperature can also be calculated by noting that f f i i i f p V pV T T , or 5 3 5 3 1.06 10 Pa 1500 cm (313.15 K) 493 K 1.01 10 Pa 1000 cm f f f i i i p V T T p V . 4. (a) With T = 283 K, we obtain CHAPTER 19 798 ( )( ) ( )( ) 3 3 100 10 Pa 3.00m 128mol. 8.31J/mol K 283K pV n RT ´ = = = × (b) We can use the answer to part (a) with the new values of pressure and temperature, and solve the ideal gas law for the new volume, or we could set up the gas law in ratio form as: f f f i i i p V T pV T (where n i = n f and thus cancels out), which yields a final volume of 3 3 100kPa 303K 3.00m 1.07 m 300kPa 283K f i f i f i T p V V p T . 5. With V = 1.0 10 –6 m 3 , p = 1.01 10 –13 Pa, and T = 293 K, the ideal gas law gives 13 6 3 23 1.01 10 Pa 1.0 10 m 4.1 10 mole. 8.31 J/mol K 293 K pV n RT Consequently, Eq. 192 yields N = nN A = 25 molecules. We can express this as a ratio (with V now written as 1 cm 3 ) N / V = 25 molecules/cm 3 . 6. The initial and final temperatures are 5.00 C 278 K i T and 75.0 C 348 K f T , respectively. Using the ideal gas law with i f V V , we find the final pressure to be 348K 1.00 atm 1.25 atm 278K f f f f f i i i i i p V T T p p pV T T ....
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This document was uploaded on 04/22/2011.
 Spring '11
 Mass

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