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# Chapter-19 - Chapter 19 1(a Eq 19-3 yields n = Msam/M =...

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Chapter 19 1. (a) Eq. 19-3 yields n = M sam / M = 2.5/197 = 0.0127 mol. (b) The number of atoms is found from Eq. 19-2: N = nN A = (0.0127)(6.02 × 10 23 ) = 7.64 × 10 21 . 2. 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. 3. With V = 1.0 × 10 –6 m 3 , p = 1.01 × 10 –13 Pa, and T = 293 K, the ideal gas law gives ( 29 ( 29 ( 29 ( 29 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. 19-2 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 . 4. (a) We solve the ideal gas law pV = nRT for n : ( 29 ( 29 ( 29 ( 29 6 3 8 100Pa 1.0 10 m 5.47 10 mol. 8.31J/mol K 220K pV n RT - - × = = = × (b) Using Eq. 19-2, the number of molecules N is ( 29 ( 29 6 23 1 16 A 5.47 10 mol 6.02 10 mol 3.29 10 molecules. N nN - - = = × × = × 5. Since (standard) air pressure is 101 kPa, then the initial (absolute) pressure of the air is p i = 266 kPa. Setting up the gas law in ratio form (where n i = n f and thus cancels out — see Sample Problem 19-1), we have f f f i i i p V T pV T = which yields 807

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CHAPTER 19 ( 29 2 3 2 3 1.64 10 m 300K 266kPa 287 kPa 1.67 10 m 273K f i f i f i T V p p V T - -  × = = =  × . Expressed as a gauge pressure, we subtract 101 kPa and obtain 186 kPa. 6. (a) With T = 283 K, we obtain ( ) ( ) ( ) ( ) 3 3 100 10 Pa 2.50m 106mol. 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 in Sample Problem 19-1 (where n i = n f and thus cancels out): f f f i i i p V T pV T = which yields a final volume of ( 29 3 3 100kPa 303K 2.50m 0.892 m 300kPa 283K f i f i f i T p V V p T   = = =     . 7. (a) In solving pV = nRT for n , we first convert the temperature to the Kelvin scale: (40.0 273.15)K 313.15 K T = + = . And we convert the volume to SI units: 1000 cm 3 = 1000 × 10 –6 m 3 . Now, according to the ideal gas law, ( 29 ( 29 ( 29 ( 29 5 6 3 2 1.01 10 Pa 1000 10 m 3.88 10 mol. 8.31J/mol K 313.15K pV n RT - - × × = = = × (b) The ideal gas law pV = nRT leads to ( 29 ( 29 ( 29 ( 29 5 6 3 2 1.06 10 Pa 1500 10 m 493K. 3.88 10 mol 8.31J/mol K pV T nR - - × × = = = × We note that the final temperature may be expressed in degrees Celsius as 220°C. 8. The pressure p 1 due to the first gas is p 1 = n 1 RT / V , and the pressure p 2 due to the second gas is p 2 = n 2 RT / V . So the total pressure on the container wall is ( 29 1 2 1 2 1 2 . n RT n RT RT p p p n n V V V = + = + = + 40
The fraction of P due to the second gas is then ( 29 ( 29 2 2 2 1 2 1 2 / 0.5 0.2. / 2 0.5 p n RT V n p n n RT V n n = = = = + + + 9. (a) Eq. 19-45 (which gives 0) implies Q = W . Then Eq. 19-14, with T = (273 + 30.0)K leads to gives Q = –3.14 × 10 3 J, or | Q | = 3.14 × 10 3 J.

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Chapter-19 - Chapter 19 1(a Eq 19-3 yields n = Msam/M =...

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