Chapter 18

# Chapter 18 - In doing the numerical calculations for the exercises and problems for this chapter the values of the ideal-gas constant have been

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In doing the numerical calculations for the exercises and problems for this chapter, the values of the ideal-gas constant have been used with the precision given on page 501 of the text, K. mol atm L 0.08206 K mol J 3145 . 8 = = R Use of values of these constants with either greater or less precision may introduce differences in the third figures of some answers. 18.1: a) mol. 3 . 56 ) mol kg 10 400 ( kg) 225 . 0 ( 3 tot = × = = - M m n b) Of the many ways to find the pressure, Eq. (18.3) gives Pa. 10 6.81 atm 67.2 L) (20.0 K) K)(291.15 mol atm L 06 mol)(0.082 3 . 56 ( 6 × = = = = V nRT p 18.2: a) The final temperature is four times the initial Kelvin temperature, or 4(314.15 K) –273.15= C 983 ° to the nearest degree. kg. 10 24 . 5 ) K 15 . 314 )( K atm/mol L 08206 . 0 ( ) L 60 . 2 )( atm 30 . 1 )( kg/mol 10 00 . 4 ( b) 4 3 tot - - × = × = = = RT MpV nM m 18.3: For constant temperature, Eq. (18.6) becomes atm. 96 . 0 ) 390 . 0 110 . 0 )( atm 40 . 3 ( ) ( 2 1 1 2 = = = V V p p 18.4: a) Decreasing the pressure by a factor of one-third decreases the Kelvin temperature by a factor of one-third, so the new Celsius temperatures is 1/3(293.15 K) – 273.15= C 175 ° - rounded to the nearest degree. b) The net effect of the two changes is to keep the pressure the same while decreasing the Kelvin temperature by a factor of one- third, resulting in a decrease in volume by a factor of one-third, to 1.00 L.

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18.5: Assume a room size of 20 ft X 20 ft X 20 ft C. 20 of re temperatu a Assume . m 113 ft 4000 3 3 ° = = V molecules 10 8 . 2 mol 4685 ) K 293 )( K J/mol 315 . 8 ( ) m 113 )( Pa 10 01 . 1 ( so 27 3 5 × = = = × = = = A nN N RT pV n nRT pV b) 3 19 3 6 27 cm molecules/ 10 5 . 2 cm 10 113 molecules 10 8 . 2 × = × × = V N 18.6: The temperature is K. 15 . 295 C 0 . 22 = ° = T (a) The average molar mass of air is so mol, kg 10 8 . 28 3 - × = M kg. 10 07 . 1 K) K)(295.15 mol atm L 08206 . 0 ( mol) kg 10 L)(28.8 atm)(0.900 00 . 1 ( 3 3 tot - - × = × = = = M RT pV nM m (b) For Helium mol, kg 10 00 . 4 3 - × = M so kg. 10 49 . 1 K) K)(295.15 mol atm L 08206 . 0 ( mol) kg 10 L)(4.00 atm)(0.900 00 . 1 ( 4 3 tot - - × = × = = = M RT pV nM m 18.7: From Eq. (18.6), C. 503 K 776 ) cm Pa)(499 10 01 . 1 ( ) cm Pa)(46.2 10 (2.821 K) 15 . 300 ( 3 5 3 6 1 1 2 2 1 2 ° = = × × = = V p V p T T 18.8: a) ( 29 ( 29 ( 29 ( 29 ( 29 kg. 373 . 0 K 15 . 310 K mol J 3145 . 8 m 0750 . 0 Pa 10 013 . 4 mol kg 10 0 . 32 3 5 3 tot = × × = = - RT MpV m b) Using the final pressure of Pa 10 813 . 2 5 × and temperature of kg, 0.275 K, 15 . 295 = m so the mass lost is kg 098 . 0 where extra figures were kept in the intermediate calculation of tot m . 18.9:
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## This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Chapter 18 - In doing the numerical calculations for the exercises and problems for this chapter the values of the ideal-gas constant have been

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