problem18_06

University Physics with Modern Physics with Mastering Physics (11th Edition)

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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 .
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Unformatted text preview: 4 3- = M so kg. 10 49 . 1 K) K)(295.15 mol atm L 08206 . ( mol) kg 10 L)(4.00 atm)(0.900 00 . 1 ( 4 3 tot-- = = = = M RT pV nM m...
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