TAILIEUCHUNG - Physics exercises_solution: Chapter 18

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 18 | 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 R J mol K L atm mol K. Use of values of these constants with either greater or less precision may introduce differences in the third figures of some answers. a n mtoJM kg 400 x 10 kg mol mol. b Of the many ways to find the pressure Eq. gives nRT _ mol atm mol K P V L atm x 106 Pa. a The final temperature is four times the initial Kelvin temperature or 4 K 983 C to the nearest degree. MpV x 10-3kg mol 1A_4 b mtot nM ------------- -------------------- - x10 4kg. RT L atm mol K For constant temperature Eq. becomes p2 pxVV . 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 K -175 C 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 L. Assume a room size of 20 ft X 20 ft X 20 ft V 4000 ft3 113 m3. Assume a temperature of 20 C. pV nRT so n pV _ x 105 Pa 113m3 RT J mol K 293K 4685 mol N nNA x 1027 molecules N x 10 27 molecules 9 3 b -------------------- x 10 molecules cm V 113 x 106 cm3 The temperature is T C . a The average molar mass of air is M x 10 3 kg mol so m nM PVm -am 88x 10-ikg mol 10-3 kg. RT L atm mol K b For Helium M x 10 3 kg mol so atm L x 10-3 kg mol 1 49 x 10-4 atm mol K . g. mot nM -M tot RT From Eq. T 1 2 TfpV1 K f x 10 I 776 K 503 C. 1 p1V1 J x 105 Pa 499cm3 J a

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