chap18 - In doing the numerical calculations for the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 . ( 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 . ( ) 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 . ) 390 . 110 . )( 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. 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 . 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 . ( 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 . ( 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 . K 15 . 310 K mol J 3145 . 8 m 0750 . Pa 10 013 . 4 mol kg 10 . 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 . where extra figures were kept in the intermediate calculation of tot m ....
View Full Document

Page1 / 27

chap18 - In doing the numerical calculations for the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online