Soal Kimfis.docx

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10. Hitunglah cm , c , crms , Z dan λ untuk gas N2 dalam atmosfer bumi pada ketinggian 20 km, dimana T = 217 K dan pN2 = 0,05 atm. (1 atm = 101,325 Pa, Navo = 6.02 x 1023; R = 8,314 J/mol.K) Jawab : 411 m/s Jawab Dik : h = 20 km T= 217 k Nav = 6,02 x 10 23 / m2 M N2 = 28 g/mol = 0,028 kg/mol P = 0,05 atm x 101,325 Pa = 5 x 103 Pa R = 8,314 J/mol K ̅ Crms , z , λ = ....? Dit : Cm, C, Cm

=( =(

2 RT 1/2 ) M J 217K 1 mol k 2 kg 10,028 mol

2 8,314

= (

)

kg m2 1 s2 2

3608,276

10,028 kg

)

= √128867 m2 /s2 = 359 m/s 1

8 RT C̅ = ( π m ) 2 = (

=(

j 217 K mol K

8 .8,314

0,028 kg/mol

5412 .414 kg m2 /s2 1 0,028 kg

)2

= √193300,5m2 /s 2

1

)2

= 439,66 Z

1

1

= P (2πmkT) 2 1

1

= (5.103 N/m2) (

)2

kg 10−23 J )( 217 K) )(1,38 x mol k

2 (3,14)(0,028

= (5.103 N/m2) ( √0,0184897 x 10−23 = (5.103 N/m2) (0,14 x 1012) = 7.108 s-1  λ

̅ C

=Z

m

405

= 7 x 108ss−1 = 5,79 x 10 -9 m 24. Hitunglah kecepatan yang paling mungkin pada molekul oksigen dengan molekul karbon dioksida pada suhu 20oC Jawab: Cm(O2) = 390,19 m/s dan Cm(CO2) = 417,13 m/s Jawab Dik: T=20oC = 293 K Dit: Cm (O2) dan Cm (CO2)=...? Penye: Cm (O2)= (

2 RT 1/2

)

M

= √152250,13

M

= √110727,36

1/2 J 2 (8,314 ⁄mol K)(293 K) =( ) 0.032 kg/mol

4872,004 1/2

=(

0,032

)

= 390,19 m/s

2 RT 1/2

Cm (CO2)= (

M O2= 32 g/mol = 0,032 kg/mol

)

=

1/2 J 2 (8,314 ⁄mol K)(293 K) ( ) 0.044 kg/mol

= 332,76 m/s

4872,004 1/2

=(

0,044

)

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