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TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT

TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT

 

12/ mass Hence Initial volume 45 1/₂ = 1.12 g moles of H₂ where - For reversible TVI Final volume = 250cm 3 = 161.4 x 1 T2 = = 161.4 x 10.4 (0.250.4 T2V₂ r = ²p Cv Hence for №₂2₂ = 1.4-1 1.12g =1000cm3 8-1 103cm3 Initial pressure - you tors x 0.53 atm Hence initial temperature '4₁' = PV n R 0.53 atm x 1L 0.0821 xo.oy mule Latmk mol - - 28 g/mol 1 = maps as N2. molar mass as №₂ 0.04 mole - XIL 161.4K adiabatic process 16³cm3 161.4 = 1L =0-25L latm 760 toss T₁ = 161.4K T₂ = ? and Cp = C₁ TR , Y₁ =1L V2=0.25L =) Cv=Cp-R = 3.5R-R=2.5R 3.5R = 3.5=1.4 3 5R-R 2.5 T2X (0.25 ) 1.4-1 = 281.2K 00574 Hence binal temperature T₂ = 281.2K
Transcribed Image Text:12/ mass Hence Initial volume 45 1/₂ = 1.12 g moles of H₂ where - For reversible TVI Final volume = 250cm 3 = 161.4 x 1 T2 = = 161.4 x 10.4 (0.250.4 T2V₂ r = ²p Cv Hence for №₂2₂ = 1.4-1 1.12g =1000cm3 8-1 103cm3 Initial pressure - you tors x 0.53 atm Hence initial temperature '4₁' = PV n R 0.53 atm x 1L 0.0821 xo.oy mule Latmk mol - - 28 g/mol 1 = maps as N2. molar mass as №₂ 0.04 mole - XIL 161.4K adiabatic process 16³cm3 161.4 = 1L =0-25L latm 760 toss T₁ = 161.4K T₂ = ? and Cp = C₁ TR , Y₁ =1L V2=0.25L =) Cv=Cp-R = 3.5R-R=2.5R 3.5R = 3.5=1.4 3 5R-R 2.5 T2X (0.25 ) 1.4-1 = 281.2K 00574 Hence binal temperature T₂ = 281.2K
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