Question

O butelie de volum V=20 l,ce contine hidrogen,are temperatura T1=250K presiunea de 15 atm. In butelie se introduce o cantitate suplimentara de hidrogen astfel incat presiunea devine p2=60 x 10(la puterea a 5) Pa si temperatura 280K . Caculeaza masa de hidrogen care a fost introdusa in butelie .(R=8,31 J/mol x K)

Answer 1

p2=60 x 10(la puterea a 5) Pa => P2 = 60 atm

P1V = n1RT1 => n1 = P1V/RT1

=> 15x20/0,082x250 = 14,63 moli initiali

P2V = n1RT2 => n2 = P2V/RT2

=> n2 = 60x20/0,082x280 = 52,26 moli

n2 - n1 = 52,23 - 14,63 = 37,635 moli H2 au fost introdusi.

niu = nr moli = masa-miu(H2)

=> m = niuxmiu(H2) = 37,635x2 = 75,27 g H

Answer 2

V=20 l=2*10⁻² m³

p₁=15 atm=15*10⁵ Pa

μ=2 g/mol=2*10⁻³ kg/mol

mi=μυi=μp₁V/RT₁=2*10⁻³*2*10⁻²*15*10⁵/8,314*250=0,028 kg

mf=μυf=μp₂V/RT₂=2*10⁻³*2*10⁻²*60*10⁵/8,314/280=0,103 kg

Δm=mf-mi=0,103-0,028=0,075095 kg≅75 g