GAS TROUBLE

0.5%...

CONFIRM PLZ

yup !!!!
post the soln. please

Let the number of moles of Hydrogen gas escaped be n₁ and the number of moles of Oxygen gas escaped be n₂ ..
since the pressure remains the same so n₁ + n₂ = y

and acc to the other condition given we have
2n₁ + 32n₂ = 16y ..

solving these equations u will get
n₁ = 16y/30 and n₂ = 14y/30

now we have to find
\frac{\frac{16y*2}{30}}{\frac{16y*2}{30}+\frac{14y*32}{30}} * 100

so from here u will get 6.67%

See ...

let n_o & n_H be the initial moles of oxygen and hydrogen in the container...

let n'_o & n'_H be the amount of moles escaping... of oxygen & hydrogen respectively...

let volume be V

so P_i = (n_o + n_H)RTV

now ATQ,

n_CH4 = (32 n'_o + 2 n'_H)16

n_CH4 = 2 n'_o + n'_H8

so final moles after addition of methane..

= (n_o - n'_o) + (n_H - n'_H) +(2 n'_o + n'_H8)

= n_o + n_H + n'_o - 7n'_H8

so final pressure..

P_f = (n_o + n_H + n'_o - 7n'_H8)R TV

so P_i = P_f

(n_o + n_H)RTV = (n_o + n_H + n'_o - 7n'_H8)R TV

solving we get...

n'_o = 7n'_H8

now m = M n (m=weight , M = molar mass , n= moles)

so

m_om_H = M_o n_oM_H n_H

m_om_H = 32 x 72 x 8 = 14

or m_Hm_o = 1 14

now..

m_Hm_H + m_o = 1 1+ 14 = 115

or

m_Hm_H + m_o x 100 = 115 x 100

or %hydrogen by weight in the gas escaping=100/15= 6.67%

answer = 6.67 %... [1]

THANKS A LOT GUYS.......IT CAME IN OUR EXAM TODAY.....WHAT LUCK !!!!!!