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[Post New]10 Apr 2008 18:47:41 IST
    Subject: help
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sweet08 (72)

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1 mole of N2O4(g) at 300K is kept in a  closed  container under 1 atm .
it is heated to 600K  when 20% by mass of N2O4 (g)  decomposes to N2O(g).
the resultant pressure is :
ans--2.4 atm
    
[Post New]11 Apr 2008 16:49:05 IST
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srujana (3045)

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Volume\ remains\ constant\ throughout.(Closed\ container)\\\ \\ P_i=1\ atm\\\ \\ T_i=300\ K\\\ \\ n_i=1\\\ \\ Amount\ of\ N_2O_4=92g\\\ \\ \\\ \\ Amount\ of\ N_2O_4\ decomposed=\frac{20}{100}\times 92=18.4g=amount\ of\ NO_2\\\ \\ Amount\ of\ N_2O_4\ left=73.6g\\\ \\ n_f=\frac{73.6}{92}+\frac{18.4}{\frac{92}{2}}\\\ \\ Given\ T_f=600K,\\\ \\ From\ gas\ laws\ \frac{P_i}{n_i\times T_i}=\frac{P_f}{n_f\times T_f}\\\ \\ Solve\ for\ P_f, we\ get\ P_f=2.4\ atm
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[Post New]11 Apr 2008 17:16:09 IST
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bansalshivanii (10)

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