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One "mole" of N(2)O(4)(g) at 300 K is ke...

One "mole" of `N_(2)O_(4)(g)` at `300 K` is kept in a closed container under `1` atm. It is heated to `600 K`, when `20%` by mass of `N_(2)O_(4)(g)` decomposes to `NO_(2)(g)`. The resultant pressure is

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`N_(2)O_(4) (g) hArr 2NO_(2)(g)`
`{:("Initial Conc.",1mol,0mol),("Final Conc.",1(1-0.2)=0.8mol,2(0.2)0.4mol):}`
Total no. of moles after decomposition `= 0.8 + 0.4 = 1.2 mol`
Now, `P_(1) = 1` atm , `P_(2) = ?`
`n_(1) = 1 mol` , `n_(2) = 1.2` mol
`T_(1) = 300 K` , `T_(2) = 600 K`
According to ideal gas equation, `PV = nRT`.
`:. (P_(1)V)/(P_(2)V) = (n_(1)RT_(1))/(n_(2)RT_(2))` or `((1 atm))/(P_(2)) = ((1 mol) xx (300 K))/((1.2 mol) xx (600 K))`
or `P_(2) = ((1 atm) xx (1.2 mol) xx (600 K))/((1 mol) xx (300 K)) = 2.4` atm
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