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Question

One mole of N2O4(g) at 300 K is kept in a closed container under 1 atomsphere. It is heated to 600K when 20% by mass of N2O4 decomposes to NO2(g). The resultant pressure (in atm) is .


Solution

intialMole  N2O412NO20  ,V=1 mole×R×300 K1 atm  (i)
at equilibriumMole  N2O410.22NO22×0.2   nT=0.8+0.4=1.2Pf×V=nT×R×600 so by substituting volume of (i)
PT=2.4 atm

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