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 One mole of N2O4 is kept in a close container at 300K and 1 atm pressure. When it is heated to 600 K 20% of it decomposes to NO2 (g). Calculate the resultant pressure exerted on the walls of the container
  • a)
    3.4 atm
  • b)
    1.2 atm
  • c)
    2.0 atm
  • d)
    3.6 atm
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
One mole of N2O4is kept in a close container at 300K and 1 atm pressur...
The answer is 2.4 atm
N2O4    ⇔  2NO2

1mole  0   Initial moles

-20% of 1=0.2  +2(20% of 1)=0.40  At equilibrium

1-0.2= 0.8  0.4  Moles at equilibrium

Total moles at equilibrium = 0.8+0.4 = 1.2 moles

1 mole vapour pressure = 1 atm at 300 K

Applying PV = nRT

 1xV = 1x R x 300 .... (1)

When n=1.2 moles, T = 600 K

 P x V = 1.2 x R x 600 .... (2)

Dividing (2) by (1),

PxV/(1xV) = (1.2 x R x 600)/ (1 x R x 300)

Therefore, P= 2.4 atm

Hence, resultant pressure of mixture is 2.4 atm 
This question is part of UPSC exam. View all Class 11 courses
Most Upvoted Answer
One mole of N2O4is kept in a close container at 300K and 1 atm pressur...
2.4 atm
Community Answer
One mole of N2O4is kept in a close container at 300K and 1 atm pressur...
1.2atm
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One mole of N2O4is kept in a close container at 300K and 1 atm pressure. When it is heated to 600 K 20% of it decomposes to NO2(g). Calculate the resultant pressure exerted on the walls of the containera)3.4 atmb)1.2 atmc)2.0 atmd)3.6 atmCorrect answer is option 'B'. Can you explain this answer?
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