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arvind1990 (305)

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A teacher enters in class from front door,student enters from back door.There are 12 equidistant row of benches in class. Teacher releases N2O from first bench student releases weeping gas from last bench.At which row,student starts weeping and laughing simultaneously???
    
master_purav (1343)

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Its based on Graham's Law of Diffusion...

Wait a moment till i find the solution...

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master_purav (1343)

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Its 9th bench...

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arvind1990 (305)

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How did u get it???
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master_purav (1343)

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Use Graham's Law

rate of diffusion fo a gas is inversely proportional to its molecular weight.

Take ratio of rates

But the time for the diffusion will be same for both.

So u are left with ratio of volumes

Volume is length * cross section area

Cross section area same. So cancelled in numerator and denominator

Take length in units of bench

solve the eq. get the ans.

Sorry i could not post the entire solution. Coz its a bit long and difficult to type as it has many subscripts etc...

"If you win, you shall not have to explain and if you lose, you wont be there to explain"
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master_purav (1343)

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By the way, it should be specified in the ques that weeping gas is C6H11OBr

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khyatigupta (49)

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hiiiii
 since mol wt. of N2O gas = 44g/mol
and mol wt. of tear gas(C6H11OBr)=176g/mol
 
R=rate of diffusion of N2O
r=rate of diffusion of tear gas
M=mol wt of N2O
m=mol wt of tear gas
 
Acc to the formula
R/r=sqrt(m/M)
R/r=sqrt(176/44)
R/r=2
 
R=2*r
 
This means rate of diffusion of N2O gas is twice the rate of diffusion of tear gas
Thts why 9th row will have the tendency of recieving both the gases...
 
Its very easy n short method.hope u like it......
 
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khyatigupta (49)

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It is also very easy to find out the row
U jst draw 13 rows on a paper  then for every two rows frm the frnt cross one row frm the end n then u will see tht they will meet at the 9th row. ...thnks
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deepak_agarwal (539)

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people have done really nice job above

a 2nd year IIT DELHI student, doing B.Tech in chemical engineering
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