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1.30 Lit N₂ gas at 2atm and 300K in a container is exposed to 4g of solid surface. After complete adsorption the pressure of N₂ is reduced by 30% calculate the value of x/m

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0.56

0.32

0.43

0.22

answer is A.

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Detailed Solution

Pressure is reduced by 30% $= \frac{2 \times 30}{100} = 0.6 atm$

$PV = \frac{WRT}{M} 0 .6 \times 1 .3 = \frac{W}{28} \times 0 .0821 \times 300, wt = 0 .8732 \frac{x}{m} = \frac{0 .8732}{4} = 0 .22$

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