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Q. A gaseous mixture was prepared by taking equal moles of $CO$ and $N_2.$ If the total pressure of the mixture was found $1$ atmosphere, the partial pressure of the nitrogen $(N_2)$ in the mixture is

AIPMTAIPMT 2011States of Matter

Solution:

The partial pressure of gas $\propto$ Mole fraction of gas ( $X$ gas)
$X _{ CO }=\frac{ n _{ CO }}{ n _{ CO }+ n _{ N _{2}}}=\frac{ n _{ CO }}{2 n _{ CO }} \left[\because n _{ CO }= n _{ N _{2}}\right]$
$\Rightarrow X _{ CO }=\frac{1}{2}=0.5$
Partial pressure of $N _{2}= X _{ N _{2}} \cdot P _{ T }=\left[1- X _{ CO }\right] \cdot P$
$ =[1-0.5] \times[1]=0.5\, atm$