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Ideal Gas Law

'n' mol of ...

Question

'n' mol of $N_{2}$ and 0.05 mol of Ar are enclosed in a vessel of capacity 2 L at 1 atm and 27 $^{o}$ C. Find 'n'?
[R = 0.082 L atm mol $K^{- 1}$ ].

0.3

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0.03

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0.06

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Solution

The correct option is C 0.03
PV = nRT
$\Rightarrow 2 \times 1 = n \times 0.082 \times (27 + 273)$
$\Rightarrow n = \frac{2}{0.082 \times 300}$
$\Rightarrow n = 0.0813$
Total number of moles, n = $n_{N_{2}} + n_{A r}$
$\Rightarrow 0.0813 = n_{N_{2}} + 0.05$
Therefore, $n_{N_{2}} = 0.0813 - 0.05 = 0.0313 m o l e s$
Hence, the correct option is C.

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'n' mol of N2 and 0.05 mol of Ar are enclosed in a vessel of capacity 2 L at 1 atm and 27oC. Find 'n'? [R = 0.082 L atm mol K−1].

The correct option is C 0.03PV = nRT⇒2×1=n×0.082×(27+273)⇒n=20.082×300⇒n=0.0813Total number of moles, n = nN2+nAr⇒0.0813=nN2+0.05Therefore, nN2=0.0813−0.05=0.0313molesHence, the correct option is C.

'n' mol of $N_{2}$ and 0.05 mol of Ar are enclosed in a vessel of capacity 2 L at 1 atm and 27 $^{o}$ C. Find 'n'?
[R = 0.082 L atm mol $K^{- 1}$ ].