A litre of air weights $1.293$ grams at NTP. At what temperature will a litre of air weigth $1$ gram, the pressure being $72 c m$ ?

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Solution

$W = 1.293 g$

$V = 1 l i t$ at NTP

no. of moles $= \frac{V}{22.4} = \frac{1}{22.4} = \frac{1.293}{M . W ._{a i r}}$

$\Rightarrow M . W ._{a i r} = 29 g$

For $V = 1 l i t, W = 1 g \Rightarrow n_{a i r} = \frac{1}{29}$

$P = 72 c m$ of $H g = \frac{72}{76} a t m$

$P V = n R T$

$T = \frac{\frac{72}{76} \times 1}{0.0821 \times \frac{1}{29}} = \frac{72 \times 29}{76 \times 0.0821} = 334.6 K$

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