At constant temperature $250 m L$ of argon at a $760 m m H g$ pressure and $60 m L$ of nitrogen at $500 m m$ pressure are put together in one litre flask. The final pressure is _________ $m m H g$ .

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Solution

$P V = n R T$
For Argon, $P = 760 m m H g = 1 a t m$
$V = 0.25 L$
So, $n = \frac{1 \times 0.25}{R T} = \frac{1}{4 R T}$
For Nitrogen, $P = 500 m m H g = \frac{500}{760} a t m$
$V = 0.06 L$
So, $n = \frac{500 \times 0.06}{760 R T} = \frac{3}{76 R T}$
After mixing both the gases,
Total number of moles of both the gases, $n = \frac{1}{4 R T} + \frac{3}{76 R T} = \frac{88}{304 R T}$
$V = 1 L$
$P = \frac{n R T}{V} = \frac{88}{304 R T} \times R T = \frac{88}{304} a t m$

$P = \frac{88}{304} \times 760 m m H g = 220 m m H g$

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