The compressibility factor of nitrogen at 400 K and 800 atm is 2.00 and at 600 K and 200 atm it is 1.20. A certain mass of $N_{2}$ occupies a volume of $1 d m^{3}$ at 400 K and 800 atm. Volume occupied by same quantity of $N_{2}$ gas at 600 K and 200 atm in litre is :

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Solution

$z = \frac{P V}{n R T}$ ,

Given that same quantity of $N_{2}$ gas is present , hence same number of moles.
$∴$ $n_{400 K}$ $=$ $n_{600 K}$

$n_{400 K}$ $=$ $\frac{P V}{Z R T}$ $=$ $\frac{1 \times 800}{2 \times R \times 400}$ ..... $e q (1)$

$n_{600 K}$ $=$ $\frac{P V}{Z R T}$ $=$ $\frac{V \times 200}{1.2 \times R \times 600}$ ..... $e q (2)$

Equating eq(1) and eq(2), we obtain
V = 3.6 L

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