Calculate the volume occupied by $2.0 m o l e$ of $N_{2}$ at $200 K$ and $8.21 a t m$ pressure, if $\frac{P_{C} V_{C}}{R T_{C}} = \frac{3}{8}$ and $\frac{P_{r} V_{r}}{T_{r}} = 2.4$ .

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Solution

First we will calculate the volume occupied by two moles of nitrogen at STP. We know that one mole of every gas occupies a volume of $22.4 l i t e r s$ at STP.
Therefore volume occupied by 2 moles of nitrogen at STP will be $2 \times 22.4 = 44.8 l i t e r s$
STP stands for standard temperature and pressure. For STP condition, the temperature is $273.15 K (0^{o} C)$ and pressure is $1 b a r (10^{5} P a s c a l s o r 0.987 a t m)$
We will now use the relation
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Here,
$P_{1} = 0.987 a t m$
$V_{1} = 44.8 l i t r e s$
$T_{1} = 273.15 K$
$P_{2} = 8.21 a t m$
$T_{2} = 200 K$
$V_{2} = ?$
Substituting in above equation, we get
$\frac{0.987 \times 44.8}{273.15} = \frac{8.21 \times V_{2}}{200}$
$V_{2} = \frac{0.987 \times 44.8 \times 200}{273.15 \times 8.21}$
$V_{2} = 3.94 l i t e r s$

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