Ncert Exercises & Solutions

$G i v e n,$

$d_{1} = 5.46 g / d m^{3}$

$p_{1} = 2 b a r$

$T_{1} = 27 ° C = (27 + 273) K = 300 K$

$p_{2} = 1 b a r$

$T_{2} = 273 K$

$d_{2} = ?$

The density

$(d_{2})$ of the gas at STP can be calculated using the equation,

$d = \frac{M p}{R T}$

$\Rightarrow \frac{d_{1}}{d_{2}} = \frac{\frac{M p_{1}}{R T_{1}}}{\frac{M p_{2}}{R T_{2}}}$

$\Rightarrow \frac{d_{1}}{d_{2}} = \frac{p_{1} T_{2}}{p_{2} T_{1}}$

$\Rightarrow d_{2} = \frac{p_{2} T_{1} d_{1}}{p_{1} T_{2}}$

$= \frac{1 \times 300 \times 5.46}{2 \times 273}$

$= 3 g d m^{- 3}$

Hence, the density of the gas at STP will be 3 g

$d m^{- 3}$ .