If the density of a certain gas at $30^{\circ} C$ and $768$ torr is $1.35 {kg/m}^{3}$ , its density at STP would be:

1.48 {kg/m}^{3}

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1.58 {kg/m}^{3}

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1.25 {kg/m}^{3}

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1.3 {kg/m}^{3}

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Solution

The correct option is A $1.48 {kg/m}^{3}$
Given:
$d = 1.35 k g / m^{3}$
$T = 30 + 273 = 303 K$
$P = 768 t o r r$
$R = 62.364 L t o r r K^{- 1} m o l^{- 1}$

We know that, Ideal gas equation is,
$P V = n R T$

$n = \frac{W}{M}$
where, $W$ and $M$ are the mass and molar mass respectively.

Substituting $n = \frac{W}{M}$ in the ideal gas equation we get,

$P V = \frac{W}{M} R T$

$P M = \frac{W}{V} R T$

Density, $d = \frac{W}{V} (\frac{m a s s}{v o l u m e})$

Substituting $\frac{W}{V} = d$ we get,

$P M = d R T$

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

Substituting values,
$M = \frac{1.35 k g / L \times 62.364 L t o r r K^{- 1} m o l^{- 1} \times 303 K}{768 t o r r}$

$M = 33.216 k g / m o l$

At STP,
$P = 1 a t m$
$R = 0.082057 L a t m m o l^{- 1} K^{- 1}$
$T = 273 K$
$M = 33.216 k g / m o l$

Density, $d = \frac{P M}{R T}$

$d = \frac{1 \times 33.216}{0.082057 \times 273}$

$d = 1.4827 k g / m^{3}$

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