Question

What should be the volume occupied by $16$ grams of oxygen gas ( $O_{2}$ ) at $300K$ and $8.31MPa$ , provided:
$\frac{\text{P}_{\text{c}} \text{V}_{\text{c}}}{\text{RT}_{\text{c}}} = \frac{3}{8} and \frac{\text{P}_{\text{r}} \text{V}_{\text{r}}}{\text{T}_{\text{r}}} = \text{2.21}$
​(Given: $R=8.314cm3MPaK-1mol-1$ )

• A. $125.31mL$
• B. $124.31mL$
• C. $248.62mL$
• D. None of these

Answer 1

Answer: (B)

$\frac{\left(\text{PV}\right)_{\text{m}}}{\text{RT}} = \frac{3}{8} \times \text{2.21 ;} V_{\text{m}} = \left(\frac{3}{8} \times \text{2.21}\right) \times \frac{\text{RT}}{\text{P}} ;$
$\text{V}_{\text{m}} = \frac{3}{8} \times \text{2.21} \times \frac{\text{8.314} \times 3 0 0}{\text{8.314} \times 1 0^{6}}$
$= 248.625 mL ;$
$\text{V}_{\text{O}_{2}} = \frac{1 6}{3 2} \times \text{248.625} = 124.31 mL$