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)

$\frac{{\left(\text{PV}\right)}_{\text{m}}}{\text{RT}}=\frac{3}{8}\times \text{2.21 ;}{V}_{\text{m}}=\left(\frac{3}{8}\times \text{2.21}\right)\times \frac{\text{RT}}{\text{P}};$
${\text{V}}_{\text{m}}=\frac{3}{8}\times \text{2.21}\times \frac{\text{8.314}\times 300}{\text{8.314}\times 10^6}$
$=248.625mL;$
${\text{V}}_{{\text{O}}_2}=\frac{16}{32}\times \text{248.625}=124.31mL$