Find the moles of $O_2$ having pressure $250 \ bar$ in $500 \ mL$ vessel at $300 \ K$ temperature. $[R = 8.314 \times 10^{-2} \ bar \ L \ K^{-1} \ mol^{-1}]$

(5.01) Using the ideal gas equation: $PV = nRT$
Rearranging for moles: $n = \frac{PV}{RT}$
Given values:
$P = 250 \ bar$
$V = 500 \ mL = 0.5 \ L$
$T = 300 \ K$
$R = 8.314 \times 10^{-2} \ bar \ L \ K^{-1} \ mol^{-1}$
Calculation:
$n = \frac{250 \times 0.5}{8.314 \times 10^{-2} \times 300}$
$n = \frac{125}{24.942}$
$n \approx 5.01 \ mol$
Thus,the number of moles of $O_2$ is $5.01 \ mol$.