Question

what is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°c? (given: r = 0.08205 l · atm/mol · k)
2.81 liters
4.39 liters
6.19 liters
9.67 liters

Explanation:

Step1: Convert temperature to Kelvin

$T = 227 + 273=500\ K$

Step2: Use ideal - gas law formula

The ideal - gas law is $PV = nRT$, and we want to find $V$. Rearranging for $V$ gives $V=\frac{nRT}{P}$.
We know that $n = 0.8\ mol$, $R = 0.08205\ L\cdot atm/mol\cdot K$, $T = 500\ K$, and $P = 5.3\ atm$.

Step3: Substitute values and calculate

$V=\frac{0.8\times0.08205\times500}{5.3}$
$V=\frac{0.8\times41.025}{5.3}$
$V=\frac{32.82}{5.3}\approx6.19\ L$

Answer:

6.19 liters