Question

a sample of he gas (2.35 mol) occupies 57.9 l at 300.0 k and 1.00 atm. the volume of this sample is ____l at 423 k and 1.00 atm.

Explanation:

Step1: Apply Charles's Law

Charles's Law is $V_1/T_1 = V_2/T_2$. Here, $V_1 = 57.9$ L, $T_1=300.0$ K, $T_2 = 423$ K.

Step2: Solve for $V_2$

Rearrange the formula to $V_2=\frac{V_1\times T_2}{T_1}$. Substitute the values: $V_2=\frac{57.9\times423}{300.0}$.
Calculate $57.9\times423 = 24591.7$ and then divide by $300.0$: $V_2=\frac{24591.7}{300.0}=81.6$ L.

Answer:

$81.6$ L