Question

3.25 moles of gas take up 40.0 l of space under constant pressure and temperature conditions. what volume is required to hold 1.25 moles of the gas? v₂ = ? l

Explanation:

Step1: Apply Avogadro's Law

$\frac{V_1}{n_1} = \frac{V_2}{n_2}$
Given $V_1=40.0\ \text{L}$, $n_1=3.25\ \text{mol}$, $n_2=1.25\ \text{mol}$

Step2: Solve for $V_2$

$V_2 = V_1 \times \frac{n_2}{n_1}$
$V_2 = 40.0\ \text{L} \times \frac{1.25}{3.25}$

Step3: Calculate the value

$V_2 = 40.0 \times \frac{1.25}{3.25} \approx 15.4\ \text{L}$

Answer:

15.4 L