Question

3.00 l of a gas contains 0.965 mol. if the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?

Explanation:

Step1: Identify the gas - law relationship

At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles (Avogadro's law: $V_1/n_1 = V_2/n_2$).

Step2: Define the given values

$V_1 = 3.00\ L$, $n_1=0.965\ mol$, $n_2 = 1.80\ mol$.

Step3: Rearrange the formula to solve for $V_2$

$V_2=\frac{n_2}{n_1}\times V_1$.

Step4: Substitute the values into the formula

$V_2=\frac{1.80\ mol}{0.965\ mol}\times3.00\ L$.
$V_2=\frac{1.80\times3.00}{0.965}\ L$.
$V_2=\frac{5.4}{0.965}\ L\approx5.60\ L$.

Answer:

$5.60\ L$