Question

A 4 L sample of gas at 298 K and 2 atm contains 0.250 mol of gas. If we add another 0.250 mol of gas at the same pressure and temperature, what is the final total volume of gas?

Answer 1

Answer:

$V_2=8L$

Explanation:

Hello there!

In this case, considering the Avogadro's gas law, which treats the volume and moles in a directly proportional way via:

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

Which can be solved for the final volume, V2, as shown below:

$V_2=\frac{V_1n_2}{n_1}$

Thus, by plugging in the given data, we obtain:

$V_2=\frac{4L*(0.250mol+0.250mol)}{0.250mol}\\\\V_2=8L$

Regards!