Question

What would be the pressure exerted by 5.000 grams of chlorine gas in a 10.000 liter vessel at 100.0 c?

Answer 1

Answer:

0.2188 bar.

Explanation:

We can use the Ideal Gas Law to solve this problem.

pV = nRT

However, $n = \frac{ m}{ M}$, so

$pV = \frac{ m}{ M}RT$

$p = \frac{mRT }{ MV}$

In this problem,

T = (100.0 + 273.15) K = 373.15 K

$p = \frac{\text{5.000 g}\times \text{0.083 14 bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{373.15 K}}{ 70.91 \text{g}\cdot\text{mol}^{-1}\times \text{10.000 L}} = \text{0.2188 bar}$