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A flexible container holding 1.25 moles of gas contracts from 25.0 L to 10.0 L when some gas is removed. The pressure and temperature remain the same.

How many moles of gas are in the contracted container?



Answer :

GinnyAnswer
Certainly! Let's solve this step-by-step. We are given that:

- Initial volume, [tex]\( V_1 = 25.0 \, L \)[/tex]
- Final volume, [tex]\( V_2 = 10.0 \, L \)[/tex]
- Initial moles of gas, [tex]\( n_1 = 1.25 \, \text{moles} \)[/tex]

We need to find the final moles of gas, [tex]\( n_2 \)[/tex], after the container contracts.

Since the temperature and pressure remain constant, we can use Avogadro's Law, which states that the volume of a gas is directly proportional to the number of moles of gas when the temperature and pressure are constant. Mathematically, this relationship is given by:

[tex]\[ n_1 \times V_1 = n_2 \times V_2 \][/tex]

Rearranging this equation to solve for [tex]\( n_2 \)[/tex]:

[tex]\[ n_2 = \frac{n_1 \times V_1}{V_2} \][/tex]

Now, we can substitute the given values into the equation:

[tex]\[ n_2 = \frac{1.25 \, \text{moles} \times 25.0 \, L}{10.0 \, L} \][/tex]

[tex]\[ n_2 = \frac{31.25 \, \text{moles} \cdot L}{10.0 \, L} \][/tex]

[tex]\[ n_2 = 3.125 \, \text{moles} \][/tex]

However, this seems to make an error of not removing some gas. We need to reconsider all terms critically.

Actually using ideal gas laws as more applicable, the simple ratio [tex]\(n_2 = n_1 \times (V_2 / V_1)\)[/tex] gives straight correct:

So:

[tex]\[ n_2 = 1.25 \times \frac{10.0}{25.0} \][/tex]

[tex]\[ n_2 = 1.25 \times 0.4 \][/tex]

[tex]\[ n_2 = 0.5 \, \text{moles} \][/tex]

So, the number of moles of gas in the contracted container is [tex]\(0.5\)[/tex] moles.
amysun2025

Answer:

0.500 mol

Explanation:

Pre-Solving

We are given that a container with 1.25 moles of gas has a volume of 25.0 L. When some of the gas is removed, the volume of the container is 10.0L.

We can use Avogadro's Law in order to solve this problem. Recall that the formula for Avogadro's Law is [tex]\frac{V_1}{n_1} = \frac{V_2}{n_2}[/tex], where V is volume and n is moles.

Solving

We can substitute what we know into the equation. The [tex]V_1[/tex] and [tex]n_1[/tex] go together (if 1.25 mol is [tex]n_1[/tex], [tex]V_1[/tex] is 25.0 L), so do the [tex]V_2[/tex] and [tex]n_2[/tex].

So:

[tex]\frac{25.0 L}{1.25 mol} = \frac{10.0 L }{n_2}[/tex]

Now, cross multiply:

[tex]25.0 L * n_2 = 10.0 L * 1.25 mol[/tex]

Divide both sides by 25.0 L

[tex]n_2 = 0.500 mol[/tex] (with 3 sig figs)

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