Q. 8.49 - An Introduction to General, Organic, and Biological Chemistry

Question

A sample containing$1.50 moles of Ne$ gas has an initial volume of $8.00L$. What is the final volume, in liters, when each of the following occurs and pressure and temperature do not change?

a. A leak allows one-half of $Ne$ atoms to escape.

b. A sample of $3.50 moles of Ne$ is added to the $1.50 moles of Ne$ gas in the container.

c. A sample of $25.0 g$of $Ne$is added to the $1.50 moles of Ne$gas in the container.

Step-by-Step Solution

Answer

a. The final volume is $4L$.

b. The final volume is $26.7L$.

c. The final volume is $14.4L$.

Part (a) step 1: Given Information

We need to find the final volume.

Part (a) step 2: Simplify

Consider:

$$\begin{gathered} n_1=1.50 mole \\ {V}_1=8.00L \end{gathered}$$

Now, calculating the final volume$V_2$:

conversion factor is:

$\frac{6\times {10}^{23}atom of Ne}{1 mol of Ne}$

Contained in $1.50 mol of Ne$

$$\begin{gathered} 1.5 mole of Ne\times \frac{6\times {10}^{23}atom of Ne}{1 mol of Ne} \\ =9\times {10}^{23} atoms of Ne \end{gathered}$$

the final volume $V_2$ is calculated by Avogadro's law equation:

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

Putting the value:

$$\begin{gathered} \frac{V_1}{n_1}=\frac{V_2}{n_2} \\ {V}_2=\frac{V_1\times n_2}{n_1} \\ {V}_2=\frac{8.00L\times 0.75 mol}{1.50 mol} \\ {V}_2=\frac{6L}{1.50L} \\ {V}_2=4L \end{gathered}$$

Part (b) step 1: Given Information

We need to find the final volume.

Part (b) step 2: Simplify

Consider:

$$\begin{gathered} n_1=1.50 mole \\ {V}_1=8.00L \end{gathered}$$

Now, calculating the final volume $V_2:$

$n_2=3.50 mol+1.50mol=5.00 mol of Ne$

the final volume $V_2$ is calculated by Avogadro's law equation:

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

Putting the value:

$$\begin{gathered} \frac{V_1}{n_1}=\frac{V_2}{n_2} \\ {V}_2=\frac{V_1\times n_2}{n_1} \\ {V}_2=\frac{8.00L\times 5.00 mol}{1.50 mol} \\ {V}_2=\frac{40L}{1.50L} \\ {V}_2=26.7L \end{gathered}$$

Part (c) step 1: Given Information

We need to find the final volume.

Part (c) step 2: Simplify

Consider:

$$\begin{gathered} n_1=1.50 mole \\ {V}_1=8.00L \end{gathered}$$

Now, calculating the final volume $V_2$:

conversion factor is:

$\frac{1 mol of Ne}{20.2 g of Ne}$

$$\begin{gathered} 25.0 g of Ne\times \frac{1 mol of Ne}{20.2 g of Ne} \\ =1.20 mol of Ne \end{gathered}$$

$n_2=1.20 mol of Ne+1.50 mol of Ne=2.70 mol of Ne$

the final volume $V_2$ is calculated by Avogadro's law equation:

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

Putting the value:

$$\begin{gathered} \frac{V_1}{n_1}=\frac{V_2}{n_2} \\ {V}_2=\frac{V_1\times n_2}{n_1} \\ {V}_2=\frac{8.00L\times 2.70 mol}{1.50 mol} \\ {V}_2=\frac{21.6L}{1.50L} \\ {V}_2=14.4L \end{gathered}$$