Question

A sample containing 4.80 g and O₂ gas has an initial volume of 15.0 L. What is the final volume in liters, when each of the following occurs and pressure and temperature do not change?

a. A sample of 0.500 moles of O₂ is added to the 4.80 g of O₂ in the container.

b. A sample of 2.00 g of O₂ is removed.

c. A sample of 4.00 g of O₂ is added to the 4.80 g of O₂ gas in the container.

Short answer

a. The final volume is 50 L.

b. The final volume is 8.75 L.

c. The final volume is 27.5 L.

Step-by-step solution

Part (a) Given Information

Initial volume = 15.0 L

Initial mass = 4.80 g

Part (a) Explanation

Avogadro law tells us that the two gases will have the same volume if they have the same number of moles at a certain pressure and temperature condition.

This can be expressed in terms of relation which is also known as Avogadro's relation:

$\frac{{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{n}}_2}$ ......(1)

Here, V₁ is the initial volume, V₂ is the final volume, n₁ is the initial number of moles and n₂ represents the final number of moles.

Now, the mass of O₂ gas is given, thus we can calculate the number easily using the molar mass of O₂ which can be calculated as follows:

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} \\ \mathrm{n}=\frac{4.80\mathrm{g}}{32.00\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}=0.15\mathrm{mole} \end{gathered}$$

On rearranging, equation (1) we will get,

$\frac{{\mathrm{V}}_1}{{\mathrm{n}}_1}\times {\mathrm{n}}_2={\mathrm{V}}_2$

Thus, the calculation for the final volume is as follows:

$15\mathrm{L}\times \frac{0.500\cancel{\mathrm{mole}}}{0.15\cancel{\mathrm{mole}}}=15\times 3.33=50\mathrm{L}$

Hence, the final volume is 50 L.

Part (a) Given Information

Initial volume = 15.0 L

Initial moles = 0.15 moles

Part (b) Explanation

Avogadro law tells us that the two gases will have the same volume if they have the same number of moles at a certain pressure and temperature condition.

This can be expressed in terms of relation which is also known as Avogadro's relation:

$\frac{{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{n}}_2}$ ......(1)

Here, V₁ is the initial volume, V₂ is the final volume, n₁ is the initial number of moles and n₂ represents the final number of moles.

Now, the 2.00 g f O₂ gas is removed from the sample, thus we can calculate the number easily using the molar mass of O₂ which can be calculated as follows:

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} \\ \mathrm{n}=\frac{2.80\mathrm{g}}{32.00\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}=0.0875\mathrm{mole} \end{gathered}$$

On rearranging, equation (1) we will get,

$\frac{{\mathrm{V}}_1}{{\mathrm{n}}_1}\times {\mathrm{n}}_2={\mathrm{V}}_2$

Thus, the calculation for the final volume is as follows:

$15\mathrm{L}\times \frac{0.0875\cancel{\mathrm{mole}}}{0.15\cancel{\mathrm{mole}}}=8.75\mathrm{L}$

Hence, the final volume of O₂ gas is 8.75 L.

Part (c) Given Information

Initial volume = 15.0 L

Initial moles = 0.15 moles

Part (c) Explanation

Avogadro law tells us that the two gases will have the same volume if they have the same number of moles at a certain pressure and temperature condition.

This can be expressed in terms of relation which is also known as Avogadro's relation:

$\frac{{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{n}}_2}$............(1)

Here, V₁ is the initial volume, V₂ is the final volume, n₁ is the initial number of moles and n₂ represents the final number of moles.

Now, 4.00 g of O₂ gas is added to the sample , thus we can calculate the number easily using the molar mass of O₂ which can be calculated as follows:

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} \\ \mathrm{n}=\frac{8.80\mathrm{g}}{32.00\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}=0.275\mathrm{mole} \end{gathered}$$

On rearranging, equation (1) we will get,

$\frac{{\mathrm{V}}_1}{{\mathrm{n}}_1}\times {\mathrm{n}}_2={\mathrm{V}}_2$

Thus, the calculation for the final volume is as follows:

$15\mathrm{L}\times \frac{0.0275\cancel{\mathrm{mole}}}{0.15\cancel{\mathrm{mole}}}=27.5\mathrm{L}$

Hence, the final volume of O₂ gas is 27.5 L.