AP Chemistry - Phases of Matter

Assume air contains 21% oxygen and 79% nitrogen.

If air is compressed to 5.5atm, what is the partial pressure of the oxygen?

Explanation

Use Dalton's law of partial pressure:

$P_{O_2}=P_{total}*y_{O_2}$

Where $P_{O_2}$ is the partial pressure of oxygen and $y_{O_2}$ is the mole fraction of oxygen. Plug in known values and solve.

A gas in a container is at is compressed to a volume of . What is the new pressure of the container?

$P_{2}=2.10, atm$

$P_{2}=0.525, atm$

$P_{2}=472.5, atm$

$P_{2}=1.05, atm$

Explanation

Boyle's Law is:

$P_{1} V_{1}=P_{2} V_{2}$

The initial volume ( $V_{1}$ ) and pressure ( $P_{1}$ ) of the gas is given. The volume changes to a new volume ( $V_{2}$ ). Our goal is to find the new pressure ( $P_{2}$ ). Solving for the new pressure gives:

$\frac{P_{1} V_{1}}{V_{2}}=\frac{P_{2} V_{2}}{V_{2}}\rightarrow P_{2}=\frac{P_{1} V_{1}}{V_{2}}$

$P_{2}=\frac{(1.05, atm)(30, mL)}{15, mL}=2.10, atm$

Notice the answer has 3 significant figures.

Assume air contains 21% oxygen and 79% nitrogen.

If air is compressed to 5.5atm, what is the partial pressure of the oxygen?

Explanation

Use Dalton's law of partial pressure:

$P_{O_2}=P_{total}*y_{O_2}$

Where $P_{O_2}$ is the partial pressure of oxygen and $y_{O_2}$ is the mole fraction of oxygen. Plug in known values and solve.

A $\small 4.5L$ container of gas has a pressure of $\small 3.0atm$ at a temperature of $\small 100^oC$ . The container is expanded to $\small 6L$ , and the temperature is increased to $\small 200^oC$ .

What is the final pressure of the container?

Explanation

In this case, two variables are changed between the initial and final containers: volume and temperature. Since we are looking for the final pressure on the container, we can use the combined gas law in order to solve for the final pressure:

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

When using the ideal gas law, remember that temperature must be in Kelvin, not Celsius, so we will need to convert.

Use the given values to solve for the final pressure.

$\frac{(3atm)(4.5L)}{373K} = \frac{(P_{2})(6L)}{473K}$

$P_{2} = 2.9atm$

A gas in a container is at is compressed to a volume of . What is the new pressure of the container?

$P_{2}=2.10, atm$

$P_{2}=0.525, atm$

$P_{2}=472.5, atm$

$P_{2}=1.05, atm$

Explanation

Boyle's Law is:

$P_{1} V_{1}=P_{2} V_{2}$

$\frac{P_{1} V_{1}}{V_{2}}=\frac{P_{2} V_{2}}{V_{2}}\rightarrow P_{2}=\frac{P_{1} V_{1}}{V_{2}}$

$P_{2}=\frac{(1.05, atm)(30, mL)}{15, mL}=2.10, atm$

Notice the answer has 3 significant figures.

What is the final pressure of the container?

Explanation

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

When using the ideal gas law, remember that temperature must be in Kelvin, not Celsius, so we will need to convert.

Use the given values to solve for the final pressure.

$\frac{(3atm)(4.5L)}{373K} = \frac{(P_{2})(6L)}{473K}$

$P_{2} = 2.9atm$

A gas in a container is at is compressed to a volume of . What is the new pressure of the container?

$P_{2}=2.10, atm$

$P_{2}=0.525, atm$

$P_{2}=472.5, atm$

$P_{2}=1.05, atm$

Explanation

Boyle's Law is:

$P_{1} V_{1}=P_{2} V_{2}$

$\frac{P_{1} V_{1}}{V_{2}}=\frac{P_{2} V_{2}}{V_{2}}\rightarrow P_{2}=\frac{P_{1} V_{1}}{V_{2}}$

$P_{2}=\frac{(1.05, atm)(30, mL)}{15, mL}=2.10, atm$

Notice the answer has 3 significant figures.

What is the final pressure of the container?

Explanation

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

When using the ideal gas law, remember that temperature must be in Kelvin, not Celsius, so we will need to convert.

Use the given values to solve for the final pressure.

$\frac{(3atm)(4.5L)}{373K} = \frac{(P_{2})(6L)}{473K}$

$P_{2} = 2.9atm$

Assume air contains 21% oxygen and 79% nitrogen.

If air is compressed to 5.5atm, what is the partial pressure of the oxygen?

Explanation

Use Dalton's law of partial pressure:

$P_{O_2}=P_{total}*y_{O_2}$

Where $P_{O_2}$ is the partial pressure of oxygen and $y_{O_2}$ is the mole fraction of oxygen. Plug in known values and solve.

A gas is initially in a 5L piston with a pressure of 1atm.

If pressure changes to 3.5atm by moving the piston down, what is new volume?

Explanation

Use Boyle's Law:

Plug in known values and solve for final volume.

Phases of Matter

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