Gas Laws - AP Chemistry

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Question

A sample of gas at a constant temperature has an initial pressure of $14.5\text{L}$ at a pressure of $720\text{mmHg}$ . If the volume of gas is decreased to $10.5\text{L}$ , what is its pressure?

Answer

Since we are given the volume and the pressure of this sample of gas, we will need to use Boyle's Law, which states that the pressure and volume of a gas, at a constant temperature, are inversely related. As thus, we can then write the following equation:

$\text{P}_1\text{V}_1=\text{P}_2\text{V}_2$

Since all the answer choices are in units of atmospheres, we will need to convert the given units into atmospheres.

$720\text{mmHg}\cdot\frac{1\text{ atm}}{760\text{mmHg}}=0.947\text{atm}$

Plug in the given pressures and volume into the equation, and solve for $\text{P}_2$ .

$(14.5)(0.947)=\text{P}_2(10.5)$

$\text{P}_2=1.31\text{atm}$

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Question

What is the pressure of a $5.20\text{L}$ cylinder filled with moles of helium gas at $317\text{K}$ ?

Answer

Recall the ideal gas law:

$\text{PV=nRT}$ ,

where $\text{P=Pressure}$ ,

$\text{V=Volume}$ ,

$\text{n=moles of gas}$ ,

$\text{R=Ideal gas constant}$ ,

and $\text{T=temperature}$ .

Since the question asks for pressure, rearrange the equation to solve for $\text{P}$ .

$\text{P}=\frac{\text{nRT}}{V}$

Plug in the given values. Remember that $\text{R=0.0821}\text{L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mole}^{-1}$

$\text{P}=\frac{(0.681)(0.0821)(317)}{5.20}=3.41\text{atm}$

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Question

Consider the following chemical reaction:

$6\text{Li}(s)+\text{N}_2(g)\rightarrow2\text{Li}_3\text{N}(s)$

How many grams of lithium is needed to react with $12.6\text{L}$ of nitrogen gas measured at $745\text{mmHg}$ and $28.5^\circ \text{C}$ ?

Answer

Start by finding the number of moles of nitrogen gas by using the ideal gas law:

$\text{PV}=\text{nRT}$

Rearrange the equation to solve for the variable $\text{n}$ :

$\text{n}=\frac{\text{PV}}{\text{RT}}$

In order to use the ideal gas law, the pressure must have units of atmospheres and the temperature must have units of Kelvin.

Convert the given pressure into atmospheres:

$745\text{mmHg}\cdot\frac{1\text{atm}}{760\text{mmHg}}=0.980\text{atm}$

$28.5\text{C}+273.15=301.65\text{K}$

Now, substitute in all the given values in order to find the number of moles of nitrogen gas.

$\text{n}=\frac{(0.980)(12.6)}{(0.0821)(301.65)}=0.499\text{ moles of nitrogen gas}$

Next, use the stoichiometric ratio given by the chemical equation to find the number of moles of lithium needed to react completely with the nitrogen gas.

$0.499\ \text{moles of N}_2\cdot\frac{6\text{ moles of Li}}{1\text{ mole of N}_2}=2.99\text{ moles of Li}$

Finally, convert the number of moles of lithium to number of grams of lithium.

$2.99\text{ moles of Li}\cdot\frac{6.94\text{g of Li}}{1\text{ mole of Li}}=20.8\text{g of Li}$

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Question

What is the pressure exerted by $0.50 mol N_{2}$ in a container at ?

Answer

$P=\frac{nRT}{V}$

$P=\frac{(0.50 mol N_{2}) (0.0821 L\cdot atm) (303K)}{13.0 L}=0.947 atm$

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Question

Avogadro's Laws relate which of the following variables?

Answer

Avogadro's law states that volume is proportional to moles of gas. Temperature and pressure are fixed. Boyle law states that volume of gas is inversely proportional to pressure. Charles' law states that volume of gas is proportional to temperature.

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Question

Ideal gas law calculation

$CO(g)+2H_{2}(g)\rightarrow CH_{3}OH(g)$

The given equation shows the synthesis of methanol. Find the volume in liters of $H_{2}(g)$ required to synthesize of methanol at $395^{\circ}K$ and a pressure of .

Answer

Use the ideal gas equation

Rearrange to solve for the volume

$V_{H2}=\frac{nRT}{P}$

$103.6gCH_{3}OH(g)\frac{1molCH_{3}OH}{32.04gCH_{3}OH}=3.23mol CH_{3}OH\frac{2molH_{2}}{1molCH_{3}OH}=6.47molH_{2}$

$V_{H2}=\frac{6.47molH_{2}(.08206\frac{Latm}{molK})395K}{\frac{725mmHg}{760}}$

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Question

What is the volume occupied by of helium gas at a pressure of $0.67\text{atm}$ and a temperature of degrees celsius?

Answer

Recall the Ideal Gas Law:

Since the question is asking for volume, rearrange the equation to the following:

$V=\frac{nRT}{P}$

Also recall that for the Ideal Gas Law equation to work, the temperature must be given in degrees Kelvin, and we will also need the number of moles of gas.

Start by converting the mass of gas into moles:

$15.3g(\frac{1 \text{mole He}}{4.00g})=3.825\text{ moles of He}$

Don't round the values just yet.

Next, convert the temperature into degrees Kelvin.

Now, plug in all the given values into the equation to solve for volume:

$V=\frac{(3.825)(0.082057)(301.25)}{0.67}=141.545L$

The answer should have significant figures.

The volume is .

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Question

A cylinder contains of oxygen gas at a pressure of $4.10\text{atm}$ and a temperature of $405\text{K}$ . How many grams of gas are in the cylinder?

Answer

Recall the Ideal Gas Law:

$\text{PV}=nRT$

Because we want the amount of gas in the cylinder, we must find the number of moles of oxygen present first. Rearrange the Ideal Gas Law to solve for the number of moles of oxygen.

$n=\frac{PV}{RT}$

Plug in the given information to solve for the number of moles.

$n=\frac{(4.10\text{ atm})(29.1\text{L})}{(0.08206\frac{L\cdot atm}{mole\cdot K})(405\text{K})}=3.58997\text{ moles of O}_2$

Now, convert the number of moles to the number of grams by using the molar mass of oxygen gas.

$3.58997\text{ moles of O}_2(\frac{32.00\text{ g O}_2}{1\text{ mole O}_2})=114.88g$

Make sure that your answer has significant figures.

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Question

A gas fills a container, where heat is added raising its temperature from to and it increases its pressure from to . What is the gas’ new volume?

Answer

Use the combined gas law:

$\frac{P1T1}{V1}=\frac{P2T2}{V2}$

Solve for since that is the new volume for which we need to solve:

$V2=P2T2\frac{V1}{T1}P1$

$V2=\frac{1.7atm455K2.5L}{(310K1.1atm)}$

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Question

An ideal gas with a pressure of is expanded to times its original volume, what is the new pressure?

Answer

Use Boyle’s law:

You want to solve for the new pressure so isolate :

$P2=\frac{P1V1}{V2}$

It expands to its original volume, so

Plug in everything we know:

$P2=3\frac{V1}{(4*V1)}$

Cancel from both the numerator and denominator

$P2=\frac{3}{4}atm$

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Question

An ideal gas with a pressure of is compressed to $\frac{1}{12}$ of its original volume, what is the new pressure?

Answer

An ideal gas with a pressure of is is compressed to $\frac{1}{12}$ of its original volume, what is the new pressure?

Use Boyle’s law:

You want to solve for the new pressure so isolate :

$P2=\frac{P1V1}{V2}$

It compresses to $\frac{1}{12}$ its original volume so, $V2=\frac{V1}{12}$

Plug in everything we know:

$P2=\frac{6V1}{(\frac{1}{12}*V1)}$

Cancel from the numerator and denominator

$P2=\frac{6}{\frac{1}{12}}$

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Question

A sample of gas at a constant temperature has an initial pressure of $14.5\text{L}$ at a pressure of $720\text{mmHg}$ . If the volume of gas is decreased to $10.5\text{L}$ , what is its pressure?

Answer

Since we are given the volume and the pressure of this sample of gas, we will need to use Boyle's Law, which states that the pressure and volume of a gas, at a constant temperature, are inversely related. As thus, we can then write the following equation:

$\text{P}_1\text{V}_1=\text{P}_2\text{V}_2$

Since all the answer choices are in units of atmospheres, we will need to convert the given units into atmospheres.

$720\text{mmHg}\cdot\frac{1\text{ atm}}{760\text{mmHg}}=0.947\text{atm}$

Plug in the given pressures and volume into the equation, and solve for $\text{P}_2$ .

$(14.5)(0.947)=\text{P}_2(10.5)$

$\text{P}_2=1.31\text{atm}$

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Question

What is the pressure of a $5.20\text{L}$ cylinder filled with moles of helium gas at $317\text{K}$ ?

Answer

Recall the ideal gas law:

$\text{PV=nRT}$ ,

where $\text{P=Pressure}$ ,

$\text{V=Volume}$ ,

$\text{n=moles of gas}$ ,

$\text{R=Ideal gas constant}$ ,

and $\text{T=temperature}$ .

Since the question asks for pressure, rearrange the equation to solve for $\text{P}$ .

$\text{P}=\frac{\text{nRT}}{V}$

Plug in the given values. Remember that $\text{R=0.0821}\text{L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mole}^{-1}$

$\text{P}=\frac{(0.681)(0.0821)(317)}{5.20}=3.41\text{atm}$

Compare your answer with the correct one above

Question

Consider the following chemical reaction:

$6\text{Li}(s)+\text{N}_2(g)\rightarrow2\text{Li}_3\text{N}(s)$

How many grams of lithium is needed to react with $12.6\text{L}$ of nitrogen gas measured at $745\text{mmHg}$ and $28.5^\circ \text{C}$ ?

Answer

Start by finding the number of moles of nitrogen gas by using the ideal gas law:

$\text{PV}=\text{nRT}$

Rearrange the equation to solve for the variable $\text{n}$ :

$\text{n}=\frac{\text{PV}}{\text{RT}}$

In order to use the ideal gas law, the pressure must have units of atmospheres and the temperature must have units of Kelvin.

Convert the given pressure into atmospheres:

$745\text{mmHg}\cdot\frac{1\text{atm}}{760\text{mmHg}}=0.980\text{atm}$

$28.5\text{C}+273.15=301.65\text{K}$

Now, substitute in all the given values in order to find the number of moles of nitrogen gas.

$\text{n}=\frac{(0.980)(12.6)}{(0.0821)(301.65)}=0.499\text{ moles of nitrogen gas}$

Next, use the stoichiometric ratio given by the chemical equation to find the number of moles of lithium needed to react completely with the nitrogen gas.

$0.499\ \text{moles of N}_2\cdot\frac{6\text{ moles of Li}}{1\text{ mole of N}_2}=2.99\text{ moles of Li}$

Finally, convert the number of moles of lithium to number of grams of lithium.

$2.99\text{ moles of Li}\cdot\frac{6.94\text{g of Li}}{1\text{ mole of Li}}=20.8\text{g of Li}$

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Question

What is the pressure exerted by $0.50 mol N_{2}$ in a container at ?

Answer

$P=\frac{nRT}{V}$

$P=\frac{(0.50 mol N_{2}) (0.0821 L\cdot atm) (303K)}{13.0 L}=0.947 atm$

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Question

Avogadro's Laws relate which of the following variables?

Answer

Avogadro's law states that volume is proportional to moles of gas. Temperature and pressure are fixed. Boyle law states that volume of gas is inversely proportional to pressure. Charles' law states that volume of gas is proportional to temperature.

Compare your answer with the correct one above

Question

Ideal gas law calculation

$CO(g)+2H_{2}(g)\rightarrow CH_{3}OH(g)$

The given equation shows the synthesis of methanol. Find the volume in liters of $H_{2}(g)$ required to synthesize of methanol at $395^{\circ}K$ and a pressure of .

Answer

Use the ideal gas equation

Rearrange to solve for the volume

$V_{H2}=\frac{nRT}{P}$

$103.6gCH_{3}OH(g)\frac{1molCH_{3}OH}{32.04gCH_{3}OH}=3.23mol CH_{3}OH\frac{2molH_{2}}{1molCH_{3}OH}=6.47molH_{2}$

$V_{H2}=\frac{6.47molH_{2}(.08206\frac{Latm}{molK})395K}{\frac{725mmHg}{760}}$

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Question

What is the volume occupied by of helium gas at a pressure of $0.67\text{atm}$ and a temperature of degrees celsius?

Answer

Recall the Ideal Gas Law:

Since the question is asking for volume, rearrange the equation to the following:

$V=\frac{nRT}{P}$

Also recall that for the Ideal Gas Law equation to work, the temperature must be given in degrees Kelvin, and we will also need the number of moles of gas.

Start by converting the mass of gas into moles:

$15.3g(\frac{1 \text{mole He}}{4.00g})=3.825\text{ moles of He}$

Don't round the values just yet.

Next, convert the temperature into degrees Kelvin.

Now, plug in all the given values into the equation to solve for volume:

$V=\frac{(3.825)(0.082057)(301.25)}{0.67}=141.545L$

The answer should have significant figures.

The volume is .

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Question

A cylinder contains of oxygen gas at a pressure of $4.10\text{atm}$ and a temperature of $405\text{K}$ . How many grams of gas are in the cylinder?

Answer

Recall the Ideal Gas Law:

$\text{PV}=nRT$

Because we want the amount of gas in the cylinder, we must find the number of moles of oxygen present first. Rearrange the Ideal Gas Law to solve for the number of moles of oxygen.

$n=\frac{PV}{RT}$

Plug in the given information to solve for the number of moles.

$n=\frac{(4.10\text{ atm})(29.1\text{L})}{(0.08206\frac{L\cdot atm}{mole\cdot K})(405\text{K})}=3.58997\text{ moles of O}_2$

Now, convert the number of moles to the number of grams by using the molar mass of oxygen gas.

$3.58997\text{ moles of O}_2(\frac{32.00\text{ g O}_2}{1\text{ mole O}_2})=114.88g$

Make sure that your answer has significant figures.

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Question

A gas fills a container, where heat is added raising its temperature from to and it increases its pressure from to . What is the gas’ new volume?

Answer

Use the combined gas law:

$\frac{P1T1}{V1}=\frac{P2T2}{V2}$

Solve for since that is the new volume for which we need to solve:

$V2=P2T2\frac{V1}{T1}P1$

$V2=\frac{1.7atm455K2.5L}{(310K1.1atm)}$

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