Gases - College Chemistry

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Question

Per Graham's law of effusion, how does the molar mass relate to both the rate and time of effusion?

Answer

$\frac{Rate : 1}{Rate : 2} = \sqrt{\frac{Molar: Mass, , 2}{Molar: Mass: 1} }$

This equation explicitly shows how the rate of effusion is inversely proportional to the molar mass of a gas in a gaseous solution.

Because $Rate = \frac{Amount , , , of , , Substance}{Time}$ , time relates to molar mass by:

$\frac{\frac{Amount , , of , , Substance , , 1}{Time: 1}}{\frac{Amount, , of , , Subtance, , 2}{Time : 2}}=\sqrt{\frac{Molar, , Mass, , 2}{Molar, , Mass, , 1}}$

Simplifying this equation, we see that:

$\frac{Time : 2}{Time: 1}=\sqrt{\frac{Molar, , Mass, , 2}{Molar, , Mass, , 1}}$

As a result, time relates directly to molar mass

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Question

At the same temperature, an unknown gas effuses at a rate that is times that of oxygen. Find the molar mass, in grams per mole, of the unknown gas.

Answer

Recall Graham's Law of Effusion for two gases, A and B:

$\frac{\text{Rate}_A}{\text{Rate}_B}=\sqrt{\frac{\text{Molar Mass of B}}{\text{Molar mass of A}}}$

From the equation, we know the following:

$\frac{\text{Rate}_\text{unknown}}{\text{Rate}_\text{oxygen}}=\sqrt{\frac{\text{Molar Mass of oxygen}}{\text{Molar mass of uknown}}}=0.291$

Thus, we can solve for the molar mass of the unknown gas. Let be the molar mass of the unknown gas.

$\sqrt{\frac{32.00}{x}}=0.291$

$\frac{32.00}{x}=0.084681$

Make sure that your answer has significant figures.

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Question

Which of the following is a true statement with regards to the relative effusion rates of oxygen and carbon dioxide?

Answer

We're being asked to compare the effusion rates of oxygen and carbon dioxide.

Remember that effusion is the spontaneous movement of a gas through a small hole from one area to another. It's worth noting that at a given temperature, the average speed of all gas molecules in a system is used to calculate the average kinetic energy of the gas particles. This dependence of kinetic energy on temperature means that at a given temperature, any gas particle will have the same kinetic energy.

In this case, we can say that the kinetic energy of oxygen molecules in one system is equal to the kinetic energy of carbon dioxide molecules in another system. Furthermore, since mass is inversely proportional to velocity, identical kinetic energies would mean that as the mass of the gas particles in a system decreases, their velocity (and thus, effusion rates) would increase.

We can use this information to solve for the relative effusion rates between oxygen and carbon dioxide. By setting their kinetic energies equal to each other, we can derive an expression that relates their relative speeds to their relative masses.

$KE_{O_{2}}=KE_{CO_{2}}$

$\frac{1}{2}m_{O_{2}}v^{2}_{O_{2}}=\frac{1}{2}m_{CO_{2}}v^{2}_{CO_{2}}$

$m_{O_{2}}v^{2}_{O_{2}}=m_{CO_{2}}v^{2}_{CO_{2}}$

$\frac{v^{2}_{O_{2}}}{v^{2}_{CO_{2}}}=\frac{m_{CO_{2}}}{m_{O_{2}}}$

$\frac{v_{O_{2}}}{v_{CO_{2}}}=\sqrt{\frac{m_{CO_{2}}}{m_{O_{2}}}}$

Generally speaking, this expression shows how the velocity of any two gasses depends on their mass. In this case, the gasses are oxygen and carbon dioxide.

We can use the periodic table of the elements to find out the mass of each gas, and use that information to calculate the relative effusion rates.

$v_{O_{2}}=v_{CO_{2}} \sqrt{\frac{m_{CO_{2}}}{m_{O_{2}}}}$

$v_{O_{2}}=v_{CO_{2}} \sqrt{\frac{44:\frac{grams}{mol}}{32:\frac{grams}{mol}}}=1.17v_{CO_{2}}$

This shows that oxygen will effuse at a rate that is about faster than carbon dioxide.

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Question

A sample of Ne(g) effusses through a tiny hole in 60.7 s. An unknown gas, under identical conditions, effusses in 45.6 s.

What is the molar mass of the unknown gas?

Answer

To solve this problem use Graham's Law of Effusion

$\frac{\textrm{rate of effusion of A}}{\textrm{rate of effusion of B}}=\sqrt{\frac{M_{\textrm{B}}}{M_{\textrm{A}}}}$

$\textrm{A}=\textrm{Ne}$

$\textrm{B}=\textrm{unknown gas}$

By plugging in the values we can rewrite the equation as

$\frac{60.7\textrm{ s}}{45.6\textrm{ s}}=\sqrt{\frac{M_{\textrm{unknown}}}{20.2\frac{\textrm{g}}{\textrm{mol}}}}$

$(\frac{60.7\textrm{ s}}{45.6\textrm{ s}})^{2}=\frac{M_{\textrm{unknown}}}{20.2\frac{\textrm{g}}{\textrm{mol}}}$

$M_{\textrm{unknown}}=35.8\frac{\textrm{g}}{\textrm{mol}}$

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Question

Suppose that gas A effuses at a rate that is twice that of gas B. If the mass of gas A is halved and the mass of gas B is doubled, which of the following correctly describes the new relative effusion rates of these two gasses?

Answer

For this question, we're given the relative effusion rates for two gasses. We're then told how the mass of each of these gasses is changed, and then we're asked to determine the new relative effusion rates of the two gasses.

First, we can recall the expression that describes the dependence of the effusion rates of two gasses on their mass. Since we're told that the rate of gas A is twice that of gas B, we can write the following expression.

$\frac{Rate_{A}}{Rate_{B}}=\sqrt\frac{Mass_{B}}{Mass_{A}}=2$

Furthermore, since we're told that the mass of gas B is doubled and the mass of gas A is halved, we can determine how the rate will change.

$\sqrt\frac{2}{0.5}=\sqrt{4}=2$

Thus, we can see that the rate will change by a factor of two. Hence, the new rate will be $2\cdot2=4$ . Thus, gas A will now effuse at a rate times that of gas B.

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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

One flask is at STP and another is at $100^{\circ}C$ . What is the pressure at $100^{\circ}C$ ?

Answer

The pressure of the flask at $100^{\circ}C$ is .

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

Because the volume between the flasks and the moles in each flask are constant, we can cancel out and .

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

At STP, conditions are and .

$\frac{1.00 atm}{273.15K}=\frac{P_{2}}{373.15K}$

$P_{2}= 1.37 atm$

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Question

A 3.00 L container at 273 K is filled with 1.00 mol Cl2(g), which behaves non-ideally.

Using the van der Waals equation, calculate pressure exerted by the gas.

$a=6.49\frac{\textrm{L}^{2}\textrm{atm}}{\textrm{mol}^{2}}$

$b=0.0562\frac{\textrm{L}}{\textrm{mol}}$

Answer

Recall the van der Waals equation for non-ideal gases

$\textrm{(P + }\frac{n^{2}a}{\textrm{V}})(\textrm{V - }nb)=nR\textrm{T}$

Rewrite the equation using the known values and solve for P

$[\textrm{P + }\frac{(1\textrm{ mol})^{2}(6.49\frac{\textrm{L}^{2}\textrm{ atm}}{\textrm{mol}^{2}})}{(3.00\textrm{ L})^2}][\textrm{3.00 L - (1 mol)}(0.0562\frac{\textrm{L}}{\textrm{mol}})]=(1\textrm{ mol})^{2}(0.0821\frac{\textrm{atm}}{\textrm{mol K}})(273K)$

$(\textrm{P + 0.721 atm})(2.94\textrm{ L})=22.4\textrm{ atm L}$

$\textrm{P = 6.90 atm}$

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Question

A mixture of helium, nitrogen, and neon has a total pressure of $815\text{ mmHg}$ at a temperature of $255\text{K}$ . If the partial pressure of helium is $201\text{ mmHg}$ and the partial pressure of nitrogen is $351\text{ mmHg}$ , what mass of neon is present in the mixture?

Answer

Since we have the total pressure and the pressures of helium and nitrogen, we can find the pressure of neon in this container.

$\text{P}_{tot}=\text{P}_{He}+\text{P}_{N}_2+\text{P}_{Ne}$

$\text{P}_{tot}-\text{P}_{He}-\text{P}_{N}_2=\text{P}_{Ne}$

$\text{P}_{Ne}=815-201-351=263\text{ mmHg}$

Next, convert this pressure into atmospheres.

$263\text{ mmHg}(\frac{1\text{ atm}}{760\text{mmHg}})=0.34605\text{ atm}$

Now, recall the Ideal Gas Law:

Rearrange the equation to solve for moles of gas:

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

Using the pressure of neon, find the moles of neon that is present in the container.

$n=\frac{(0.34605\text{ atm})(2.00L)}{0.08206 \frac{L \cdot atm}{mol \cdot K}(255K)}=0.03307\text{moles of Ne}$

Finally, convert the moles of neon into grams of neon using the molar mass.

$0.3307\text{ moles of Ne}(\frac{20.18\text{ g}}{1\text{ mole of Ne}})=0.667\text{ g}$

Make sure your answer has significant digits.

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Question

Suppose that hydrochloric acid and aluminum are reacted to produce hydrogen gas. This hydrogen gas is collected by displacement of water at $25^{o}C$ and a total pressure of . If the volume of the gas collected is , then how many moles of hydrogen were collected?

Note: The vapor pressure of water at $25^{o}C$ is .

Answer

For this question, we're told that a chemical reaction is producing hydrogen gas. In doing so, the gas is displacing a certain amount of water vapor. We're asked to determine the number of moles of gas produced in this process.

The amount of water displaced by the production of hydrogen gas will be the amount (volume) of hydrogen gas produced. To measure its pressure, we take the reading of the total pressure and then subtract from it the pressure of water vapor (since it contributes to the total pressure). Using this information, we can use the ideal gas equation to determine the number of moles of hydrogen gas.

$P_{total}=P_{water}+P_{H_{2}}$

$P_{H_{2}}=P_{total}-P_{water}$

$P_{H_{2}}=700:torr-23.8:torr=676.2:torr$

$676.2:torr\cdot \frac{1:atm}{760:torr}=0.890:atm$

Next, we can use the ideal gas equation to solve for our answer.

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

$n=\frac{(0.890:atm)(400\cdot 10^{-3}:L)}{(0.08206:\frac{L\cdot atm}{mol\cdot K})(298.15:K)}$

$n=1.46\cdot10^{-2}:moles\cdot\frac{1000:mmoles}{1:mole}=14.6:mmoles$

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