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College Chemistry : Solutions, States of Matter, and Thermochemistry

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Example Questions

Example Question #1 : Non Ideal Gas Behavior

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

Possible Answers:

1.00 atm

0.732 atm

1.37 atm

1.73 atm

0.841 atm

Correct answer:

1.37 atm

Explanation:

The pressure of the flask at $100^{\circ}C$ is 1.37 atm .

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

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 1.00 atm and 273.15 K .

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

$P_{2}= 1.37 atm$

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Example Question #4 : Gases

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.

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

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

Possible Answers:

$6.42\textrm{ atm}$

$7.47\textrm{ atm}$

$7.43\textrm{ atm}$

$6.90\textrm{ atm}$

$3.45\textrm{ atm}$

Correct answer:

$6.90\textrm{ atm}$

Explanation:

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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Example Question #1 : Gas Laws

What is the pressure of a $5.20\text{L}$ cylinder filled with 0.681 moles of helium gas at $317\text{K}$ ?

Possible Answers:

$\text{2.22atm}$

$\text{3.50atm}$

$\text{2.19atm}$

$\text{3.41atm}$

Correct answer:

$\text{3.41atm}$

Explanation:

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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Example Question #2 : Gas Laws

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?

Possible Answers:

$1.01\text{atm}$

$1.31\text{atm}$

$1.52\text{atm}$

$0.862\text{atm}$

Correct answer:

$1.31\text{atm}$

Explanation:

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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Example Question #1 : Gas Laws

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 $28.5^\circ \text{C}$ ?

Possible Answers:

22.4

15.7

20.8

19.2

Correct answer:

20.8

Explanation:

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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Example Question #4 : Gas Laws

What is the pressure exerted by $0.50 mol N_{2}$ in a 13.0 L container at 300 K ?

Possible Answers:

0.947 atm

3.79 atm

0.085 atm

11.5 atm

1.78 atm

Correct answer:

0.947 atm

Explanation:

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

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

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Example Question #1 : Gas Laws

Avogadro's Laws relate which of the following variables?

Possible Answers:

Volume is proportional to temperature

Volume is inversely proportional to pressure

Volume is proportional to moles of gas

Volume is inversely proportional to temperature

Temperature is proportional to moles of gas

Correct answer:

Volume is proportional to moles of gas

Explanation:

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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Example Question #1 : Gas Laws

What is the volume occupied by 15.3g of helium gas at a pressure of $0.67\text{atm}$ and a temperature of 29.0 degrees celsius?

Possible Answers:

140L

130L

150L

120L

Correct answer:

140L

Explanation:

Recall the Ideal Gas Law:

PV=nRT

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.

29.0+273.15=301.25K

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 140L .

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Example Question #2 : Gas Laws

A cylinder contains 29.1L 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?

Possible Answers:

121

102

115

Correct answer:

115

Explanation:

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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Example Question #42 : Solutions, States Of Matter, And Thermochemistry

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 103.6g of methanol at $395^{\circ}K$ and a pressure of 725mmHg .

Possible Answers:

$V_{H_{2}}=65L$

$V_{H_{2}}=219.8L$

$V_{H_{2}}=.289L$

$V_{H_{2}}=.0045L$

Correct answer:

$V_{H_{2}}=219.8L$

Explanation:

Use the ideal gas equation

PV=nRT

Rearrange to solve for the volume

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

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

$V_{H2}=\frac{6.47molH_{2}(.08206\frac{L*atm}{mol*K})395K}{\frac{725mmHg}{760}}$

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College Chemistry : Solutions, States of Matter, and Thermochemistry

Study concepts, example questions & explanations for College Chemistry

All College Chemistry Resources

Example Questions

Example Question #1 : Non Ideal Gas Behavior

Example Question #4 : Gases

Example Question #1 : Gas Laws

Example Question #2 : Gas Laws

Example Question #1 : Gas Laws

Example Question #4 : Gas Laws

Example Question #1 : Gas Laws

Example Question #1 : Gas Laws

Example Question #2 : Gas Laws

Example Question #42 : Solutions, States Of Matter, And Thermochemistry

All College Chemistry Resources

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