Gas Law Practice

One rule governs this whole page and there are no exceptions to it: every temperature is in kelvins, converted the moment you write it down. $T(\text{K})=T{(}^{\circ }\text{C})+273.15$. The gas laws are proportionalities, and a proportionality needs a scale whose zero means none of the quantity. Celsius zero is the freezing point of water, which is a fact about water and not about gases.

The relationships you need:

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}PV=nRT$

with $R=8.314\text{ kPa}\cdot \text{L}/(\text{mol}\cdot \text{K})$ when pressure is in kilopascals and volume in litres.

Conditions used on this page, because a molar volume quoted without its conditions is not a number:

Name	Temperature	Pressure	Molar volume of an ideal gas
STP	0 °C, which is 273.15 K	101.325 kPa	22.4 L/mol
SATP	25 °C, which is 298.15 K	100 kPa	24.8 L/mol

Some textbooks define STP with a pressure of 100 kPa instead, where the molar volume works out to 22.7 L/mol. Check which convention a question is using before you take a molar volume from memory.

1. A gas occupies 250.0 mL at 101.3 kPa. What volume does it occupy at 152.0 kPa, at constant temperature?

Answer 1

Temperature is constant, so it cancels from both sides of the combined law and what is left is Boyle’s law, $P_1{V}_1=P_2{V}_2$.

$V_2=\frac{P_1{V}_1}{P_2}=\frac{(101.3\text{ kPa})(250.0\text{ mL})}{152.0\text{ kPa}}=166.6\text{ mL}$

166.6 mL.

The volumes may stay in millilitres because they appear on both sides and cancel. Temperature never gets that privilege, because it appears as a ratio of absolute values.

Sanity check first, every time: the pressure went up, so the volume must come down. If your answer had been larger than 250.0 mL you inverted the fraction, and the check takes two seconds against the several minutes of a redone question.

2. A balloon has a volume of 2.00 L at 20.0 °C. What is its volume at 80.0 °C, at constant pressure? What answer would you get by using degrees Celsius directly, and why is it wrong?

Answer 2

Convert both temperatures before anything else:

$T_1=20.0+273.15=293.15\text{ K}{T}_2=80.0+273.15=353.15\text{ K}$

Pressure is constant, so $\frac{V_1}{T_1}=\frac{V_2}{T_2}$:

$V_2=V_1\times \frac{T_2}{T_1}=(2.00\text{ L})\times \frac{353.15}{293.15}=2.409\text{ L}$

2.41 L.

Using Celsius you would compute $2.00\times \frac{80.0}{20.0}=8.00$ L — four times the original volume rather than a twenty per cent increase. The answer is wrong by a factor of more than three, and nothing about 8.00 L looks obviously absurd, which is what makes this error dangerous.

The reason it fails: the ratio $\frac{80.0}{20.0}$ claims that 80 °C is “four times as hot” as 20 °C. It is not four times anything. On the absolute scale the two temperatures differ by only twenty per cent, and twenty per cent is what the balloon does. Had the first temperature been 0 °C, the Celsius method would have asked you to divide by zero, which is the scale telling you rather loudly that it is not the right one for ratios.

3. A gas occupies 500.0 mL at 25.0 °C and 98.0 kPa. What volume would it occupy at STP?

Answer 3

Both pressure and temperature change, so this is the combined law. Rearranged for $V_2$:

$V_2=V_1\times \frac{P_1}{P_2}\times \frac{T_2}{T_1}$

$T_1=298.15\text{ K}$, and STP is $T_2=273.15$ K and $P_2=101.325$ kPa.

$\begin{array}{rl}{V}_2& =(500.0\text{ mL})\times \frac{98.0}{101.325}\times \frac{273.15}{298.15}\\ & =(500.0)(0.96718)(0.91615)=443.0\text{ mL}\end{array}$

443 mL, to three significant figures — the pressure was given to three, and that is the ceiling.

Both changes push the same way here, which is a useful check. Raising the pressure squeezes the gas smaller, and lowering the temperature shrinks it further, so the answer had to come out below 500.0 mL. If only one factor had gone that way you would want to look at which effect was larger before trusting the sign of the change.

4. What mass of oxygen is contained in a 5.00 L cylinder at 25.0 °C and 850. kPa?

Answer 4

This one needs the amount of gas, not just a before and after, so it is the ideal gas law.

$T=25.0+273.15=298.15\text{ K}$

$n=\frac{PV}{RT}=\frac{(850.\text{ kPa})(5.00\text{ L})}{(8.314)(298.15)}=\frac{4250}{2478.8}=1.7146\text{ mol}$

Oxygen as a gas is $\mathrm{O}{}_2$, so $M=32.00$ g/mol:

$m=nM=(1.7146\text{ mol})(32.00\text{ g/mol})=54.87\text{ g}$

54.9 g of oxygen.

The units are the whole reason this works. Kilopascals times litres divided by kilopascal-litres per mole-kelvin, times kelvins, leaves moles. Use a pressure in atmospheres with this value of $R$ and the answer is out by a factor of about a hundred — so check that your pressure unit matches your gas constant before you press equals.

Writing the pressure as “850.” with the decimal point is deliberate: it says three significant figures rather than two.

5. What volume of hydrogen, measured at STP, is produced when 0.500 g of magnesium reacts with excess dilute hydrochloric acid?

Answer 5

$\mathrm{Mg}+2\mathrm{HCl}\to \mathrm{MgCl}{}_2+\mathrm{H}{}_2$

$n(\mathrm{Mg})=\frac{0.500\text{ g}}{24.31\text{ g/mol}}=0.020568\text{ mol}$

The mole ratio of magnesium to hydrogen is 1 to 1, so $n(\mathrm{H}{}_2)=0.020568$ mol. At STP the molar volume is 22.4 L/mol:

$V=(0.020568\text{ mol})(22.4\text{ L/mol})=0.4607\text{ L}$

0.461 L, which is 461 mL.

Half a gram of ribbon produces the better part of half a litre of gas. That is the reason gas volumes are a convenient thing to measure in a school lab — the quantity is large and easy to read, where the corresponding mass of hydrogen is only about 0.0415 g and sits right at the edge of a school balance’s resolution. This calculation is exactly what you are doing in Measuring a Gas Law, run backwards.

And note the phrase “measured at STP” doing real work. Without it the question has no answer, because the same amount of gas occupies different volumes under different conditions.

6. Propane burns: $\mathrm{C}{}_3\mathrm{H}{}_8+5\mathrm{O}{}_2\to 3\mathrm{CO}{}_2+4\mathrm{H}{}_2\mathrm{O}$. At SATP, what volume of oxygen is needed to burn 10.0 L of propane, and what volume of carbon dioxide is produced? What about the water?

Answer 6

No molar masses and no moles are needed, and seeing why is the point of the question.

At a fixed temperature and pressure, equal volumes of any gas contain equal numbers of particles — that falls straight out of $PV=nRT$, since $V$ and $n$ are proportional when $P$, $T$, and $R$ are all fixed. So for gases at the same conditions, the coefficient ratio is a volume ratio directly.

$\begin{array}{rl}V(\mathrm{O}{}_2)& =10.0\text{ L}\times \frac{5}{1}=50.0\text{ L}\\ V(\mathrm{CO}{}_2)& =10.0\text{ L}\times \frac{3}{1}=30.0\text{ L}\end{array}$

50.0 L of oxygen, producing 30.0 L of carbon dioxide.

The water is the trap. The shortcut only works for gases, and at SATP — 25 °C — water is a liquid. So the answer is not 40.0 L. The 4 moles of water per mole of propane would occupy about 40 L as a vapour at these conditions and instead occupy roughly 40 millilitres as a liquid, a thousandfold difference.

This is not a technicality invented for exams. It is why a cold car exhaust drips water on a driveway and a hot one does not, and it is why the volume of exhaust gas leaving an engine depends on how hot the pipe is.

7. A 0.250 L flask holds 0.349 g of an unknown gas at 100.0 °C and 98.5 kPa. Find its molar mass, and suggest what the gas might be.

Answer 7

$T=100.0+273.15=373.15\text{ K}$

$n=\frac{PV}{RT}=\frac{(98.5)(0.250)}{(8.314)(373.15)}=\frac{24.625}{3102.4}=7.9375\times 10^{-3}\text{ mol}$

$M=\frac{m}{n}=\frac{0.349\text{ g}}{7.9375\times 10^{-3}\text{ mol}}=43.97\text{ g/mol}$

44.0 g/mol, to three significant figures.

What it might be. Carbon dioxide has a molar mass of 44.01 g/mol, which is an excellent match. But be careful how you say it: dinitrogen monoxide is 44.02 g/mol and propane is 44.11 g/mol, and this measurement cannot separate any of the three. The honest conclusion is “a molar mass of 44.0 g/mol, consistent with carbon dioxide among others”, and identifying it would take a chemical test rather than a better balance.

This is how molar masses of gases were determined before instruments, and it is what you are doing from the other end in Measuring a Gas Law — there you know the substance and measure the molar volume; here you assume the molar volume behaviour and measure the substance.

8. Three claims from a study group. Correct each. (a) “Going from 25 °C to 50 °C doubles the Celsius temperature, so at constant pressure the volume doubles.” (b) “Pressure and volume are directly proportional — squeezing a gas raises its pressure, so they go up together.” (c) “One mole of any substance occupies 22.4 L at STP, so one mole of water occupies 22.4 L.”

Answer 8

(a) Celsius ratios are meaningless. In kelvins the change is from 298.15 K to 323.15 K:

$\frac{323.15}{298.15}=1.0839$

so the volume increases by about 8.4%, not by 100%. The student’s method overstates the change by more than a factor of ten.

The reason is that the Celsius scale has an offset zero. A ratio only means something on a scale where zero corresponds to none of the quantity, and the only such scale for temperature is the absolute one.

(b) Inversely proportional, and the student’s own evidence says so. Squeezing a gas means reducing its volume, and the pressure rises — so one goes up while the other goes down, which is the definition of inverse. Boyle’s law is $P_1{V}_1=P_2{V}_2$, meaning the product stays constant, not the ratio.

A quick test whenever you are unsure which way a relationship runs: plot it in your head. A directly proportional relationship gives a straight line through the origin; an inverse one gives a curve approaching both axes. A gas squeezed to half its volume doubles its pressure, and squeezed to a tenth it goes to ten times — that curve, not that line.

(c) The molar volume applies to gases, and only to gases. At STP, 0 °C, water is a liquid — indeed it is at or below its freezing point. One mole of water is 18.02 g, and as a liquid that occupies about 18 mL, which is roughly a thousand times smaller than 22.4 L.

The reason 22.4 L/mol works for any gas is that the volume of a gas is overwhelmingly empty space, so the size of the individual particles barely matters. In a liquid or a solid the particles are in contact, so the volume depends entirely on what the substance is, and there is no universal molar volume to quote.

The general lesson: check the state of every substance before applying a gas law to it. Half of the difficult questions in this unit are difficult only because one of the substances is not a gas.

Reference: The Gas Laws and Gases and the Atmosphere. Measuring these relationships yourself: Measuring a Gas Law.

Curriculum connection

F2.1

use appropriate terminology related to gases and atmospheric chemistry, including, but not limited to: standard temperature, standard pressure, molar volume, and ideal gas [C]

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F2.3

solve quantitative problems by performing calculations based on Boyle’s law, Charles’s law, Gay-Lussac’s law, the combined gas law, Dalton’s law of partial pressures, and the ideal gas law [AI]

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F2.4

use stoichiometry to solve problems related to chemical reactions involving gases (e.g., problems involving moles, number of atoms, number of molecules, mass, and volume) [AI]

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