$PV = nRT$

The ideal gas law has been developed from both an empirical approach and by building up a picture of gas particles from first principles in physics – the statistical mechanics approach.

It is extremely useful, but it does have limitations, which we'll discuss at the end of this section. Gases aren't "ideal," but many are ideal enough under the right conditions that the ideal gas law gives us very reliable results.

In developing the ideal gas law, we make the following assumptions that you should bear in mind while using it:

1. Small particles – Gases consist of spherical particles so small the distance between gas particles (atoms or molecules) is always much greater than the size of the particle.

Units and the molar gas law

When using the molar gas law, $PV = nRT$, you have some choices of units. The SI unit of volume is the cubic meter ($m^3$), but that unit can be cumbersome, and the use of liters as a unit of gas volume is often favored. Likewise, the SI unit of pressure is the Pascal ($Pa$), but the atmosphere ($1 \text{ atm} = 101,325\; Pa$) is used more frequently in some fields

You can adapt to the set of units you'd like to use just by changing the gas constant. Here are the constants and the units of pressure, temperature and volume that go with them. While, when using gas laws like Charles' law and the Gay-Lussac law, it's OK to use Celsius temperatures (because they appear in ratios where the units cancel), it's important to use Kelvin temperatures in the ideal gas law.

Examples – using the ideal gas law

Example 1

How many moles of gas must be present in a gas cylinder of volume 0.01 m³ pressurized to 200 KPa at a temperature of 25˚C?

Solution: Our units are mostly SI, so we should use the gas constant $R = 8.314 J\cdot mol^{-1}K^{-1}$, and convert the temperature to Kelvin: $T = (273 + 25) = 298 \; K$.

Now we rearrange the ideal gas law to solve for the number of moles. Do the rearrangement first so that we can plug numbers and units into the result an know we're on the right track:

$$PV = nRT \: \color{#E90F89}{\longrightarrow} \: n = \frac{PV}{RT}$$

The result is:

$$ \require{cancel} \begin{align} n &= \frac{(200,000 \, Pa)(0.01 \, m^3)}{(8.314 \, J \cdot mol^{-1}\cancel{K^{-1}})(298 \, \cancel{K})} \\[5pt] &= 0.807 \; \text{moles} \end{align}$$

Example 2

Calculate the temperature (in ˚C) of 3.5 moles of a gas at a pressure of 9.3 atm contained in a 10.0 liter cylinder.

Solution: We want temperature, so we need to first rearrange the ideal gas law to solve for temperature:

$$PV = nRT \; \color{#E90F89}{\longrightarrow} \; T = \frac{PV}{nR}$$

Now our units of pressure and volume are atmospheres and liters, so using $R = 0.0821 \; L \cdot atm \cdot mol^{-1}K^{-1}$ is the best way to go.

Example 3

A 1-liter water bottle is emptied and sealed on the top of a 14,000 ft. peak, at a temperature of 15˚C. At 14,000 ft., the air pressure is about 0.59 atm. (a) Calculate the number of moles of air (assume that "air" is an ideal gas). At sea level, the bottle is observed to have collapsed. This collapse is due to the difference between the outside air pressure at sea level (P = 1 atm.) and the low pressure inside the bottle. (b) Using this information, calculate the volume of the bottle at sea level.

Solution: For part (a), we'll rearrange $PV = nRT$ to find the number of moles:

$$PV = nRT \: \color{#E90F89}{\longrightarrow} \: n = \frac{PV}{nR}$$

Plugging in the information,

$$n = \frac{(0.59 \, \cancel{atm})(1 \, \cancel{L})}{(0.0821 \, \cancel{L} \, \cancel{atm} \, mol^{-1} \, \cancel{K^{-1}})(288 \, \cancel{K})}$$

gives

$$n = 0.025 \; \text{moles}$$

Now for part (b) we want the volume from PV = nRT:

$$PV = nRT \: \color{#E90F89}{\longrightarrow} \: V = \frac{nRT}{P}$$

Plugging in the new pressure and the rest of the information gives us:

$$ \begin{align} V &= \frac{0.025 \cancel{mol}\left( 0.0821 \, \frac{L \cdot \cancel{atm}}{\cancel{mol} \cdot \cancel{K}}\right)(288 \, \cancel{K})}{1 \, \cancel{atm}} \\[5pt] &= 0.59 \, L \end{align}$$

Example 4

What volume does one mole of gas occupy at STP?

Solution: As usual, we'll rearrange the ideal gas law to solve for volume:

$$PV = nRT \: \color{#E90F89}{\longrightarrow} \: V = \frac{nRT}{V}$$

Plugging in n = 1 mole, T = 273.15 K and P = 1 atm (and the appropriate gas constant, R = 0.0821 L·atm/mol·K), we get

Practice problems

1. To what volume must 2.5 moles of nitrogen gas (N₂) be compressed in order to achieve a final pressure of 500 KPa at T = 300 K ?

   Solution

   First let's convert the pressure from KPa to atmospheres:

   $$500 \, \cancel{KPa} \left( \frac{1 \, atm}{101.325 \, \cancel{KPa}} \right) = 4.935 \, atm$$

   Now we can use the ideal gas law, rearranged to solve for volume:

   $$PV = nRT \: \color{#E90F89}{\longrightarrow} \: V = \frac{nRT}{V}$$

   $$ \begin{align} V &= \frac{(2.5 \, \cancel{mol}) \left(0.0821 \frac{L \cancel{atm}}{\cancel{mol} \cancel{K}} \right)(300 \cancel{K})}{4.935 \cancel{atm}} \\[5pt] &= \bf{12.5 \, L} \end{align}$$

   The gas will have to be compressed to a volume of 12.5 L




2. The circumference of a basketball is 30 inches and it is inflated to a pressure of 8 pounds per square inch (psi) at room temperature (298 K). How much air (in moles) is in a properly inflated basketball ? [Hint: 1 atm. = 14.7 psi, and 1 inch = 2.54 cm].

   Solution

   First let's convert units so we're working in liters and atmospheres. The volume of a sphere is $V = \frac{4}{3} \pi r^3,$ and the circumference is $c = 2 \pi r,$, so the volume as a function of circumference is

   $$V = \frac{4}{3} \pi r^3 = \frac{4 \pi}{3} \left( \frac{c}{2 \pi} \right)^3 = \frac{c^3}{6 \pi^2}$$

   Now convert the circumference to dm because 1 L = 1 dm³:

   $$30 \cancel{in} \left( \frac{2.54 \cancel{cm}}{1 \cancel{in}} \right) \left( \frac{1 \, dm}{10 \cancel{cm}} \right) = 7.62 \, dm$$

   Now the volume is

   $$V = \frac{c^3}{6 \pi^2} = 7.47 \, L$$

   Next we need the pressure in atmospheres:

   $$8 \cancel{psi} \left( \frac{1 \, atm}{14.7 \cancel{psi}} \right) = 0.544 \, atm$$

   Now we can calculate the number of moles of air:

   $$ \begin{align} n &= \frac{PV}{RT}\\[5pt] &= \frac{(0.544 \cancel{atm})(7.47 \cancel{L})}{(0.0821 \, \frac{\cancel{L} \cancel{atm}}{mol \, \cancel{K}})(298 \cancel{K})}\\[5pt] &= \bf{0.166 \, mol} \end{align}$$

   There will be about 0.17 moles of gas inside the basketball.




3. Calculate the pressure, in Pa, of 25 moles of gas compressed into a volume of 1 liter at T = 298 K.

   Solution

   Because we want the pressure in Pascals, one way to approach the problem is to use $R = 8.314 \, Pa \, m^3/mol \cdot K.$ Then we'll have to convert the volume, 1 L, to m³, which is just 0.001 m³. So we have

   $$ \begin{align} P &= \frac{nRT}{V}\\[5pt] &= \frac{(25 \cancel{mol})(8.314 \, \frac{Pa \cancel{m^3}}{\cancel{mol} \cancel{K}})(298 \cancel{K})}{0.001 \cancel{m^3}}\\[5pt] &= \bf{61,931,300 \, Pa}\\[5pt] &\approx 611 \, atm \end{align}$$




4. Two gas cylinders are connected by a valve, as shown. The volume of cylinder 1 is 24 L, and the volume of cylinder 2 is 18 L. If cylinder A contains 1.1 moles of gas at T = 300K and valve A is opened: (a) Calculate the pressure of the two cylinders together after their temperature is stabilized at 300K, and (b) What will happen if valve B is opened?

   

   Solution

   We'll first want to calculate the pressure in tank 1:

   $$ \begin{align} P &= \frac{nRT}{V} = \frac{(0.0821 \frac{\cancel{L} \, atm}{\cancel{mol} \cancel{K}})(1.1 \cancel{mol})(300 \cancel{K})}{24 \cancel{L}} \\[5pt] &= 1.1289 \; \text{atm} \end{align}$$

   Now because the temperature and number of moles will remain the same when valve A is opened, and R is a constant, we can write $P_1 V_1 = P_2 V_2,$ and solve for P₂:

   $$ \begin{align} P_2 &= \frac{P_1 V_1}{V_2}\\[5pt] &= \frac{(1.1289 \, atm)(24 \cancel{L})}{(24 + 18) \cancel{L}}\\[5pt] &= \bf{0.6451 \, atm}, \end{align}$$

   where the final volume, V₂ is the sum of the volumes of the two tanks.

   If valve B is opened and we assume that the outside pressure is 1 atm., then air from outside will flow in to equalize the pressures. Eventually the gases inside and outside will mix, though this can take longer than you might think.




5. Two gas cylinders are connected by a valve, as in the drawing above. They are maintained at a temperature of 30˚C. The first cylinder, which has a volume of 24 L, contains 12 moles of argon (Ar) gas. The second contains 12 moles of Neon (Ne) gas. Calculate the pressure of each gas in its own cylinder with valve A closed, then calculate the pressure in both cylinders after valve A is opened.

   Solution

   First, recognize that 30˚C is 303 K. Then we can calculate the individual pressures, call them P₁ and P₂:

   $$ \begin{align} P_1 &= \frac{nRT}{V} = \frac{(0.0821 \frac{\cancel{L} \, atm}{\cancel{mol} \cancel{K}})(12 \cancel{mol})(303 \cancel{K})}{24 \cancel{L}} \\[5pt] &= 12.44 \; \text{atm} \\[5pt] P_2 &= \frac{(0.0821 \frac{\cancel{L} \, atm}{\cancel{mol} \cancel{K}})(12 \cancel{mol})(303 \cancel{K})}{18 \cancel{L}} \\[5pt] &= 16.58 \; \text{atm} \end{align}$$

   Now to calculate the combined pressure, we understand that we're treating each of these gases as an ideal gas (a good approximation for inert gases), so we can just add the number of moles and sum the volumes:

   $$ \begin{align} P_{total} &= \frac{nRT}{V} = \frac{(0.0821 \frac{\cancel{L{} \, atm}{\cancel{mol} \cancel{K}})(12 \cancel{mol})(303 \cancel{K})}{(24 + 18) \cancel{L} \\[5pt] &= \bf{14.22 \, atm}, \end{align}$$

   It makes sense that the final pressure can't be larger than that of the high-pressure tank, and it has to be greater than the low pressure tank.




6. A commercial cylinder of Neon (Ne) is pressurized to 151 atm. The cylinder has a diameter of 11.75 cm and a height of 140 cm. Calculate the mass (in Kg) of Ne in such a tank, assuming a constant temperature of 298 K. Unless the cylinder is pumped out with a vacuum pump, some residual gas will remain inside when it is "empty." How much gas, in grams, will remain in the tank when it's "empty" ?

   Solution

   First, let's calculate the volume of the tank in liters. Recall that 1 L = 1000 cm³.

   $$ \begin{align} V &= \pi r^2 h = \pi \left( \frac{11.75}{2} \right)^2 \cdot 140 \, cm^3 \\[5pt] &= 15,181 \, cm^3 = 15.181 \, L \end{align}$$

   Now the number of moles is:

   $$ \begin{align} n &= \frac{PV}{RT} = \frac{(151 \cancel{atm})(15.181 \cancel{L})}{(0.0821 \frac{\cancel{L} \cancel{atm}}{mol \cancel{K}})(298 \cancel{K})} \\[5pt] &= 93.69 \; \text{mol} \end{align}$$

   Now use the formula weight of Neon (20.1797 g/mol) to calculate the mass:

   $$ \begin{align} m &= 93.69 \cancel{mol} \left( \frac{20.1797 \, g}{\cancel{mol}} \right) \\[5pt] &= 1,890 \, g = 1.89 \, Kg \end{align}$$

   It's a surprisingly small mass of gas that creates so much pressure in such a volume.




7. Let's say you need a source of O₂ gas that has a pressure of at least 20 atm. If you are limited to batches of 1.5 Kg of O₂, what minimum volume of tank will you need to achieve that pressure at room temperature (298 K) ?

   Solution

   Ultimately, we just want to calculate a volume using the ideal gas law. In order to do that we'll need the number of grams of O₂ in 1.5 Kg of O₂.

   $$ \begin{align} m &= 1500 \cancel{g} \left( \frac{1 \, mol}{32 \cancel{g}} \right) \\[5pt] &= 46.875 \, mol \; O_2 \end{align}$$

   Now the volume is:

   $$ \begin{align} V &= \frac{nRT}{P} = \frac{(46.875 \cancel{mol})(0.0821 \frac{L \cancel{atm}}{\cancel{mol} \cancel{K}})(298 \cancel{K})}{20 \cancel{atm}} \\[5pt] &= 57.3 \; \text{L} \end{align}$$

   So our tank would need to have a volume less than 57.3 Liters.




8. What mass of krypton (Kr) gas is in a container of volume V = 0.22 m³ at a temperature of 20˚C and a pressure of 500 KPa ? Would you expect this calculation to be more or less accurate if you changed the gas to helium (He) ?

   Solution

   The units of our information suggest that we use the gas constant R = 8.314 m³·Pa·mol^-1K^-1. First calclulate the number of moles of krypton, which we're treating like an inert gas:

   $$ \begin{align} n &= \frac{PV}{RT} = \frac{(500,000 \cancel{Pa})(0.22 \cancel{m^3})}{(8.314 \frac{\cancel{m^3} \cancel{Pa}}{mol \cancel{K}})(293 \cancel{K})} \\[5pt] &= 45.156 \, \text{mol} \end{align}$$

   Now 1 mole of Kr has a mass of 83.798 Kg, so we have

   $$m = 45.156 \cancel{mol} \left( \frac{83.798 \, g}{\cancel{mol}} \right) = 3.784 \, Kg$$

   Kr is a large atom, and therefore has some attraction to other Kr atoms, reducing the pressure that we'd expect. Therefore we're likely calculating less Kr mass in the tank than is there. We don't have the same issue with helium (He) because it has no significant attraction to other He atoms.

Non-ideal behavior

Sometimes our assumptions of ideality break down and we have to consider the effects of those non-idealities on our calculations using the ideal gas law. A good place to start thinking about them is with the assumptions we made in the first place.

The first was that gases are spherical, basically have no size, and that the distance between them is much larger than the particle size. We know that this is not strictly true. some gases like CO₂, water (H₂O) and methane (CH₄) are linear or bent or three-dimensional.

And for some gases, the attraction between "particles" is actually quite large, particularly at high pressure when the distance between them is significantly reduced. CO₂, for example, actually liquefies at room temperature and pressures above about 100 atm – definitely not ideal gas behavior.