ideal gas law problem solving

13.3 The Ideal Gas Law

Learning objectives.

By the end of this section, you will be able to:

• State the ideal gas law in terms of molecules and in terms of moles.
• Use the ideal gas law to calculate pressure change, temperature change, volume change, or the number of molecules or moles in a given volume.
• Use Avogadro’s number to convert between number of molecules and number of moles.

In this section, we continue to explore the thermal behavior of gases. In particular, we examine the characteristics of atoms and molecules that compose gases. (Most gases, for example nitrogen, N 2 N 2 , and oxygen, O 2 O 2 , are composed of two or more atoms. We will primarily use the term “molecule” in discussing a gas, but note that this discussion also applies to monatomic gases, such as helium.)

Gases are easily compressed. We can see evidence of this in Table 13.2 , where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same β β . This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.

The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in Figure 13.18 . Because atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them.

To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into an initially deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If we continue to pump air into it, the pressure increases. The pressure will further increase when the car is driven and the tires move. Most manufacturers specify optimal tire pressure for cold tires. (See Figure 13.19 .)

In many common circumstances, including, for example, room temperature air, the gas particles have negligible volume and do not interact with each other, aside from perfectly elastic collisions. In such cases, the gas is called an ideal gas, and the relationship between the pressure, volume, and temperature is given by the equation called the ideal gas law. An equation such as the ideal gas law, which relates behavior of a physical system in terms of its thermodynamic properties, is called an equation of state.

Ideal Gas Law

The ideal gas law states that

where P P is the absolute pressure of a gas, V V is the volume it occupies, N N is the number of atoms and molecules in the gas, and T T is its absolute temperature. The constant k k is called the Boltzmann constant in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value

The ideal gas law can be derived from basic principles, but was originally deduced from experimental measurements of Charles’ law (that volume occupied by a gas is proportional to temperature at a fixed pressure) and from Boyle’s law (that for a fixed temperature, the product PV PV is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of V V . The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that N N is the total number of atoms and molecules, independent of the type of gas.)

Let us see how the ideal gas law is consistent with the behavior of filling the tire when it is pumped slowly and the temperature is constant. At first, the pressure P P is essentially equal to atmospheric pressure, and the volume V V increases in direct proportion to the number of atoms and molecules N N put into the tire. Once the volume of the tire is constant, the equation PV = NkT PV = NkT predicts that the pressure should increase in proportion to the number N of atoms and molecules .

Example 13.6

Calculating pressure changes due to temperature changes: tire pressure.

Suppose your bicycle tire is fully inflated, with an absolute pressure of 7 . 00 × 10 5 Pa 7 . 00 × 10 5 Pa (a gauge pressure of just under 90 . 0 lb/in 2 90 . 0 lb/in 2 ) at a temperature of 18 . 0 º C 18 . 0 º C . What is the pressure after its temperature has risen to 35 . 0 º C 35 . 0 º C ? Assume that there are no appreciable leaks or changes in volume.

The pressure in the tire is changing only because of changes in temperature. First we need to identify what we know and what we want to know, and then identify an equation to solve for the unknown.

We know the initial pressure P 0 = 7 .00 × 10 5 Pa P 0 = 7 .00 × 10 5 Pa , the initial temperature T 0 = 18 . 0ºC T 0 = 18 . 0ºC , and the final temperature T f = 35 . 0ºC T f = 35 . 0ºC . We must find the final pressure P f P f . How can we use the equation PV = NkT PV = NkT ? At first, it may seem that not enough information is given, because the volume V V and number of atoms N N are not specified. What we can do is use the equation twice: P 0 V 0 = NkT 0 P 0 V 0 = NkT 0 and P f V f = NkT f P f V f = NkT f . If we divide P f V f P f V f by P 0 V 0 P 0 V 0 we can come up with an equation that allows us to solve for P f P f .

Since the volume is constant, V f V f and V 0 V 0 are the same and they cancel out. The same is true for N f N f and N 0 N 0 , and k k , which is a constant. Therefore,

We can then rearrange this to solve for P f P f :

where the temperature must be in units of kelvins, because T 0 T 0 and T f T f are absolute temperatures.

1. Convert temperatures from Celsius to Kelvin.

2. Substitute the known values into the equation.

The final temperature is about 6% greater than the original temperature, so the final pressure is about 6% greater as well. Note that absolute pressure and absolute temperature must be used in the ideal gas law.

Making Connections: Take-Home Experiment—Refrigerating a Balloon

Inflate a balloon at room temperature. Leave the inflated balloon in the refrigerator overnight. What happens to the balloon, and why?

Example 13.7

Calculating the number of molecules in a cubic meter of gas.

How many molecules are in a typical object, such as air in a tire? We can use the ideal gas law to give us an idea of how large N N typically is.

Calculate the number of molecules in a cubic meter of air at standard temperature and pressure (STP), which is defined to be 0 º C 0 º C and atmospheric pressure.

Because pressure, volume, and temperature are all specified, we can use the ideal gas law PV = NkT PV = NkT , to find N N .

1. Identify the knowns.

2. Identify the unknown: number of molecules, N N .

3. Rearrange the ideal gas law to solve for N N .

4. Substitute the known values into the equation and solve for N N .

This number is undeniably large, considering that a gas is mostly empty space. N N is huge, even in small volumes. For example, 1 cm 3 1 cm 3 of a gas at STP has 2 . 68 × 10 19 2 . 68 × 10 19 molecules in it. Once again, note that N N is the same for all types or mixtures of gases.

Moles and Avogadro’s Number

It is sometimes convenient to work with a unit other than molecules when measuring the amount of substance. A mole (abbreviated mol) is defined to be the amount of a substance that contains as many atoms or molecules as there are atoms in exactly 12 grams (0.012 kg) of carbon-12. The actual number of atoms or molecules in one mole is called Avogadro’s number ( N A ) ( N A ) , in recognition of Italian scientist Amedeo Avogadro (1776–1856). He developed the concept of the mole, based on the hypothesis that equal volumes of gas, at the same pressure and temperature, contain equal numbers of molecules. That is, the number is independent of the type of gas. This hypothesis has been confirmed, and the value of Avogadro’s number is

Avogadro’s Number

One mole always contains 6 . 02 × 10 23 6 . 02 × 10 23 particles (atoms or molecules), independent of the element or substance. A mole of any substance has a mass in grams equal to its molecular (molar) mass, which can be calculated by multiplying the number of moles of the substance by its atomic mass. The atomic masses of elements are given in the periodic table of elements and in Appendix A

Check Your Understanding

The active ingredient in a Tylenol pill is 325 mg of acetaminophen ( C 8 H 9 NO 2 ) ( C 8 H 9 NO 2 ) . Find the molar mass of acetaminophen, and from this, the number of moles and the number of molecules of acetaminophen in a single pill.

We first need to calculate the molar mass (the mass of one mole) of acetaminophen. To do this, we need to multiply the number of atoms of each element by the element’s atomic mass.

Then we need to calculate the number of moles in 325 mg.

Then use Avogadro’s number to calculate the number of molecules.

Example 13.8

Calculating moles per cubic meter and liters per mole.

Calculate: (a) the number of moles in 1 . 00 m 3 1 . 00 m 3 of gas at STP, and (b) the number of liters of gas per mole at STP.

Strategy and Solution

(a) We are asked to find the number of moles per cubic meter, and we know from Example 13.7 that the number of molecules per cubic meter at STP is 2 . 68 × 10 25 2 . 68 × 10 25 . The number of moles can be found by dividing the number of molecules by Avogadro’s number. We let n n stand for the number of moles,

(b) Using the value obtained for the number of moles in a cubic meter, and converting cubic meters to liters, we obtain

This value is very close to the accepted value of 22.4 L/mol. The slight difference is due to rounding errors caused by using three-digit input. Again this number is the same for all gases. In other words, it is independent of the gas.

The (average) molar weight of dry air (approximately 80% N 2 N 2 and 20% O 2 O 2 ) at STP is M = 28.8 g/mol. M = 28.8 g/mol. Thus the mass of one cubic meter of air is 1.28 kg. The density of dry room temperature air is about 10% lower. If a living room has dimensions 5 m × 5 m × 3 m, 5 m × 5 m × 3 m, the mass of air inside the room is around 90 kg, which is the typical mass of a human.

The density of air at standard conditions ( P = 1 atm ( P = 1 atm and T = 20 º C ) T = 20 º C ) is 1.20 kg/m 3 1.20 kg/m 3 . At what pressure is the density 0.60 kg/m 3 0.60 kg/m 3 if the temperature and number of molecules are kept constant?

The best way to approach this question is to think about what is happening. If the density drops to half its original value and no molecules are lost, then the volume must double. If we look at the equation PV = NkT PV = NkT , we see that when the temperature is constant, the pressure is inversely proportional to volume. Therefore, if the volume doubles, the pressure must drop to half its original value, and P f = 0 . 50 atm . P f = 0 . 50 atm .

The Ideal Gas Law Restated Using Moles

A very common expression of the ideal gas law uses the number of moles, n n , rather than the number of atoms and molecules, N N . We start from the ideal gas law,

and multiply and divide the equation by Avogadro’s number N A N A . This gives

Note that n = N / N A n = N / N A is the number of moles. We define the universal gas constant R = N A k R = N A k , and obtain the ideal gas law in terms of moles.

Ideal Gas Law (in terms of moles)

The ideal gas law (in terms of moles) is

The numerical value of R R in SI units is

In other units,

You can use whichever form of R R is most convenient for a particular problem.

Example 13.9

Calculating number of moles: gas in a bike tire.

How many moles of gas are in a bike tire with a volume of 2 . 00 × 10 – 3 m 3 ( 2 . 00 L ) , 2 . 00 × 10 – 3 m 3 ( 2 . 00 L ) , a pressure of 7 . 00 × 10 5 Pa 7 . 00 × 10 5 Pa (a gauge pressure of just under 90 . 0 lb/in 2 90 . 0 lb/in 2 ), and at a temperature of 18 . 0 º C 18 . 0 º C ?

Identify the knowns and unknowns, and choose an equation to solve for the unknown. In this case, we solve the ideal gas law, PV = nRT PV = nRT , for the number of moles n n .

2. Rearrange the equation to solve for n n and substitute known values.

The most convenient choice for R R in this case is 8 . 31 J/mol ⋅ K, 8 . 31 J/mol ⋅ K, because our known quantities are in SI units. The pressure and temperature are obtained from the initial conditions in Example 13.6 , but we would get the same answer if we used the final values.

The ideal gas law can be considered to be another manifestation of the law of conservation of energy (see Conservation of Energy ). Work done on a gas results in an increase in its energy, increasing pressure and/or temperature. This increased energy can also be viewed as increased internal kinetic energy, given the gas’s atoms and molecules.

The Ideal Gas Law and Energy

Let us now examine the role of energy in the behavior of gases. When you inflate a bike tire by hand, you do work by repeatedly exerting a force through a distance. This energy goes into increasing the pressure of air inside the tire and increasing the temperature of the pump and the air.

The ideal gas law is closely related to energy: the dimensions on both sides are those of energy, with units of joules when using SI units. The right-hand side of the ideal gas law in PV = NkT PV = NkT is NkT NkT . This term is proportional to the amount of translational kinetic energy of N N atoms or molecules at an absolute temperature T T , as we shall see formally in Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature . The left-hand side of the ideal gas law is PV PV , which also has the units of joules. Pressure is force per unit area, so pressure multiplied by volume is force times displacement, or energy. The important point is that there is energy in a gas related to both its pressure and its volume. The energy can be changed when the gas is doing work as it expands—something we explore in Heat and Heat Transfer Methods —similar to what occurs in gasoline or steam engines and turbines.

Problem-Solving Strategy: The Ideal Gas Law

Step 1 Examine the situation to determine that an ideal gas is involved. Most gases are nearly ideal.

Step 2 Make a list of what quantities are given, or can be inferred from the problem as stated (identify the known quantities). Convert known values into proper SI units (K for temperature, Pa for pressure, m 3 m 3 for volume, molecules for N N , and moles for n n ).

Step 3 Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written list is useful.

Step 4 Determine whether the number of molecules or the number of moles is known, in order to decide which form of the ideal gas law to use. The first form is PV = NkT PV = NkT and involves N N , the number of atoms or molecules. The second form is PV = nRT PV = nRT and involves n n , the number of moles.

Step 5 Solve the ideal gas law for the quantity to be determined (the unknown quantity). You may need to take a ratio of final states to initial states to eliminate the unknown quantities that are kept fixed.

Step 6 Substitute the known quantities, along with their units, into the appropriate equation, and obtain numerical solutions complete with units. Be certain to use absolute temperature and absolute pressure.

Step 7 Check the answer to see if it is reasonable: Does it make sense?

Liquids and solids have densities about 1000 times greater than gases. Explain how this implies that the distances between atoms and molecules in gases are about 10 times greater than the size of their atoms and molecules.

Atoms and molecules are close together in solids and liquids. In gases they are separated by empty space. Thus gases have lower densities than liquids and solids. Density is mass per unit volume, and volume is related to the size of a body (such as a sphere) cubed. So if the distance between atoms and molecules increases by a factor of 10, then the volume occupied increases by a factor of 1000, and the density decreases by a factor of 1000.

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• Authors: Paul Peter Urone, Roger Hinrichs
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PV = k 1 V / T = k 2 P / T = k 3 V / n = k 4 P / n = k 5 1 / nT = 1 / k 6

P 3 V 3 / n 3 T 3 = k 1 k 2 k 3 k 4 k 5 / k 6

k 1 = L-atm k 2 = L / K k 3 = atm / K k 4 = L / mol k 5 = atm / mol 1 / k 6 = 1 / mol-K

PV / nT = R

    1 V  ∝  ––     P V ∝ T V ∝ n

    nT V  ∝  –––     P

    nRT V  =  –––     P

(1.000 atm) (22.414 L) = (1.000 mol) (R) (273.15 K)

(660.0 mmHg / 760.0 mmHg/1.00 atm) (11.0 L) = (n) (0.08206 L atm / mol K) (298 K) Note the conversion from mmHg to atm and from Celsius to Kelvin.

    (0.868421 atm) (11.0 L) n = –––––––––––––––––––––––––––     (0.08206 L atm / mol K) (298 K) n = 0.391 mol

(720.0 mmHg / 760.0 mmHg/1.00 atm) (6.20 L) = (n) (0.08206 L atm / mol K) (301 K) Note the conversion from mmHg to atm and from Celsius to Kelvin.

n = 0.238 mol

(2.43 atm) (V) = (0.400 mol) (0.08206 L atm / mol K) (284 K)

    (0.400 mol) (0.08206 L atm / mol K) (284 K) V = –––––––––––––––––––––––––––––––––––––     2.43 atm V = 3.84 L

(780 mmHg) (6.00 L) = (0.300 mol) (62.3638 L mmHg / mol K) (284 K) Note the different value and unit for R, to be in agreement with using mmHg for the pressure unit. Alternatively, you could convert 780. mmHg to atm and then use 0.08206 L atm / mol K for the value of R. A variety of values for R can be found here .

T = PV / nR T = [(780.) (6.00)] / [(0.300) (62.3638)] T = 250. K The problem does not specify the final unit, but Celsius is most often requested. 250 − 273 = −23 °C

PV = nRT (P) (3.5 L) = (2.3 mol) (0.08206 L atm / mol K) (313 K) P = [(2.3) (0.08206) (313)] / 3.5 P = 16.9 atm

740.0 mm Hg ÷ 760.0 mm Hg/atm = 0.973684 atm

(0.973684 atm) (2.850 L) = (n) (0.08206 L atm / mol K) (295.0 K) To four sig figs, the answer is 0.1146 mol

This is a very common use of this law and the odds are very good you will see this type of question on a test. The key is to remember the units on molar mass: grams per mole. We know from the problem statement that 2.1025 grams of the gas is involved and we know (by the above calculation that 0.1146329 mole of the gas is present. So all we have to do is divide the grams of gas by how many moles it is: 2.1025 g / 0.1146329 mol = 18.34 g/mol

(1) You have to know the grams of gas involved. Usually the problem will just give you the value, but not always. You might have to calculate it. (2) You are going to have to calculate the moles of gas. Use PV = nRT and solve for n. Make sure to use L, atm and K. (3) Divide grams by moles and there's your answer.

PV = nRT (0.00100 atm) (1.00 L) = (n) (0.08206 L atm/mol K) (333 K) n = 0.000036595 mol

(1.17 x 10¯ 6 g/cm 3 ) (1000 cm 3 / 1.00 L) = 1.17 x 10¯ 3 g/L

1.17 x 10¯ 3 g / 0.000036595 mol = 31.97 g/mol The molecular weight of oxygen gas (O 2 ) is 31.9994 g/mol The unknown gas is oxygen. Hydrazine (N 2 H 4 ) weighs a bit more than 32, but it is a liquid at 60 °C

(1.29 g/L) (5.00 L) = 6.45 g

543.251 g − 6.45 g = 536.801 g

566.107 g − 536.801 g = 29.306 g

PV = nRT (1.00 atm) (5.00 L) = (n) (0.08206) (273 K) n = 0.22319 mol

29.306 g / 0.22319 mol = 131.3 g/mol

(5.00 x 10 9 molecules) / (6.022 x 10 23 molecules/mole) = 8.303 x 10¯ 15 mol

PV = nRT (P) (1.00 m 3 ) = (8.303 x 10¯ 15 mol) (8.20575 x 10¯ 5 m 3 atm / mol K) (298 K) I looked up the value for R here . P = 2.03 x 10¯ 16 atm

take 1 liter of volume, that is 0.001 m 3 . so there are 5e6 molecules. 6.022e23 molecules/mole (Avogadro's constant) 1 mole is 6.022e23 molecules, so (5e6 molecules) x (1 mol/6.022e23 molecules) = 8.303e-18 mol Ideal gas law PV = nRT n = number of moles R = gas constant = 0.08206 (L atm)/(mol K) T = temperature in Kelvins P = absolute pressure in atm V = volume in liters P = nRT/V = (8.303e-18) (0.08206) (298) / 1 = 2.03e-16 atm

(a) Ne (b) Ar (c) Kr (d) Xe (e) He

PV = nRT (1.16 atm) (1.00 L) = (n) (0.08206 L atm / mol K) (315 K) n = 0.044876 mol 0.906 g / 0.044876 mol = 20.19 g/mol

PV = nRT PV = (m/M)RT Since density = m/V, we have this (after substituting and rearranging): PVM = mRT M = mRT / PV M = [(0.906 g) (0.08206 L atm / mol K) (315 K)] / [(1.00 L) (1.16 atm)] = 20.19 g/mol Neon.

At STP , 1 mol of gas has a volume of 22.414 L Using the Combined Gas Law, convert the volume to 315 K and 1.16 atm: (22.414) (315/273) (1.00/1.16) = 22.295 L Now, you need to calculate densities until you get the correct answer: Ne---> 20.180 g / 22.295 L = 0.905 g/L For comparison's sake, here is argon: Ar ---> 39.95 g / 22.295 L = 1.79 g/L

PV = nRT (560 torr / 760 torr/atm) (0.125 L) = (n) (0.08206 L atm / mol K) (296 K) n = 0.003792 mol 0.105 g / 0.003792 mol = 27.7 g/mol Nitrogen has a molar mass of 28.0 g/mol

(a) CO (b) H 2 O (c) CH 4 (d) NO (e) They would all occupy the same volume.

V = nRT / P Since RT/P are the same for each gas, the greatest volume will be for the gas with the greatest number of moles.

CO ---> 28.0 g/mol H 2 O ---> 18.0 g/mol CH 4 ---> 16.0 g/mol NO ---> 30.0 g/mol 1.0 g of methane, CH 4 , will have the greatest volume at STP.

12.8 g / 4.0026 g/mol = 3.1979 mol

(1.00 atm) (x) = (3.1979 mol) (0.08206 L atm / mol K) (273.15 K) x = 71.6799 L To three sig figs, 71.7 L Using the more common 273 K for standard temperature results in 71.6 L. Make sure to use the value for standard temperature that your teacher uses. That is the value he/she uses to calculate the answers on the test.

PV = nRT (1.00 atm) (58.4 L) = (n) (0.08206 L atm mol¯ 1 K¯ 1 ) (275.50 K) n = 2.58321 mol

10.3 g / 2.58321 mol = 3.99 g/mol (to three sig figs)

PV = nRT (1.00 atm) (4.27 L) = (n) (0.08206 L atm mol¯ 1 K¯ 1 ) (358.0 K) n = 0.14535 mol 10.3 g / 0.14535 mol = 70.9 g/mol Look for 70.9 on the periodic table and can't find it. What to do? Aha! Remember that diatomic gases exist. Look for 35.45 (half of 70.9) and find chlorine. The diatomic gas chlorine, Cl 2 , is the answer.

1.00 L = 1000 mL = 1000 cm 3 (1.14 g/cm 3 ) (1000 cm 3 ) = 1140 g 1140 g / 2.04 g/mol = 558.824 mol You could use any volume you want. I used 1.00 L to make the PV = nRT calculation just a tiny bit simpler.

PV = nRT (2.01 x 10 9 atm) (1.00 L) = (558.824 mol) (0.08206 L atm / mol K) (T) T = 4.38 x 10 7 K (to three sig figs)

(104 kPa / 101.325 kPa/atm) (V) = (1.05 mol) (0.08206 L atm / mol K) (300. K) V = 25.184 L To three sig figs, the answer is 25.2 L

PV = nRT where P = 104000 Pa, n = 1.05 mol, R = 8.314 Pa m 3 /mol K, T = 300. K. V = nRT/p = (315) (8.31447 m 3 ) / 104000 = 0.025183 m 3 Note how only the final unit that survives in the answer is shown. V = 0.0252 m 3

(1.05 mol) (6.022 x 10 23 molecules mol¯ 1 ) = 6.3231 x 10 23 molecules

(4/3) (3.14159) (1.5 x 10¯ 10 m) 3 = 1.4137 x 10¯ 29 m 3

(6.3231 x 10 23 molecules) (1.4137 x 10¯ 29 m 3 / molecule) = 0.00000894 m 3

0.00000894 m 3 / 0.025183 m 3 = 0.000355 or about 1/2817 of the total volume

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Ideal Gas Law Formula and Examples

The ideal gas law is the equation of state for an ideal gas that relates pressure, volume, gas quantity, and absolute temperature . Although the law describes the behavior of an ideal gas, it approximates real gas behavior in many cases. Uses of the ideal gas law including solving for an unknown variable, comparing initial and final states, and finding partial pressure. Here is the ideal gas law formula, a look at its units, and a discussion of its assumption and limitations.

Ideal Gas Formula

The ideal gas formula takes a couple of forms. The most common one uses the ideal gas constant:

• P is gas pressure .
• V is the volume of gas.
• n is the number of moles of gas.
• R is the ideal gas constant , which is also the universal gas constant or the product of the Boltzmann constant and Avogadro’s number .
• T is the absolute temperature .

There are other formulas for the ideal gas equation:

Here, P is pressure, ρ is density, R is the ideal gas constant, T is absolute temperature, and M is molar mass.

P = k B ρT/ μ M u

Here, P is pressure, k B is Boltzmann’s constant, ρ is density, T is absolute temperature, μ is the average particle mass, and M u is the atomic mass constant.

The value of the ideal gas constant, R, depends on the other units chosen for the formula. The SI value of R is exactly 8.31446261815324 J⋅K −1 ⋅mol −1 . Other SI units are pascals (Pa) for pressure, cubic meters (m 3 ) for volume, moles (mol) for gas quantity, and kelvin (K) for absolute temperature. Of course, other units are fine, so long as they agree with one another and you remember the T is absolute temperature. In other words, convert Celsius or Fahrenheit temperatures to Kelvin or Rankine.

To summarize, here are the two most common sets of units:

• R is 8.314 J⋅K −1 ⋅mol −1
• P is in pascals (Pa)
• V is in cubic meters (m 3 )
• n is in moles (mol)
• T is in kelvin (K)
• R is 0.08206 L⋅atm⋅K −1 ⋅mol −1
• P is in atmospheres (atm)
• V is in liters (L)

Assumptions Made in the Ideal Gas Law

The ideal gas law applies to ideal gases . What this means is that the gas has the following properties:

• Particles in a gas move randomly.
• Atoms or molecules have no volume.
• The particles do not interact with one another. They are neither attracted to one another nor repelled by each other.
• Collisions between gas particles and between the gas and the container wall are perfectly elastic. No energy is lost in a collision.

Ideal Gas Law Uses and Limitations

Real gases do not behave exactly the same as ideal gases. However, the ideal gas law accurately predicts the behavior of monatomic gases and most real gases at room temperature and pressure. In other words, you can use the ideal gas law for most gases at relatively high temperatures and low pressures.

The law does not apply when mixing gases that react with one another. The approximation deviates from true behavior at very low temperatures or high pressures. When temperature is low, kinetic energy is low, so there is a higher likelihood of interactions between particles. Similarly, at high pressure, there are so many collisions between particles that they don’t behave ideally.

Ideal Gas Law Examples

For example, there are 2.50 g of XeF 4  gas in a 3.00 liter container at 80°C. What is the pressure in the container?

First, write down what you know and then convert units so they work together in the formula:

P=? V = 3.00 liters n = 2.50 g XeF 4  x 1 mol/ 207.3 g XeF 4  = 0.0121 mol R = 0.0821 l·atm/(mol·K) T = 273 + 80 = 353 K

Plugging in these values gives the answer:

P = 00121 mol x 0.0821 l·atm/(mol·K) x 353 K / 3.00 liter

Pressure = 0.117 atm

Here are more examples:

• Solve for the number of moles .
• Find the identity of an unknown gas.
• Solve for density using the ideal gas law.

French engineer and physicist Benoît Paul Émile Clapeyron gets credit for combining Avogadro’s law, Boyle’s law, Charles’s law, and Gay-Lussac’s law into the ideal gas law in 1834. August Krönig (1856) and Rudolf Clausius (1857) independently derived the ideal gas law from kinetic theory .

Formulas for Thermodynamic Processes

Here are some other handy formulas:

• Clapeyron, E. (1834). “Mémoire sur la puissance motrice de la chaleur.”  Journal de l’École Polytechnique  (in French). XIV: 153–90.
• Clausius, R. (1857). “Ueber die Art der Bewegung, welche wir Wärme nennen”. Annalen der Physik und Chemie (in German). 176 (3): 353–79. doi: 10.1002/andp.18571760302
• Davis; Masten (2002).  Principles of Environmental Engineering and Science . New York: McGraw-Hill. ISBN 0-07-235053-9.
• Moran; Shapiro (2000). Fundamentals of Engineering Thermodynamics (4th ed.). Wiley. ISBN 0-471-31713-6.
• Raymond, Kenneth W. (2010). General, Organic, and Biological Chemistry: An Integrated Approach (3rd ed.). John Wiley & Sons. ISBN 9780470504765.

Related Posts

Chemistry Steps

General Chemistry

The ideal gas laws show the correlation of the temperature, pressure, volume, and the amount of a gas. Although, this is not how the laws were determined, I found my students grasping the concepts a lot easier using the following model.

We are going to take a pump filled with some gas and a freely moving plunger and by changing the gas parameters determine their correlation.

• Boyle’s Law

Boyle’s Law shows the correlation of the pressure and the volume of a gas. To illustrate it, we change the volume of the gas by pushing down the plunger. This decreases the volume, and because the gas molecules have less space, the pressure is increasing:

The observation is that the volume of a fixed quantity of gas at constant temperature is inversely proportional to its pressure.

This relationship can be written as:

\[{\rm{V}} \sim \,\frac{{\rm{1}}}{{\rm{P}}}\]

To bring in the equal sign, we introduce a constant:

                                                        PV = constant                               

This can be explained using the example of a car dealership income. The income depends on the number of sales which we can represent as:

Income ∼ number of cars

However, we cannot say income = number of cars sold, so to switch an equal sign, we need to introduce a constant. This can be the price of the car transforming the equation to:

Income   = price x number of cars

So, for the gas pressure and volume, we are not interested too much in the constant, but rather in its linkage of pressure and volume at positions 1 and 2. Because the P x V product is constant, we can write that:

P 1 V 1 = constant = P 2 V 2

This is the practical implication of the Boyle’s law that is used for solving gas problems.

For example ,

The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 liters.

First, write down what you gave and what needs to be determined. If nothing is mentioned about any parameter, for example, the moles and the temperature in this case, it is assumed that they are constant, so you don’t need to worry about them.

P 1 = 2.30 atm V 1 = 1.80 L V 2 = 1.20 L P 2 = ?

\[{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}\; = \;{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}\]

Now, rearrange to calculate P 2 :

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}\]

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{\rm{2}}{\rm{.30}}\;{\rm{atm}}\;{\rm{ \times 1}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{1}}{\rm{.20}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.45}}\;{\rm{atm}}\]

• Charle’s Law

Let’s now consider what happens if we heat up the gas leaving the plunger to move freely. The gas is going to expand, and the correlation is that the volume of a gas increases with temperature:

The volume is directly proportional to the temperature of the gas:

Therefore, for different states of gas, we can write the Charle’s law as:

For example,

What will be the final volume of a 3.50 L sample of nitrogen at 20 °C if it is heated to 200. °C?

Write down what is given and what needs to be determined first:

V 1 = 3.50 L T 1 = 20 o C T 2 = 200. o C V 2 = ?

Now, before doing anything else, remember to always convert the temperature to Kelvin when solving a gas problem:

So, T 1 = 20 + 273 = 293 K ,  T 2 = 200 + 273 = 473 K  

The question studies the correlation between the volume and the temperature of a gas, so we need to use the Charle’s law.

Write it down and rearrange it to get an expression of V 2 .

\[\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]

\[{{\rm{V}}_{\rm{2}}}\; = \;\frac{{{\rm{473}}\;\cancel{{\rm{K}}}\;{\rm{ \times }}\;{\rm{3}}{\rm{.50}}\;{\rm{L}}}}{{{\rm{293}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.65}}\;{\rm{L}}\]

Gay-Lussac’s law

To study the relationship between the pressure and the temperature of a gas, the barrel is held at a fixed position to prevent changing the volume, and the sample is heated up:

In this case, the pressure increases with the temperature and for different states of the gas, the Gay-Lussac’s law is written as:

A sample of helium gas at 1.40 atm is heated from 23.0 °C to 400.0 K. How many atmospheres is the final pressure of the helium gas?

P 1 = 1.40 atm T 1 = 23.0 °C P 2 = 400.0 K T 2 = ?

Convert the temperature to Kelvin right away!

T 1 = 23.0 + 273 = 296 K

Write down the gas law and rearrange it to get an expression of P 2 .

\[\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{\rm{1}}{\rm{.40}}\;{\rm{atm}}\;{\rm{ \times }}\;{\rm{400}}{\rm{.0}}\;\cancel{{\rm{K}}}}}{{{\rm{296}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.89}}\;{\rm{atm}}\]

• Avogadro’s Law

The Avogadro’s law of ideal gases demonstrates that the volume of a gas is directly proportional to its number of moles. In other words, the more gas, the larger the volume which again is a very intuitive observation.

For this, the plunger is again left free, and pumping some gas into the system increases the volume that the gas occupies:

This is summarized in the following formula:

Talking of the volume and the moles of a gas, remember that at STP, one mole of any gas occupies a volume of 22.4 L called the molar volume , V o . 

This is because in ideal gases, the size of molecules is very small compared to the intermolecular distances. So, most of the volume is almost like an empty, and therefore, it does not matter what gas it is, the volume is determined to be 22.4 L at 0 o C and 1 atm.

• The Ideal Gas Law

The examples above are great to demonstrate the individual gas law, however, notice that in all experiments, we assumed or set up the experiment, such that two parameters are constant, and we study the correlation of the other two. For example, in the Boyle’s law, we study a constant amount of gas at a constant temperature and find that the pressure increase as the volume is decreased.

To combine all the laws together and have the four variables (n, P, V, T) in one place, the Ideal Gas Law equation is obtained:

The R is called the ideal gas constant . Although it has different values and units, you will mostly be using this:

\[R\;{\rm{ = }}\;{\rm{0}}{\rm{.08206}}\;\frac{{{\rm{L}} \cdot {\rm{atm}}}}{{{\rm{mol}} \cdot {\rm{K}}}}\]

The ideal gas law equation is used when you need to find P, V, T, or n , for a system where they do not change .

A sample of hydrogen gas is added into a 5.80 L container at 56.0 °C. How many moles of the gas is present in the container if the pressure is 6.70 atm?

Rearrange the ideal gas law to get an expression for the moles (n):

\[{\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{PV}}}}{{{\rm{RT}}}}\]

\[{\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{6}}{\rm{.70}}\;\cancel{{{\rm{atm}}}}\; \times \;{\rm{5}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}{\rm{ }} \times \;{\rm{329}}\;\cancel{{\rm{K}}}}}\; = \;1.44\;{\rm{mol}}\]

Remember, to change the pressure to atm when the ideal gas law equation is used! This is because the units of R contain atm when the 0.08206 value is used. And this is what most problems in this chapter use.

How do I know which gas law to use?

You are probably wondering about this question now that every gas law brings a new equation. For this, there is what is called the combined gas law and as long as you remember it, you do not need to remember all the gas laws to solve a problem.

Let’s keep it for another article because there is quite a lot of information in this one.

• Gay-Lussac’s Law
• Celsius or Kelvin
• Ideal-Gas Laws
• Combined Gas Law Equation
• How to Know Which Gas Law Equation to Use
• Molar Mass and Density of Gases
• Graham’s Law of Effusion and Diffusion
• Graham’s Law of Effusion Practice Problems
• Dalton’s Law of Partial Pressures
• Mole Fraction and Partial Pressure of the Gas
• Gases in Chemical Reactions
• Gases-Practice Problems

Boyle’s Law: The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 litters.

Boyle’s Law: After changing the pressure of a gas sample from 760.0 torr to 0.800 atm, it occupies 4.30 L volume. What was the initial volume of the gas?

Charles’s Law:  What will be the final volume of a 3.50 L sample of nitrogen at 20 °C if it is heated to 200 °C?

Charles’s Law:  The volume of a gas decreased from 2.40 L to 830. mL and the final temperature is set at 40.0 °C. Assuming a constant pressure, calculate the initial temperature of the gas in kelvins.

Gay-Lussac’s Law:  A sample of helium gas at 1.40 atm is heated from 23.0 °C to 400.0 K. How many atmospheres is the final pressure of the helium gas?

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Ideal Gas Law Example Problem

Find Moles of Gas Using the Ideal Gas Law

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The ideal gas law is an equation of state the describes the behavior of an ideal gas and also a real gas under conditions of ordinary temperature and low pressure. This is one of the most useful gas laws to know because it can be used to find pressure, volume, number of moles, or temperature of a gas.

The formula for the ideal gas law is:

PV = nRT

P = pressure V = volume n = number of moles of gas R = ideal or universal  gas constant  = 0.08 L atm / mol K T = absolute temperature  in Kelvin

Sometimes, you may use another version of the ideal gas law:

PV = NkT

N = number of molecules k = Boltzmann constant = 1.38066 x 10 -23  J/K = 8.617385 x 10 -5  eV/K

Ideal Gas Law Example

One of the easiest applications of the ideal gas law is finding the unknown value, given all the others.

6.2 liters of an ideal gas is contained at 3.0 atm and 37 °C. How many moles of this gas are present?

The ideal gas law states

Because the units of the gas constant are given using atmospheres, moles, and Kelvin, it's important to make sure you convert values given in other temperature or pressure scales. For this problem, convert °C temperature to K using the equation:

T = °C + 273

T = 37 °C + 273 T = 310 K

Now, you can plug in the values. Solve ideal gas law for the number of moles

n = PV / RT

n = ( 3.0 atm x 6.2 L ) / ( 0.08 L atm /mol K x 310 K) n = 0.75 mol

There are 0.75 mol of the ideal gas present in the system.

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