Many chemists had dreamed of having an equation that describes relation of a gas molecule to its environment such as pressure or temperature. However, they had encountered many difficulties because of the fact that there always are other affecting factors such as intermolecular forces. Despite this fact, chemists came up with a simple gas equation in order to study gas behavior while putting a blind eye to minor factors.The Ideal Gas Law is very simple.
Before we go any further, we must understand that this gas law is ideal. The keyword is ideal, hence why it is italicized in the last sentence. As students, professors, and chemists, we sometimes need to understand the concepts before we can apply it, and assuming the gases are in an ideal state where it is unaffected by real world conditions will help us better understand the behavior the gases. In order for a gas to be ideal, its behavior must follow the KineticMolecular Theory whereas the NonIdeal Gases will deviate from this theory due to real world conditions.
When dealing with gas, a famous equation was used to relate all of the factors needed in order to solve a gas problem. This equation is known as the Ideal Gas Equation. As we have always known, anything ideal does not exist. In this issue, two wellknown assumptions should have been made beforehand:
An ideal gas is a hypothetical gas dreamed by chemists and students because it would be much easier if things like intermolecular forces do not exist to complicate the simple Ideal Gas Law. Ideal gases are essentially point masses moving in constant, random, straightline motion. Its behavior is described by the assumtions listed in the KineticMolecular Theory of Gases. This definition of an ideal gas contrasts with the NonIdeal Gas definition, because this equation represents how gas actually behaves in reality. For now, let us focus on the Ideal Gas.
Before we look at the Ideal Gas Equation, let us state the four gas variables and one constant for a better understanding. The four gas variables are: pressure (P), volume (V), number of mole of gas (n), and temperature (T). Lastly, the constant in the equation shown below is R, known as the the gas constant, which will be discussed in depth further later:
\[ PV=nRT \]
Another way to describe an ideal gas is to describe it in mathematically. Consider the following equation:
An ideal gas will always equal 1 when plugged into this equation. The greater it deviates from the number 1, the more it will behave like a real gas rather than an ideal. A few things should always be kept in mind when working with this equation, as you may find it extremely helpful when checking your answer after working out a gas problem.
NOTE: A trick to memorizing this formula is to think of the two words Pervert and Nerd. Thinking of them? Now mush them into one word, and make a sound so it sounds kind of like Perrvvnert. Pervnert. PVnRT.
The Ideal Gas Law is simply the combination of all Simple Gas Laws (Boyle's Law, Charles' Law, and Avogadro's Law), and so learning this one means that you have learned them all. The Simple Gas Laws can always be derived from the Ideal Gas equation.
Boyle’s Law describes the inverse proportional relationship between pressure and volume at a constant temperature and a fixed amount of gas. This law came from a manipulation of the Ideal Gas Law.
\[ P \propto \dfrac{1}{V} \]
Equation:
\[ P_1V_1=P_2V_2 \]
This equation would be ideal when working with problem asking for the initial or final value of pressure or volume of a certain gas when one of the two factor is missing.
Charles's Law describes the directly proportional relationship between the volume and temperature (in Kelvin) of a fixed amount of gas, when the pressure is held constant.
\[ V\propto \; T \]
Equation:
\[ \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2} \]
This equation can be used to solve for initial or final value of volume or temperature under the given condition that pressure and the number of mole of the gas stay the same.
Volume of a gas is directly proportional to the amount of gas at a constant temperature and pressure.
\[ V \propto \; n\]
Equation:
\[ \dfrac{V_1}{n_1}=\dfrac{V_2}{n_2} \]
Avogadro's Law can apply well to problems using Standard Temperature and Pressure (see below), because of a set amount of pressure and temperature.
Given a constant number of mole of a gas and an unchanged volume, pressure is directly proportional to temperature.
\[ P \propto \; T\]
Equation:
\[ \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2} \]
Boyle's Law, Charles' Law, and Avogradro's Law and Amontons's Law are given under certain conditions so directly combining them will not work. Through advanced mathematics (provided in outside link if you are interested), the properties of the three simple gas laws will give you the Ideal Gas Equation.
The table below lists the different units for each property.
Factor  Variable  Units 
Pressure  P  atm Torr Pa mmHg 
Volume  V  L m³ 
Moles  n  mol 
Temperature  T  K 
Gas Constant  R*  see Values of R table below 
Use the following table as a reference for pressure.
Common Units of Pressure  
Unit  Symbol  Equivalent to 1 atm 
Atmosphere  atm  1 atm 
Millimeter of Mercury  mmHg  760 mmHg 
Torr  Torr  760 Torr 
Pascal  Pa  101326 Pa 
Kilopascal  kPa*  101.326 kPa 
Bar  bar  1.01325 bar 
Millibar  mb  1013.25 mb 
*note: This is the SI unit for pressure
Here comes the tricky part when it comes to the gas constant, R. Value of R WILL change when dealing with different unit of pressure and volume (Temperature factor is overlooked because temperature will always be in Kelvin instead of Celcius when using the Ideal Gas equation). Only through appropriate value of R will you get the correct answer of the problem. It is simply a constant, and the different values of R correlates accordingly with the units given. When choosing a value of R, choose the one with the appropriate units of the given information (sometimes given units must be converted accordingly). Here are some commonly used values of R:
Values of R 
0.082057 L atm mol^{1} K^{1} 
62.364 L Torr mol^{1}^{ }K^{1} 
8.3145 m^{3} Pa mol^{1} K^{1} 
8.3145 J mol^{1} K^{1}^{*} 
*note: This is the SI unit for the gas constant
Proposing question: So, which value of R should I use?
Because of the various value of R you can use to solve a problem. It is crucial to match your units of Pressure, Volume, number of mole, and Temperature with the units of R.
Example 1 


Other things to keep in mind: Know what Standard Temperature and Pressure (STP)* is. Know how to do Stoichiometry. Know your basic equations.
Take a look at the problems below for examples of each different type of problem. Attempt them initially, and if help is needed, the solutions are right below them. Remark: The units must cancel out to get the appropriate unit; knowing this will help you double check your answer.
Example 2 

5.0 g of neon is at 256 mm Hg and at a temperature of 35º C. What is the volume? SOLUTION Step 1: Write down your given information: P = 256 mmHg V = ? m = 5.0 g R = 0.0820574 L·atm·mol^{1}K^{1} T = 35º C Step 2: Convert as necessary: Pressure: \( 256 mmHg * (1 atm/ 760 mmHg) = 0.3368 atm \) Moles: \( 5.0 g Ne * (1 mol / 20.1797 g) = 0.25 mols Ne \) Temperature: \(35º C + 273 = 308 K \) Step 3: Plug in the variables into the appropriate equation. \[ V = (nRT/P) \] \[ V = [(.25mol)(0.08206Latm/Kmol)(308K)/(.3368atm)] \] \[ V = 19L\] 
Example 3 

What is a gas’s temperature in Celsius when it has a volume of 25 L, 203 mol, 143.5 atm? SOLUTION Step 1: Write down your given information: P = 143.5 atm V= 25 L n = 203 mol R = 0.0820574 L·atm·mol^{1} K^{1} T = ? You can skip Step 2 because all units are the appropriate units. Step 3: Plug in the variables into the appropriate equation. \(T = (PV/nR)\) \(T = [(143.5atm)(25L)/(203 mol)(0.08206 L·atm/Kmol)]\) \(T = 215.4K\) Step 4: You are not done. Be sure to read the problem carefully, and answer what they are asking for. In this case, they are asking for temperature in Celcius, so you will need to convert it from K, the units you have. \(215.4 K  273 = 57.4º C\) 
Example 3 

What is the density of nitrogen gas (N2) at 248.0 Torr and 18º C? SOLUTION Step 1: Write down your given information P = 248.0 Torr V = ? n = ? R = 0.0820574 L·atm·mol^{1} K^{1} T = 18º C Step 2: Convert as necessary. \((248 Torr) * (1 atm/ 760 Torr) = 0.3263 atm\) \(18ºC + 273 = 291 K\) Step 3: This one is tricky. We need to manipulate the Ideal Gas Equation to incorporate density into the equation. *Write down all known equations: \(PV = nRT\) \(d=(m/V)\) ; where d=density, m=mass, V=Volume \(m=M * n\) ; where m=mass, M=molar mass, n=moles *Now take the density equation. \(d=(m/V)\) *Keeping in mind \(m=M * n\)...replace \((M * n)\) for \(mass\) within the density formula. \(d=(M*n)/V\) > \((d/M) = (n/V)\) *Now manipulate the Ideal Gas Equation \(PV = nRT\) \((n/V) = (P/RT)\) *Hey look! \((n/V)\) is in both equations. \((n/V) = (d/M)\) \((n/V) = (P/RT)\) *Now combine them please. \((d/M) = (P/RT)\) *Isolate density. \(d = (PM/RT)\) Step 4: Now plug in the information you have. \(d = (PM/RT)\) \(d = [(0.3263atm)(2*14.01g/mol)/(.08206Latm/Kmol)(291K)\) \(d = 0.3828 g/L\) 
Example 4 

Find the volume, in mL, when 7.00 g of O2 and 1.50 g of Cl2are mixed in a container with a pressure of 482 atm and at a temperature of 22º C. SOLUTION Step 1: Write down your given information P = 482 atm V = ? n = ? R = 0.0820574 L·atm·mol^{1} K^{1} T = 22º C + 273 = 295K 1.50g Cl_{2} 7.00g O_{2} Step 2: Find the total moles of the mixed gases in order to use the Ideal Gas Equation. \(n_{total} = n_{O2}+ n_{Cl2}\) \(= [7.0 g O2 * (1 mol O2/32.00 g O2)] + [(1.5 g Cl2 * (1 mol Cl2/70.905 g Cl2)]\) \(= 0.2188 mol O2 + 0.0212 mol Cl2\) \(= 0.24 mol\) Step 3: Now that you have moles, plug in your information in the Ideal Gas Equation. \(V= nRT/P\) \(V= (.24mol)(.08206Latm/Kmol)(295K)/(482atm)\) \(V= .0121L\) Step 4: Almost done! Now just convert the liters to milliliters. \(.0121L * (1000ml/1L) = 12.1mL\) 
Example 5 

A 3.00 L container is filled with Ne(g) at 770 mmHg at 27^{o}C. A 0.633g sample of CO_{2} vapor is then added. What is the partial pressure of CO_{2} and Ne in atm? What is the total pressure in the container in atm? SOLUTION Step 1: Write down all given information, and convert as necessary. Before: P = 770mmHg > 1.01atm V = 3.00L n_{Ne}=? T = 27^{o}C > 300K Other Unknowns: n_{(co2)}= ? \(n_{(co2)} = 0.633g CO_{2} * (1 mol/44g) = .0144mol CO_{2}\) Step 2: After writing down all your given information, find the unknown moles of Ne. \(n_{(Ne)} = PV/RT\) \(n_{(Ne)} = (1.01atm)(3.00L)/(.08206atmL/molK)(300K)\) \(n_{(Ne)} = .123 mol\) Because the pressue of the container before the CO_{2} was added contained only Ne, that is your partial pressure of Ne. After converting it to atm, you have already answered part of the question! \(P_{(Ne)} = 1.01atm\) Step 3: Now that have pressure for Ne, you must find the partial pressure for CO_{2}. Use the ideal gas equation. Ne CO_{2} \((PV/nRT)\) = \((PV/nRT)\) but because Volume (V) and Temperature (T) and the Gas Constant (R) are constants, you may remove them from the equation. \((P/n)_{Ne}\) = \((P/n)_{co2}\) \((1.01atm/.123mol)_{Ne}\) = \((P/.0144mol)_{co2}\) pressure of CO_{2} \((P) = 0.118 atm\) This is the partial pressure CO_{2}. Good job! Step 4: Now find total pressure. \(P_{(total})= P_{(Ne)} + P_{(co2)}\) \(P_{(total})= 1.01atm + 0.188atm\) \(P_{(total)}= 1.20atm\) 
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