Gases

Ideal Gases

The Ideal Gas Law

The ideal gas law, given by the equation PV=nRTPV=nRT, describes the behavior of ideal gases, in which no intermolecular forces act on the gas particles, and the particles do not take up space.

Scientists want to be able to describe the behaviors of an ideal gas, that is, a theoretical gas in which no forces are acting on the gas particles, and the particles do not take up space. Intermolecular forces between the atoms or molecules of a gas are the reason that real gases differ from ideal gases. Real gases are not ideal, but understanding the way an ideal gas would behave gives scientists a way to predict how a real gas might behave under defined conditions.

The ideal gas law is a law that describes the behavior of ideal gases. It can estimate the way many gases will behave under many different conditions. The ideal gas law is represented by the equation PV=nRT PV=nRT , where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. There are many versions of R. The one used in each calculation depends on the units being used to measure pressure. For Pa, the SI unit, R=8.314m3Pa/KmolR=8.314\,\rm{m}^{3}{\rm{Pa/K}}{\cdot}{\rm{mol}}. The units can all be converted from one to the other: 8.314m3Pa/Kmol=8.314J/Kmol8.314\;{\rm m^{3}{Pa}}/{\rm K}{\cdot}{\rm{mol}}=8.314\;{\rm{J/K}}{\cdot}{\rm{mol}} because m3Pa=J{\rm{m}^{3}{Pa}}={\rm{J}}.

Useful Values of the Gas Constant, R

8.314 m3Pa/K⋅mol
8.314 J/K⋅mol
0.082057 atm⋅L/K⋅mol

The gas constant (R) can be expressed using different units. The choice of which version to use depends on the units that appear in other factors in the calculations where it is used.

It is important to review converting units so R can be in the unit that is needed. From this equation, it is obvious that pressure is directly proportional to the number of moles of the gas. For example, the more gas there is, the more pressure it exerts. It is also directly proportional to the temperature of the gas: the hotter the gas, the more pressure it exerts. Pressure is indirectly proportional to volume: the greater the volume of a gas, the less pressure it exerts.

Other gas laws are derived from the ideal gas law and are useful for practical applications.

  • Boyle's law states that pressure of gas increases as volume decreases at constant temperature and moles of gas, represented by the equation P1V1=P2V2P_1V_1=P_2V_2.
  • Charles's law states that the volume of gas increases as temperature increases at constant pressure and moles of gas, represented by the equation V1T1=V2T2\frac{V_1}{T_1}=\frac{V_2}{T_2} .
  • Avogadro's law states that the equal volumes of gases at the same temperature and pressure have equal numbers of atoms or molecules, represented by the equation V1n1=V2n2\frac{V_1}{n_1}=\frac{V_2}{n_2}.
  • Gay-Lussac's law states that for an ideal gas with constant mass and volume, the pressure exerted on the container is proportional to its absolute temperature, represented by the equation P1T1=P2T2\frac{P_1}{T_1}=\frac{P_2}{T_2} .
Using these laws often involves standard temperature and pressure (STP), defined as 1 atmosphere (atm) of pressure and 0°C (273.15 K) temperature. A more precise definition of STP includes a pressure of 1 bar (100 kPa), but a pressure of 1 atm is more commonly used. The laws may also involve standard molar volume, defined as the volume occupied by one mole of an ideal gas at STP, equal to 22.4 L.

Gas Laws

The four gas laws describe the relationship between the pressure (P), volume (V), temperature (T), and number of moles (n) of a gas, but in each law, two of the factors remain constant while the other two factors can vary.
These laws and their equations can be used to describe and predict the behaviors of ideal gases.
Step-By-Step Example
Using the Ideal Gas Equation to Calculate the Mass of a Given Volume of Gas
Given a sample of neon gas at standard temperature and pressure (STP), what mass of gas occupies 5.85 liters?
Step 1
Rearrange the ideal gas equation to solve for n, the number of moles.
PV=nRTn=PVRT\begin{aligned}PV&=nRT\\n&=\frac{PV}{RT}\end{aligned}
Step 2
Use the given volume to calculate the number of moles of the gas.
n=PVRT=(1atm)(5.85L)(0.082057atmL/Kmol)(273.15K)=0.2610mol\begin{aligned}n&=\frac{PV}{RT}\\&=\frac{(1\;\rm{atm})(5.85\;\rm{L)}}{(0.082057\;\rm{atm}{\cdot}\rm{L/K}{\cdot}\rm{mol})(273.15\;\rm{K})}\\&=0.2610\;\rm{mol} \end{aligned}
Solution
Use the molar mass of neon, 20.18 grams per mol (g/mol), to calculate the mass.
mass Ne=(moles Ne)(molar mass Ne)=(0.2610mol)(20.18g/mol)=5.27g\begin{aligned}{\text{mass }\rm{Ne}}&=({\text{moles }\rm{Ne}})({\text{molar mass }\rm{Ne}})\\&=(0.2610\;{\rm{mol}})(20.18\;{\rm{g/mol}})\\&=5.27\;{\rm{g}}\end{aligned}

Note that the chosen value for R matched the units given for pressure and temperature. It's important to use dimensional analysis to make sure the desired unit is obtained.

The laws derived from the ideal gas law can be used in simpler cases. For example, if 15 liters of carbon dioxide gas at 25°C is heated to 65°C, calculate the new volume under constant pressure. Using the ideal gas law, under constant pressure conditions, the P is eliminated, and initial and final states may be determined via Charles's law.
V1T1=V2T2V2=V1T2T1=(15L)(65°C)25°C=39L\begin{aligned}\frac{V_1}{T_1}&=\frac{V_2}{T_2}\\V_2&=\frac{V_1T_2}{T_1}\\&=\frac{(15\;\rm{L)}(65\degree\rm{C})}{25\degree\rm{C}}\\&=39\;\rm{L}\end{aligned}

Kinetic Molecular Theory

Kinetic molecular theory, a model that describes the macroscopic behavior of gases based on their microscopic components, helps explain effusion of gases, in which gases move through small openings in solids, and diffusion of gases, in which gas particles move from an area of higher concentration to an area of lower concentration.

The ideal gas law and the laws derived from it are based on the kinetic molecular theory of gases. Kinetic molecular theory is a theory involving the relationship between temperature, pressure, and volume that states that the average kinetic energy of a gas is proportional to its temperature. This theory relies on five assumptions.

  • The size of each particle of a gas is negligibly small compared to the distances between particles.
  • Gas particles exert no attractive or repulsive forces on one another or the walls of their container.
  • Gas particles travel in a continuous, straight-line motion until they collide with other gas particles or the walls of their container.
  • When gas particles collide, there is no gain or loss of energy.
  • The average kinetic energy of all particles of a gas is proportional to the absolute temperature of the gas.

Absolute temperature, also known as thermodynamic temperature, is the temperature as compared to absolute zero, the cessation of all motion, even subatomic motion, measured in kelvins (K).

The assumption underlying kinetic molecular theory leads to gas effusion and diffusion. Effusion is the process by which gases move through small openings in solids, one particle at a time. Diffusion is the process by which gas moves from an area of higher concentration to an area of lower concentration.

Effusion and Diffusion

Effusion is the movement of gas particles through small holes, or pinholes, in a solid, one particle at a time. Diffusion is the movement of gas particles from an area of higher concentration to an area of lower concentration.
In effect, the negligibly small size of gas particles combined with their constant motion leads to effusion. The tiny particles can work their way through holes in a solid because they are very small and moving all the time. In order to do this, the diameter of the hole must be less than the gas particles' mean free path, that is, the average distance that a gas particle travels before colliding with another gas particle. If the hole's diameter is larger than the mean free path of the gas, the gas particles might move into the hole and back out again. In effect, the hole simply becomes a smaller container for the gas. However, if the hole's diameter is smaller than the mean free path of the gas, once the gas particle enters the hole, it passes through without passing back again. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its mass. Let rate1 represent the rate of effusion (or diffusion) of the first gas, rate2 represent the rate of effusion (or diffusion) of the second gas, M1 represent the molar mass of the first gas, and M2 represent the molar mass of the second gas.
rate1rate2=M2M1\frac{{\rm{rate}}_1}{{\rm{rate}}_2}=\sqrt{\frac{M_2}{M_1}}
Graham's law can be used to calculate unknowns about gases when some properties have been empirically determined.
Step-By-Step Example
Using Graham’s Law to Determine the Mass of an Unknown Gas
Suppose 1.740×104mol1.740\times10^{-4}\;\rm{mol} of argon effuses through a hole in 81.3 s. Under identical conditions, 1.177×103mol1.177\times10^{-3}\;\rm{mol} of an unknown gas effuses through the same hole in 86.9 s. What is the mass of the unknown gas?
Step 1
First, determine the rate of effusion of the argon, carrying one extra digit for the intermediate calculation.
rate1=1.740×104mol81.3s=2.140×106mol/s{\rm{rate}}_1=\frac{1.740\times10^{-4}\;\rm{mol}}{81.3\;\rm s}=2.140\times10^{-6}\;\rm{mol/s}
Step 2
Next, determine the rate of effusion of the unknown gas.
rate2=1.177×103mol86.9s=1.354×105mol/s=13.54×106mol/s{\rm{rate}}_{2}=\frac{1.177\times10^{-3}\;\rm{mol}}{86.9\;\rm s}=1.354\times10^{-5}\;\rm{mol/s}=13.54\times10^{-6}\;\rm{mol/s}
Solution
According to the periodic table, the molar mass of argon is 39.95 g/mol. Use this in Graham's law to determine the molar mass of the unknown gas.
rate1rate2=M2M1M2=M1(rate1rate2)2=(39.95g/mol)(2.140×106mol/s13.54×106mol/s)2=0.998g/mol\begin{aligned}\frac{{\rm{rate}}_1}{{\rm{rate}}_2}&=\sqrt{\frac{M_2}{M_1}}\\{M_2}&={M_1}\!\left(\frac{{\rm{rate}}_1}{{\rm{rate}}_2}\right)^{2}\\&=(39.95\;{\rm{g/mol}})\!\left(\frac{2.140\times10^{-6}\;{\rm{mol/s}}}{13.54\times10^{-6}\;{\rm{mol/s}}}\right)^{2}\\&=0.998\;{\rm{g/mol}}\end{aligned}
The molar mass of the unknown gas is 0.998 g/mol.