The Ideal Gas Law is a foundational principle in physical science that mathematically describes the relationship between the measurable properties of a gas. This relationship allows scientists and engineers to predict how gases will react when their surrounding conditions change, such as when a gas is heated, cooled, or compressed. By connecting the four primary characteristics of a gas—pressure, volume, quantity, and temperature—the law provides a tool for understanding the physical behavior of gaseous substances. It functions as a simple model that approximates the state of most real gases under ordinary conditions.
Understanding the Ideal Gas Law Equation
The mathematical expression that defines this relationship is written as \(PV = nRT\). This single equation consolidates several older, empirically derived gas laws into one unified formula. It describes the state of an “ideal” gas, a hypothetical substance composed of particles that have no volume and experience no intermolecular forces. While no real gas perfectly fits this description, the model is highly accurate for gases at low pressures and high temperatures.
Each letter in the equation represents a specific, measurable property of the gas system. \(P\) stands for the pressure, and \(V\) is the volume of the container. The term \(n\) represents the amount of gas present, measured in moles (mol). \(T\) denotes the absolute temperature, and \(R\) is the Universal Gas Constant, a proportionality factor that balances the equation.
Isolating Pressure in the Formula
To determine the pressure of a gas, \(P\), when the other three variables are known, the \(PV = nRT\) equation must be algebraically rearranged. Since \(P\) is multiplied by the volume term (\(V\)), isolating \(P\) requires dividing both sides of the equation by \(V\). This manipulation transforms the original formula into the working equation for pressure.
The resulting formula for pressure is \(P = \frac{nRT}{V}\). This structure shows that pressure is directly proportional to the amount of gas (\(n\)) and its absolute temperature (\(T\)). Conversely, pressure is inversely proportional to the volume (\(V\)), meaning decreasing the volume increases the pressure if all other factors remain constant.
Critical Units and the Universal Gas Constant
Accurately calculating pressure depends entirely on using consistent units for every variable, which is dictated by the value chosen for the Universal Gas Constant, \(R\). Temperature (\(T\)) must be expressed on the absolute Kelvin (K) scale for all calculations. Using the Celsius or Fahrenheit scale is incorrect because a temperature of \(0\) on those scales would incorrectly make the entire right side of the equation equal to zero. To convert from Celsius to Kelvin, add \(273.15\) to the Celsius temperature.
The volume (\(V\)) is typically measured in Liters (L), and the quantity of gas (\(n\)) must be in moles (mol). The constant \(R\) links these specific units together, and its numerical value changes depending on the desired unit for the final pressure. The most common value used for general chemistry problems is \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\). When this specific value of \(R\) is used, the calculated pressure (\(P\)) will automatically be in atmospheres (atm).
If the problem provides volume in milliliters (mL) or temperature in Celsius (\(°C\)), these units must be converted before starting the calculation to ensure consistency with the units embedded in the gas constant. Failure to convert the temperature to Kelvin or the volume to Liters is the most common source of error in these types of problems. Other values of \(R\), such as \(8.314 \frac{J}{mol \cdot K}\) or \(8.314 \frac{L \cdot kPa}{mol \cdot K}\), are used if the final pressure is required in Pascals (Pa) or kilopascals (kPa).
Applying the Law Through a Worked Example
To demonstrate how these concepts combine, consider a problem where you need to find the pressure exerted by \(5.0\) moles of a gas contained in a \(15.0\) Liter tank at a temperature of \(27.0\) degrees Celsius. The first step is to list all known variables and ensure their units are compatible with the chosen gas constant, \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\). The initial knowns are \(n = 5.0\) mol, \(V = 15.0\) L, and \(T = 27.0\) \(°C\).
The temperature value requires conversion from Celsius to the absolute Kelvin scale. You must add \(273.15\) to the Celsius temperature, which yields \(27.0 + 273.15 = 300.15\) K. Now all variables are in the correct units. The next step is to substitute these values into the rearranged Ideal Gas Law formula, \(P = \frac{nRT}{V}\).
Substituting the values into the equation looks like this: \(P = \frac{(5.0 \text{ mol}) \cdot (0.0821 \frac{L \cdot atm}{mol \cdot K}) \cdot (300.15 \text{ K})}{15.0 \text{ L}}\). Notice how the units of moles, Liters, and Kelvin will all cancel out, leaving only the unit of atmospheres. This is a good way to double-check that the correct constant and units were used.
The calculation of the numerator is \(5.0 \cdot 0.0821 \cdot 300.15\), which equals approximately \(123.26\). This value is then divided by the volume of \(15.0\) L. Performing the final division, \(123.26 / 15.0\), results in a pressure of \(8.217\). The final answer is \(8.217 \text{ atm}\), representing the pressure of the gas inside the container.