The Combined Gas Law provides a mathematical tool for understanding how gases behave under changing conditions. This principle establishes a direct relationship between the three primary measurable properties of a fixed amount of gas: pressure, volume, and temperature. By synthesizing the findings of Boyle, Charles, and Gay-Lussac, the Combined Gas Law allows prediction of a gas’s state when two of its properties are altered simultaneously. Its utility lies in its ability to model gas behavior accurately, whether predicting the expansion of a balloon or calculating the capacity of an industrial gas tank. Mastering this law involves applying a straightforward algebraic formula to solve for any missing value.
Defining the Combined Gas Law Variables
The mathematical representation of this principle is \(P_1V_1/T_1 = P_2V_2/T_2\), a formula that relates the initial state of a gas to its final state. In this equation, ‘P’ represents the pressure exerted by the gas. Pressure is commonly measured in units such as atmospheres (atm), kilopascals (kPa), or millimeters of mercury (mmHg).
The letter ‘V’ stands for the volume occupied by the gas, which is the amount of space it fills, typically measured in liters (L) or milliliters (mL). The final variable, ‘T,’ represents the absolute temperature of the gas, a measure of the average kinetic energy of its particles.
The subscripts ‘1’ and ‘2’ differentiate between the gas’s condition at two distinct points in time. Subscript ‘1’ denotes the initial state (pressure, volume, and temperature) before any change occurs, while subscript ‘2’ refers to the final state after conditions have been altered.
The units for pressure and volume must remain consistent throughout the calculation. For instance, if \(P_1\) is in atmospheres, \(P_2\) will also be calculated in atmospheres. This consistency ensures that the ratios remain valid on both sides of the equation.
Mandatory Unit Conversions for Calculation
The Combined Gas Law requires the use of an absolute temperature scale, meaning temperature must always be expressed in Kelvin (K). The Celsius scale is unsuitable because its zero point is based on the freezing point of water, which could lead to division by zero or negative values in the equation.
The Kelvin scale is the absolute thermodynamic temperature scale, beginning at absolute zero. This ensures all Kelvin temperatures are positive, allowing the temperature ratio in the gas law equation to accurately reflect the proportional change in kinetic energy.
Converting a Celsius temperature (\(C\)) to Kelvin (\(K\)) uses the formula \(K = C + 273.15\). For example, \(25.0^\circ \text{C}\) converts to \(298.15\text{ K}\). This conversion must be performed on both the initial temperature (\(T_1\)) and the final temperature (\(T_2\)) if they are provided in Celsius.
The units for pressure and volume offer flexibility, provided they match on both sides of the equation. If \(V_1\) is in milliliters and \(P_1\) is in kilopascals, the calculated values for \(V_2\) and \(P_2\) will emerge in milliliters and kilopascals, respectively.
Methodology for Isolating the Unknown Variable
Solving a Combined Gas Law problem requires isolating the unknown variable, which could be any of the six terms in the equation. The first step is identifying the missing variable, such as a final state property (\(P_2\)) or an initial condition (\(V_1\)). Standard algebraic rules must then be applied to isolate it.
The process typically begins by cross-multiplying to remove fractions, transforming \(P_1V_1/T_1 = P_2V_2/T_2\) into the linear form \(P_1V_1T_2 = P_2V_2T_1\). Rearranging this linear form is simpler than manipulating the fractional form directly. The goal is to move all known variables away from the unknown using inverse operations.
For example, if the final volume (\(V_2\)) is the unknown, isolate \(V_2\) by dividing both sides by \(P_2\) and \(T_1\). The resulting formula is \(V_2 = (P_1V_1T_2) / (P_2T_1)\).
If the unknown is \(T_2\), the equation is rearranged to \(T_2 = (P_2V_2T_1) / (P_1V_1)\). This algebraic preparation ensures that substituting numerical values yields the correct answer.
Solving a Combined Gas Law Problem
To illustrate the complete process, consider a gas sample that initially occupies \(5.0 \text{ L}\) at \(1.5 \text{ atm}\) and \(27.0^\circ \text{C}\). The gas is moved to new conditions where the volume is \(2.0 \text{ L}\) and the temperature is \(127.0^\circ \text{C}\); the task is to find the final pressure (\(P_2\)). The first step is to perform the mandatory temperature conversions.
The initial temperature \(T_1\) of \(27.0^\circ \text{C}\) converts to \(27.0 + 273.15 = 300.15 \text{ K}\). The final temperature \(T_2\) of \(127.0^\circ \text{C}\) converts to \(127.0 + 273.15 = 400.15 \text{ K}\). The known values are \(P_1 = 1.5 \text{ atm}\), \(V_1 = 5.0 \text{ L}\), \(V_2 = 2.0 \text{ L}\), and the calculated temperatures.
Since the goal is to find \(P_2\), the linear Combined Gas Law formula \(P_1V_1T_2 = P_2V_2T_1\) must be algebraically rearranged. Dividing both sides by \(V_2\) and \(T_1\) yields the working equation: \(P_2 = (P_1V_1T_2) / (V_2T_1)\).
Substituting the known numerical values into this equation allows for the final calculation: \(P_2 = (1.5 \text{ atm} \times 5.0 \text{ L} \times 400.15 \text{ K}) / (2.0 \text{ L} \times 300.15 \text{ K})\). The volume (L) and temperature (K) units cancel out, leaving the final answer in atmospheres (atm). Performing the arithmetic yields \(P_2 \approx 5.0 \text{ atm}\).