The pressures of ideal gases increase as they are squeezed into smaller spaces and decrease as they expand into larger areas. This relationship is called Boyle’s Law, after Robert Boyle. It is written mathematically as k = P x V or, more simply, k = PV, where k represents the constant relationship, P represents pressure and V represents volume. [2] X Research source Pressures may be given using one of several possible units. One is the pascal (Pa), defined as a force of one newton applied over a square meter. Another is the atmosphere (atm), defined as the pressure of Earth’s atmosphere at sea level. A pressure of 1 atm is equal to 101,325 Pa. [3] X Research source The temperatures of ideal gases increase as their volumes increase and decrease as their volumes decrease. This relationship is called Charle’s Law after Jacques Charles. It is written mathematically as k = V / T, where k represents the constant relationship between the volume and temperature, V again represents the volume, and T represents temperature. [4] X Research source [5] X Research source Temperatures for gases in this equation are given in degrees Kelvin, which are found by adding 273 to the number of degrees Celsius in the gas’ temperature. These two relationships can be combined into a single equation: k = PV / T, which can also be written as PV = kT.
Conventional mass is measured in grams or, if there is a sufficiently large mass, kilograms. Because of how lightweight gases usually are, they are also measured with another form of mass called molecular mass or molar mass. Molar mass is defined as the sum of the atomic weights of each atom in the compound the gas is composed of, with each atom compared against the standard value of 12 for carbon’s molar mass. [6] X Research source Because atoms and molecules are too small to work with, quantities of gases are defined in moles. The number of moles present in a given gas can be found by dividing the mass by the molar mass and can be represented by the letter n. We can replace the arbitrary k constant in the gas equation with the product of n, the number of moles (mol), and a new constant R. The equation can now be written nR = PV/T or PV = nRT. [7] X Research source The value of R depends on the units used to measure the gases’ pressures, volumes, and temperatures. For volume in liters, temperature in degrees Kelvin, and pressure in atmospheres, its value is 0. 0821 L atm/K mol. This may also be written 0. 0821 L atm K-1 mol -1 to avoid using the division slash with units of measure. [8] X Research source
Dalton’s Law can be written in equation form as Ptotal = P1 + P2 + P3 … with as many addends after the equal sign as there are gases in the mixture. The Dalton’s Law equation can be expanded on when working with gases whose individual partial pressures are unknown, but for which we do know their volumes and temperatures. A gas’ partial pressure is the same pressure as if the same quantity of that gas were the only gas in the container. For each of the partial pressures, we can rewrite the ideal gas equation so that instead of the form PV = nRT, we can have only P on the left side of the equal sign. To do this, we divide both sides by V: PV/V = nRT/V. The two V’s on the left side cancel out, leaving P = nRT/V. We can then substitute for each subscripted P on the right side of the partial pressures equation: Ptotal =(nRT/V) 1 + (nRT/V) 2 + (nRT/V) 3 …
Our partial pressure equation becomes Ptotal = Pnitrogen + Poxygen + Pcarbon dioxide. Since we’re trying to find the pressure each gas exerts, we know the volume and temperature, and we can find how many moles of each gas is present based on the mass, we can rewrite this equation as : Ptotal =(nRT/V) nitrogen + (nRT/V) oxygen + (nRT/V) carbon dioxide
For our first gas, nitrogen (N2), each atom has an atomic weight of 14. Because the nitrogen is diatomic (forms two-atom molecules), we have to multiply 14 by 2 to find that the nitrogen in our sample has a molar mass of 28. We then divide the mass in grams, 10 g, by 28, to get the number of moles, which we’ll approximate as 0. 4 mol of nitrogen when rounding to the nearest tenth. For our second gas, oxygen (O2), each atom has an atomic weight of 16. Oxygen is also diatomic, so we multiply 16 by 2 to find the oxygen in our sample has a molar mass of 32. Dividing 10 g by 32 gives us approximately 0. 3 mol of oxygen in our sample. Our third gas, carbon dioxide (CO2), has 3 atoms: one of carbon, with an atomic weight of 12; and two of oxygen, each with an atomic weight of 16. We add the three weights: 12 + 16 + 16 = 44 as the molar mass. Dividing 10 g by 44 gives us approximately 0. 2 mol of carbon dioxide.
For simplicity’s sake, we’ve left out the units of measure accompanying the values. These units will cancel out after we do the math, leaving only the unit of measure we’re using to report the pressures in.
For the partial pressure of nitrogen, we multiply 0. 4 mol by our constant of 0. 0821 and our temperature of 310 degrees K, then divide by 2 liters: 0. 4 * 0. 0821 * 310/2 = 5. 09 atm, approximately. For the partial pressure of oxygen, we multiply 0. 3 mol by our constant of 0. 0821 and our temperature of 310 degrees K, then divide by 2 liters: 0. 3 *0. 0821 * 310/2 = 3. 82 atm, approximately. For the partial pressure of carbon dioxide, we multiply 0. 2 mol by our constant of 0. 0821 and our temperature of 310 degrees K, then divide by 2 liters: 0. 2 * 0. 0821 * 310/2 = 2. 54 atm, approximately. We now add these pressures to find the total pressure: Ptotal = 5. 09 + 3. 82 + 2. 54, or 11. 45 atm, approximately.
The Kelvin temperature will still be 310 degrees, and, as before, we have approximately 0. 4 mol of nitrogen, 0. 3 mol of oxygen, and 0. 2 mol of carbon dioxide. Likewise, we’ll still report the pressures in atmospheres, so we’ll use the value of 0. 0821 L atm/K mol for the R constant. Thus, our partial pressures equation still looks the same at this point: Ptotal =(0. 4 * 0. 0821 * 310/2) nitrogen + (0. 3 *0. 0821 * 310/2) oxygen + (0. 2 * 0. 0821 * 310/2) carbon dioxide.
Adding 0. 4 + 0. 3 + 0. 2 = 0. 9 mol of gas mixture. This further simplifies the equation to Ptotal = 0. 9 * 0. 0821 * 310/2.
There are 0. 4 mol of nitrogen, so 0. 4/0. 9 = 0. 44 (44 percent) of the sample, approximately. There are 0. 3 mol of nitrogen, so 0. 3/0. 9 = 0. 33 (33 percent) of the sample, approximately. There are 0. 2 mol of carbon dioxide, so 0. 2/0. 9 = 0. 22 (22 percent) of the sample, approximately. While the above approximate percentages add to only 0. 99, the actual decimals are repeating, so the sum would actual be a repeating series of 9s after the decimal. By definition, this is the same as 1, or 100 percent.
Multiplying 0. 44 * 11. 45 = 5. 04 atm, approximately. Multiplying 0. 33 * 11. 45 = 3. 78 atm, approximately. Multiplying 0. 22 * 11. 45 = 2. 52 atm, approximately.
