5.3
Applications of the Ideal Gas Law: Molar Mass, Density, and Volume
The volume occupied by one mole of a substance is its molar volume. The ideal gas law, PV = nRT, suggests that the volume of a given quantity of gas a…
All ideal gases conform, in behavior, to a particular relationship between pressure, volume, moles, and temperature as dictated by the ideal gas law.
In this equation, R is the ideal gas constant. Rearranging the equation allows any one of the variables to be calculated as long as the other three are known.
For example, what is the volume of one mole of an ideal gas under standard temperature and pressure conditions? Abbreviated as STP, these conditions are 0 °C or 273 K and 1 atm.
Rearranging the equation and substituting in the values for n (1 mole), temperature (273 K), pressure (1 atm), and the ideal gas constant (0.08206 L·atm/mol·K), one mole of an ideal gas occupies a volume of 22.4 liters. This is the molar volume at STP, which is also a good approximation for many common gases.
At higher temperatures and lower pressures, the gas expands and its molar volume is larger than it is at standard conditions. At lower temperatures and higher pressures, the molar volume is smaller.
Another useful quantity of a gas is its density. Recall that the number of moles, n, is equal to the mass of the gas divided by its molar mass. Substituting this relationship into the ideal gas equation, and then rearranging, yields an expression for mass over volume or density.
From this equation, the density of a gas is directly proportional to its molar mass. This is why helium balloons float away when released outside. The molar mass, and thus density, of helium, is much less than that of air, which is primarily nitrogen and oxygen.
Also, notice that density and temperature are inversely related. This is observed when piloting a hot air balloon. Turning on the burner heats up the air molecules within the balloon and they move faster.
The pressure in the balloon increases, but the balloon is designed so that some of the air escapes. This makes the air in the balloon less dense than the surrounding air. Because of this difference in density, the balloon ascends.
Conversely, turning off the burner and opening the vent, allows the warm to escape. As the balloon contracts, outside air enters, increasing the density in the balloon to that of the surroundings. Then, because of the weight in the basket, the balloon descends.
The equation, when rearranged, also allows us to calculate the molar mass of an unknown gas.
Suppose an unknown gas with a mass of 12.5 grams occupies a volume of 6.08 liters and exerts a pressure of 1.2 atm at 40.0 °C.
The density of the gas is known from the given mass and volume. Then, the temperature in degree Celsius is converted to units of kelvin and substituted into the equation along with the values for pressure and the gas constant.
Solving for M yields a molar mass of 44 g/mol. Therefore, carbon dioxide is the unknown gas.
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Q1: What is the molar volume of an ideal gas at standard temperature and pressure?
At standard temperature and pressure (STP)—0°C or 273 K and 1 atm—one mole of any ideal gas occupies approximately 22.4 liters. This standard molar volume applies regardless of the gas's chemical identity. Using the ideal gas law and rearranging for volume, you can calculate that 0.5 moles occupies 11.2 liters and 2 moles occupies 44.8 liters at STP.
Q2: How does temperature affect the molar volume of a gas?
Molar volume is inversely related to temperature. At higher temperatures and lower pressures, gases expand and their molar volume increases beyond the standard 22.4 liters. Conversely, at lower temperatures and higher pressures, molar volume decreases. This relationship is fundamental to understanding gas behavior and is directly derived from the ideal gas law equation.
Q3: Why is gas density directly proportional to molar mass?
Rearranging the ideal gas law and substituting the definition of molar mass (mass divided by moles) yields the density equation: d = PM/RT. This shows density is directly proportional to molar mass. Helium balloons float because helium's molar mass is much less than air's, making helium less dense. Carbon dioxide from fire extinguishers is denser than air because CO₂ has a higher molar mass than nitrogen and oxygen.
Q4: How does temperature inversely affect gas density?
The density equation d = PM/RT shows that density is inversely proportional to temperature. Heating gas molecules increases their speed and causes expansion, reducing density. Hot air balloons exploit this principle: turning on the burner heats air inside the balloon, lowering its density below surrounding air, causing ascent. Turning off the burner allows warm air to escape, increasing density and causing descent.
Q5: How can you determine the molar mass of an unknown gas using the ideal gas law?
Rearrange the ideal gas law to isolate molar mass: M = dRT/P, where density equals mass divided by volume. Given an unknown gas's mass, volume, pressure, and temperature, calculate its density first. Convert temperature to Kelvin, then substitute all values into the equation. For example, a 12.5-gram gas sample occupying 6.08 liters at 1.2 atm and 40°C yields a molar mass of 44 g/mol, identifying it as carbon dioxide.
Q6: What variables does the ideal gas law relate to calculate gas properties?
The ideal gas law, PV = nRT, relates four fundamental variables: pressure (P), volume (V), number of moles (n), and temperature (T), with R as the ideal gas constant (0.08206 L·atm/mol·K). Rearranging this equation allows you to calculate any one variable if the other three are known. This universal relationship applies to all ideal gases regardless of their chemical identity, making it essential for solving gas behavior problems.
Q7: Why does a hot air balloon rise when the burner is turned on?
Turning on the burner heats air molecules inside the balloon, causing them to move faster and the pressure to increase. Some air escapes through the balloon's design, making the air inside less dense than the surrounding atmosphere. This density difference creates buoyancy, allowing the balloon to ascend. Conversely, turning off the burner and opening the vent allows warm air to escape, increasing internal density to match surroundings, causing descent.