Video: Density of a Mixture of Gasses at a Particular Pressure and Temperature

Find the density of air at a pressure of 1.00 atm and a temperature of 20.0°C, assuming that air is 78% N₂, 21% O₂, and 1% Ar. Find the density of the atmosphere on Venus, assuming that the atmosphere is 96% CO₂ and 4% N₂, with a temperature of 737 K and a pressure of 92.0 atm.

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Video Transcript

Find the density of air at a pressure of 1.00 atmospheres and a temperature of 20.0 degrees Celsius, assuming that air is 78 percent nitrogen, 21 percent oxygen, and one percent argon. Find the density of the atmosphere on Venus, assuming that the atmosphere is 96 percent carbon dioxide and four percent nitrogen, with a temperature of 737 kelvin and a pressure of 92.0 atmospheres.

We’ll start by solving for the density of air and we’ll call this density 𝜌 sub 𝑎. 𝜌 sub 𝑎 will depend on the densities of the gases that make up air in the proportions that those gases are present. We can write that the density of air is equal to 0.78 times the density of nitrogen plus 0.21 times the density of oxygen plus 0.01 times the density of argon.

If we can solve for the densities of these three gases then, we’ll be able to solve for 𝜌 sub 𝑎. To solve for these values, we’ll use the ideal gas law, which says the pressure of an ideal gas times its volume is equal to the number of moles of the gas times the gas constant times its temperature.

When solving for the density 𝜌 of an object, we know that in general that density is equal to the mass of the object divided by the volume that it occupies. What we’ll do now is work with the ideal gas law to reform it so it expresses gas density. As a first step, we’ll divide both sides by the pressure 𝑃 so that volume is by itself on the left.

We can then recognize that 𝑛 the number of moles of the gas is equal to the mass of the ideal gas divided by its molar mass. We can plug that expression in for 𝑛. And then, if we divide both sides of this equation by the mass 𝑚, we find that the volume of the gas divided by its mass is equal to 𝑅𝑇 over the molar mass times the pressure.

As we consider the left side of this equation, we see that it’s equal to the inverse of density. So to make it more like density, we’ll invert both sides of our ideal gas equation. We now have an expression that gives the density the mass per volume of an ideal gas in terms of its molar mass, the pressure of the gas, the gas constant, and its temperature.

We can now calculate the density for the three atmospheric gases that make up air. We’ll start by figuring out the molar masses of these three gases by consulting the periodic table of elements. From the table, we find that the molar mass of atmospheric nitrogen is 28 grams per mole, the molar mass of atmospheric oxygen is 32 grams per mole, and the molar mass of argon is 40 grams per mole.

We can now insert this information into our equation for 𝜌 sub 𝑎. In this expression, we’ve inserted the molar mass for each of our three atmospheric gases in units of kilograms per mole and we factored out from each term the pressure as well as the gas constant and temperature. The pressure we’re told is 1.00 atmospheres and the temperature which we’re given in units of degrees Celsius we convert to the kelvin scale by adding to 273.15 to 20.0.

We now only need to look up the gas constant 𝑅 and plug that in to solve for the density of air. When we do, we find that 𝑅 is 0.0821 liter atmospheres per mole kelvin. The one unit change we want to make to 𝑅 before plugging it into our equation is to convert the units of volume from liters to cubic meters. The purpose of that change is so that the final density of air that we calculate can be in units of kilograms per cubic meter.

With that goal in mind, we’ll develop a new gas constant we’ll call it 𝑅 sub 𝑚 for 𝑅 modified, which is equal to the original gas constant with its volume units converted from liters into cubic meters. And it’s this modified gas constant 𝑅 sub 𝑚 that we’ll plug into our equation. When we do calculate 𝜌 sub 𝑎, we find it’s 1.20 kilograms per cubic meter. That’s the density of air at the given pressure and temperature.

Now that we’ve solved for the density of air, we want to solve next for the density of the atmosphere on Venus. We can call that 𝜌 sub 𝑣. In the problem statement, we’re told the temperature and pressure of Venus’s atmosphere. And we’re also told that 96 percent of the atmosphere is carbon dioxide and four percent is atmospheric nitrogen. Like before then, we can write the density of this atmosphere in terms of the weighted densities of its constituent parts.

When it comes to calculating the densities of carbon dioxide and nitrogen, we can we reuse the molar mass of nitrogen we found earlier. And when we look up the molar mass of carbon dioxide, find it’s 44 grams per mole. Once again, we’ll plug in our molar masses in units of kilograms per mole and factor out the pressure, gas constant, and temperature of our ideal gas.

We’ll then plug in the given pressure and temperature for Venus’s atmosphere. And we’ll once again compute a modified gas constant 𝑅 sub 𝑚 based on 𝑅 and changing its volume units from liters into cubic meters, again with the goal that 𝜌 sub 𝑣 when we calculate it, we’ll have the units of kilograms per cubic meter.

Entering this expression on our calculator, we find that 𝜌 sub 𝑣 is 65.9 kilograms per cubic meter. That’s the density of the atmosphere of Venus.

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