The Ideal Gas Law: Density - Video Tutorials & Practice Problems
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Under certain conditions of pressure and temperature, the Ideal Gas Law can determine the density of a gas.
The Ideal Gas Law (Density)
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concept
The Ideal Gas Law: Density
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Recall that density represents the amount of mass per unit of volume. And when it comes to our density formula, it's density equals mass over volume. Now gases are much less dense than solids and liquids. So when it comes to their density, we're gonna say their density is in grams per liter. M here is still mass of the gas in grams, and then here volume, v will be in liters. So again, when it comes to density it's mass over volume, and gas is being much less dense, don't use grams per milliliter or grams per centimeters cubed, but instead use grams per liter.
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example
The Ideal Gas Law: Density Example 1
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Here it says an unknown gas sample has a density of 1.70 grams per liter. If a if a sample has a volume of a 120 ml, what is its mass in grams? Alright. So here we're given 2 values. Well, we wanna start out with the 120 ml's because it just has 1 unit by itself, easier to manipulate. So we're gonna start out with 120 milliliters, realize that we need to isolate our grams, which are found here, and in order to isolate those grams I need to cancel out the liters. So that tells me that I need to convert my 120 ml's into liters first. So remember, 1 milli is 10 to the negative 3 liters. Now that I have liters, I can bring in my density which is 1.7 grams per 1 liter. So So here, liters cancel out and I'll be left with grams at the end. When I punch that in that gives me 0.204 grams of my unknown gas. Here my number has 3 sig figs because 1.70 has 3 sig figs, and this has 4 sig figs. Remember, we wanna go with the least number of sig figs when when it's reasonable. Here, 0.204 grams is a reasonable answer for unknown gas.
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concept
The Ideal Gas Law: Density
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1m
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Now we can say that the ideal gas law can be used to determine the density of a gas under certain pressure and temperature conditions. Here we're going to say that the new derived version of the ideal gas law, which we can apply to density is d equals p times m over r t. And to help us remember this order, just remember dreams push me over rough times. Here dreams, d stands for density. Push me is pressure times molar mass over r times t, rough times. So if you can remember this phrase it's a great way to remember this version of the ideal gas law when it relates to density. Now if you wanna look and see how we derive this formula, you can click on to the next video, but only do so if your professor really cares on how you derive these different types of formula. If they don't, then just remember this phrase. After that, go to the next series of videos and let's put this formula to practice.
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concept
The Ideal Gas Law: Density
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1m
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So if you clicked on this video that means you wanna see how we derive this new version of the ideal gas law. So it all begins with our density formula. Remember, density equals m over v. Remembering this, we move on to the ideal gas law where we're dealing with molar mass. Remember, molar mass equals mrt overpv, and we just said that m over v represents density. So take them out and put in density. We'd still have r t n over p. We need to isolate density, so just use algebra to isolate it. Multiply both sides by pressure, so molar mass times pressure equals d r of t. Isolate our density by dividing out our t, and we see now that density equals p times m over r t. So that's how we went about and isolated and derived this new formula for density. Now this is how we derived it, but just remember, dreams push me over rough times. Knowing that helps you to write the formula very quickly. Now that we've seen this formula, click on to the next video and let's put it to work.
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example
The Ideal Gas Law: Density Example 2
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2m
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here we're told that a gaseous compound of nitrogen and hydrogen is found to have a density of 0.977 g per leader at 0.69474 atmospheres and 3 73.15 kelvin. What is the molecular formula of the compound? We'll realize here in this question they're giving us density. They're giving us pressure and they're giving us temperature. With this information, I could find the more mass of this unknown gas. So here we're going to say that dreams push me over rough times and we're going to say that we have density. We have pressure, we have temperature. And we always know what our is. We just need toe isolate our Mueller Mass. So multiply both sides by rt. So are our tee times d equals p times Mueller Mass divide both sides by pressure. So Mueller mass equals d r t overpay. We'll take the information given to us. Our density is 0.977 g per leader. Our is our gas constant temperatures already in Kelvin and pressure is already in atmospheres. Doing this will see that we isolate So Adam atmospheres air gone. Kelvin's air gone leaders are gone, we'll have grams per mole. So we plug this into our formula. We're going to get here as our mass 06 g per mole. So here's our Moeller Mass. All we do now is we look at the different compounds and we would just calculate their Mueller masses and see which one comes closest to this 43.6 g per mole. Now, if you did this correctly, you would see that the answer would have to be Option C. It's the one with the mass closest to our answer. It has a Mueller Mass. Approximately equal to 43.38 g per mole. So option C would be the answer for this particular question. So realize they're giving us density, pressure and temperature. That means we can use the density form of the ideal gas law to help us find Moeller Mass and then use that information to compare to the molar masses off all these gasses
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Problem
Problem
Consider two containers of gases at the same temperature. One has helium at a pressure of 1.00 atm. The other contains carbon dioxide with the same density as the helium gas. What is the pressure of the carbon dioxide gas sample?
A
0.063 atm
B
0.091 atm
C
0.133 atm
D
1.51 atm
E
2.71 atm
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Problem
Problem
Determine the molecular formula of a gaseous compound that is 49.48% carbon, 5.19% hydrogen, 28.85% nitrogen, and 16.48% oxygen. At 27°C, the density of the gas is 1.5535 g/L and it exerts a pressure of 0.0985 atm.