Van der Waals Equation - Video Tutorials & Practice Problems
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The Van der Waals Equation is used when dealing with real, non-ideal gases.
Understanding the Van der Waals Equation
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concept
Van der Waals Equation
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The van der Waals equation is an equation used for real gases. It combines the effects of attractive forces and gas volume to describe non ideal behavior, so the behavior of real gases. We're gonna say deviations from this ideal gas behavior happens at high pressures and low temperatures. Remember, an ideal gas is imaginary and ideal gases behave as though they are alone. This is not possible if the pressure is incredibly high inside the container. At high pressures, it forces gas molecules to come closer together so they can't be alone. And at lower temperatures, that also causes gases start to condense downward. This also forces them to become in closer contact with one another. Now, with the van der Waals equation we have 2 coefficients, which we call variables or van der Waals constants. The polarity coefficient is the van der Waals constant with the letter a, and it corrects for the attractive forces felt between gas molecules. The size coefficient, it is the Van der Waals constant b that corrects the volume of gas molecules. Now with the van der Waals constant b, what we need to realize is as we increase the molecular weight of a gas, then this causes an increase for this Van der Waals constant. So the greater the molecular weight of a gas, the greater its Van Der Waals constant b. Now that we've seen the whole idea of these coefficients, and we know that the Van der Waals equation is used for real gases, click on the next video and let's take a look at the formula involved.
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concept
Van der Waals Equation
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Now remember that the Van der Waals equation is used for real gases. You'll notice that there's no purple box because you're not expected to memorize it. If you do have to use it on an exam it's usually embedded within the question or on a formula sheet. So if we take a look here we say that for the van der Waals equation it is pressure plus your moles squared times your van der Waals constant a divided by your volume squared. Because this portion contains our Van der Waals constant a, it is a correction for attractive forces, or just the polarity, general polarity of gas molecules. Then that's gonna be times volume minus your moles n times your Van Waals constant b. Because b is involved here, this is a correction for the volume of gases, which can be tied to their size. Now this equation may not look like it, but it's connected to your ideal gas law. The van der Waals equation is used for real gases. The ideal gas law is used for ideal gases, but ideal gases are imaginary. If ideal gases existed, postulate 1 says that their volume is so small that it's insignificant and not important, it's negligible. So since they're so small that means their size would be equal to 0. And let's look. If if event if an ideal gas is involved, b is 0, so n times 0 is 0. This would mean that this entire part here is just v. And then if we're dealing with an ideal gas, an ideal gas has completely elastic collisions, it has no attractive forces or repulsive forces. So its polarity would be equal to 0. So take that 0 and plug it in. So n squared times 0 divided by v squared, all that just becomes 0. So all you have left is p. So if we're dealing with an ideal gas, the Van Waals equation becomes simplifies into the ideal gas law. So that's the connection they have to one another. Now if we take a look here at the columns, we have all these gases. The second column are our a values and they're in units of atmospheres times liters squared over mole squared. And then here we have b. Here, their units are liters over moles. With polarity or attractive forces, it's harder to see a pattern. But remember b is our size coefficient. We said as the molecular weight increases your value for b as you as you see them. More or less there's a few deviations of course because it's chemistry, but the general trend is as you increase the weight of a gas, you should see your b value increase as well. Alright. So that's how our van der Waals equation is connected to the ideal gas law, and this long equation is the van der Waals equation. Again, don't worry too much about memorizing it, it's more important to understand the ideas of the coefficient a and b.
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example
Van der Waals Equation Example 1
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Here the example question states, using the van der Waals equation, determine the pressure of 20 grams oxygen gas in 2 150 ml graduated flask when the temperature is 50 degrees Celsius. Alright. So we're going to say the van der Waals equation is pressure +2timesaovervsquaredtimes volume minus n times b equals n r t. Now all we gotta do is plug in the values that we have. So we need to have the moles of oxygen. So remember oxygen is diatomic, So one mole of o 2 weighs 32 grams of o 2. So when we do that we're gonna get our moles of o 2 comes out as 0.625 moles of o 2. Volume has to be in liters so that's 0.250 liters, and then temperature needs to be in Kelvin. So at 273.15 that gives us 323.15 Kelvin. Since we're dealing with o 2 in the chart up above, we would see that the a constant, constant, it comes out to 1.360, the b constant comes out to 0.0318. With this information, we plug it into the formula. Alright. So let's see. We're looking for pressure so we don't know it. So this is gonna be 0.625 squared times 1.360 divided by volume squared, which is 0.250 squared. And this is gonna be times volume, which is 0.250, minus moles, which is 0.625 times 0.0318, and this is gonna equal my moles r and t. So moles so I'm gonna actually move all this out of the way, so we have space to write this out. So here our moles again come out to 0 point 625. So the Van der Waals equation related to the ideal gas law, so r again is 0.08206 and then temperature is 323.15 Kelvin. When I multiply these 3 together on this side it comes out to v. Let's see, we're gonna multiply those out together that comes out to be 16.57356. When I when I do this value minus these 2 multiplying, what I get is I'm going to get 0.230125. Then when I work all of this out in here this comes out to be 8.5 plus p. So now I need to isolate my p, so I'm gonna divide both sides here by 0.230125, So p plus 8.5 equals 72.00. Subtract 8.5 from both sides, so p equals 63.5 atmospheres. So that would be the pressure of o 2 when utilizing the Van der Waals equation.
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Problem
Problem
Which gaseous compound is expected to have the largest value for the Van der Waals constant b?