Van der Waals Equation - Online Tutor, Practice Problems & Exam Prep

The Van der Waals equation refines the Ideal Gas Law to account for real gas behavior, which deviates from the ideal at high pressures and low temperatures. This equation introduces two constants: 'a' for attractive forces (polarity) between gas molecules and 'b' for the volume occupied by the gas molecules (size). The 'b' constant increases with the molecular weight of a gas, reflecting its larger volume. The Van der Waals equation serves as a bridge between real and ideal gases, reverting to the Ideal Gas Law when the volume and attractive forces of the gas molecules are negligible. Understanding the implications of the 'a' and 'b' constants is more crucial than memorizing the equation, as it's typically provided when needed for calculations.

Van der Waals Equation

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 going to 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 to 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 two 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 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.

Van der Waals Equation

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.

More attractive forces or just the polarity, general polarity of gas molecules? Then that's going to be times volume minus your moles N times your van der 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 one 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 an ideal gas is involved, B is 0, so N * 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 zero and plug it in. So N2×0/V2, all that just becomes 0. So all you have left is P. So if we're dealing with an ideal gas, the van der Waals equation 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 liter 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 set. As the molecular weight increases, your value for B increases. So if you were to look and look at all the weights of each of these gases, it would make sense in terms of the numbers 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 & B.

Van der Waals Equation Example 1

Here are the example question states using the van der Waals equation, determine the pressure of 20 grams oxygen gas in 250ML graduated flask when the temperature is 50�C. All right, so we're going to say the Van der Waals equation is

P + N 2 A V 2 ⁢ ( V - N ⁢ B ) = N R T

Now all we got to do is plug in the values that we have.

So we need to have the moles of oxygen O. Remember, oxygen is diatomic, so one mole of O₂ weighs 32 grams O₂. So when we do that, we're going to get our moles up. O₂ comes out as 0.625 moles of O₂. Volume has to be in liter, so that's 0.250 liters. And then temperature needs to be in Kelvin, so add 273.15. That gives us 323.5 Kelvin.

Since we're dealing with O₂ in the chart up above, we would see that the A constant it comes out to 1.360. The B constant comes out to 0.0318. With this information, we plug it into the formula. All right, so let's see. We're looking for pressure, so we don't know it. So this is going to be, oh,

0.625 2 ⁢ * 1.360 / 0.250 2

And this is going to be times volume which is 0.250 minus moles which is 0.625 * 0.0318 and this is going to equal my moles R&T.

So moles I'm going to actually move all this out of the way so we have space to write this out. So here our moles again come out to 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 three together on this side, it comes out to be, let's see, we're going to multiply those out together. That comes out to be 16.57356.

When I when I do this value minus these two 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 85 plus P So now I need to isolate my P So I'm going to divide both sides here by 0.230125. So P + 8.5 equals 72.00. Subtract 8.5 from both sides so P = 63.5 atmospheres, so that would be the pressure of O₂ when utilizing the van der Waals equation.

Which gaseous compound is expected to have the largest value for the Van der Waals constant b?

• A. O₂
• B. CH₃CH₃
• C. HBr
• D. S₈
• E. Ne

Here’s what students ask on this topic:

What is the Van der Waals equation?

The Van der Waals equation is a mathematical formula used to describe the behavior of real gases, taking into account the volume occupied by gas molecules and the attractive forces between them, which are not considered in the ideal gas law. It is an improvement over the ideal gas law and is expressed as:

( P + a V 2 ) ( V m - b ) = R T Where:

• P is the pressure of the gas,
• V_m is the molar volume of the gas (the volume per mole of gas),
• T is the temperature of the gas in Kelvin,
• R is the ideal gas constant,
• a is a measure of the attraction between the particles,
• b is the volume occupied by one mole of the particles themselves.

The constants a and b are specific to each gas, reflecting how different gases have different sized molecules and strengths of intermolecular forces. The Van der Waals equation corrects for the non-ideal behavior of gases, especially at high pressures and low temperatures, where deviations from the ideal gas law are most pronounced.

How can I find the values of a and b in the van der Waals equation?

To find the values of a and b in the van der Waals equation for a specific gas, you typically refer to empirical data. The van der Waals equation is an equation of state for a fluid composed of particles with non-zero size and a pairwise attractive force, represented as:

( P + a n 2 V 2 ) ( V - n b ) = n R T

Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. The constants a and b are unique to each gas and are determined experimentally. They account for the intermolecular forces and the volume occupied by the gas particles, respectively.

To find these values, you would look them up in physical chemistry textbooks, scientific databases, or reputable online resources. These constants are typically derived from experimental measurements of the gas properties at various temperatures and pressures and then fitted to the van der Waals equation. Once known, a and b can be used to predict the behavior of the gas under non-ideal conditions.

What does the van der Waals equation account for?

The van der Waals equation is a mathematical expression that improves upon the ideal gas law by accounting for two significant factors that affect real gases: intermolecular forces and the finite size of gas particles.

Firstly, it introduces a term to correct for the intermolecular attractions between gas molecules. In an ideal gas, these forces are assumed to be negligible, but in real gases, they can have a noticeable impact, especially under high pressure. The van der Waals equation adds an attraction term that effectively reduces the pressure in the ideal gas equation to account for these forces.

Secondly, it corrects for the volume occupied by the gas particles themselves. The ideal gas law assumes that the particles have no volume, but real gas particles do occupy space. The van der Waals equation adds a volume correction term to the ideal gas law, which effectively increases the volume over which the gas particles are free to move.

By including these two corrections, the van der Waals equation provides a more accurate description of the behavior of real gases under various conditions than the ideal gas law.

What advantage does the van der Waals equation give us over the ideal gas law?

The van der Waals equation provides a more accurate description of the behavior of real gases compared to the ideal gas law. While the ideal gas law assumes that gas particles have no volume and do not exert any attractive forces on each other, these assumptions are not true for real gases. The van der Waals equation modifies the ideal gas law by introducing two constants, a and b. The a term accounts for the attractive forces between gas molecules, which become significant at high pressures and low temperatures. The b term corrects for the finite volume that gas molecules occupy, which becomes important at high pressures where the volume of the gas particles is no longer negligible compared to the volume of the container. By including these factors, the van der Waals equation can more accurately predict the behavior of gases under non-ideal conditions, such as high pressures and low temperatures, where deviations from ideal behavior are most pronounced.