Table 10.3 shows that the van der Waals b parameter has units of L / mol. This implies that we c

Table 10.3 shows that the van der Waals b parameter has...

Table 10.3 shows that the van der Waals $b$ parameter has units of $\mathrm{L} / \mathrm{mol}$. This implies that we can calculate the size of atoms or molecules from $b$. Using the value of $b$ for $\mathrm{Xe},$ calculate the radius of a Xe atom and compare it to the value found in Figure $7.7,$ that is, $140 \mathrm{pm}$. Recall that the volume of a sphere is $(4 / 3) \pi r^{3}$.

Key Concepts

Molar Volume and Unit Conversion

This concept involves converting the per-mole measurement given by the b parameter into the volume occupied by a single molecule using Avogadro's number. It is essential for transitioning from macroscopic gas parameters to microscopic dimensions, thus allowing the calculation of atomic or molecular size.

Volume of a Sphere

The formula for the volume of a sphere, V = (4/3)?r³, is a fundamental geometric relationship that relates the radius of a spherical object to its volume. This formula is used to approximate the volume of atoms or molecules (assumed to be spherical) when calculating their radius from an experimentally determined volume such as the 'b' parameter.

van der Waals Equation

This concept refers to an adjustment to the ideal gas law that introduces two parameters (a and b) to account for intermolecular forces and the finite size of molecules, respectively. The van der Waals equation helps to explain the behavior of real gases by incorporating the effects of molecular volume and attraction, making it easier to understand deviations from ideality.

Excluded Volume Parameter

The 'b' parameter in the van der Waals equation represents the excluded volume, or the volume effectively occupied by the gas molecules themselves. It is typically given in units such as L/mol, and when converted appropriately, provides an estimate of the actual size of individual atoms or molecules, which is crucial for determining physical dimensions from macroscopic measurements.

Transcript

00:01 In this problem, we are asked to do some calculations, and let me find where i am here, and get the rest of my problem step down.

00:11 We are asked to reference table 10 .3 and look up the vanderbals, the vanderval's b, parameter b for xe, for xenon.

00:37 And it is 0 .0 510 liters per mole.

00:54 That's what we're given.

00:55 And we're also reminded that the volume of a sphere is 4 thirds pi r cubed.

01:09 What else are we given here? and we're asked to find, we're told that this is leaders per mole, we're told that this is leaders per mole right there.

01:20 It implies we can calculate the size of the size of the atoms from b.

01:26 Then we're asked to compare our answer.

01:35 We're going to solve for r in the equation above to figure 7 .7 number, which is 140 picometers.

01:52 Let us begin.

01:54 This is actually a very simple problem to do.

01:57 So the first thing we're going to do is convert liters to centimeters cubed in order to do that, all i need to do is move the decimal 3 to the left.

02:10 So i get 51 .0 centimeters cubed per mole.

02:25 Okay, what am i going to do next? then we are going to figure out our volume as follows.

02:51 Our volume of xenon is going to equal 4 thirds pi r cubed.

03:04 We're going to multiply that by 6 .02 times 10 to the 23rd.

03:12 Particles, in this case atoms, per mole.

03:22 And when we're done with this part, i got 2 .5 -225 times 10 to the 24th r cubed equals 51 .0 centimeters cubed per mole and this is r -cube per mole.

04:03 So my moles crosses out.

04:08 Then i stuck this into my calcule.

04:09 Then i stuck this into my and i got r cubed equals 51 centimeters cubed divided by 2 .5 .225 times 10 to the 24th and that equaled 2 .78 times 10 to the minus 8 centimeters.

04:51 No, it was 2 .72...

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