It turns out that the Van Dar Waals constant b is equal to four times the total volume actually occupied by the molecules of a mole of gas. Using this figure, calculate the fraction of the volume in a container actually occupied by Ar atoms:

a) at STP
b) at 100 atm pressure and 0 degrees Celsius

((Assume for simplicity that the ideal-gas equation still holds.))

Ar

a= 1.34
b = 0.0322

I'm extremely confused about this question, are they asking for the volume when they state fraction of the volume.

The units of b are liters/mole. So the actual molecules in a liter of Argon occupy b/4 or 0.008 liters.

At STP, the (mostly empty) space occupied by gas is 22.4 liters.

For the fraction, take the ratio.

Then do the same thing for the 100 atm case. The volume occupied (including space between atoms) will be 100 times less than at 1 atm STP, or 0.224 liters.

Therefore, they are asking for the volume

(a) at STP

v = 22.4 L
T = 273 K
P = 1 atm

Would I use Van der Waals equation for (a) or for (b)? I think I would have to use this equation for b

(a) V = nRT/ P

Can I use this for question a

Okay, then what equation can I use for (a)

Instead of using the ideal-gas equation,
can I use the mole-fraction

By stating "For the fraction, take the ratio. " does this apply to the mole-fraction

You should take the ratio of the computed volumes.

0.008/22.4 in the first case

It doesn't make sense to talk about the mole fraction of solid spheres

Okay but wait we can find the number of moles using PV = nRT? Right?

V = 22.4 L
P = 1 atm
R= 0.0831 L-atm
T = 273 K

Don't worry, I can help you with that! In this question, when they mention the "fraction of the volume," they are referring to the ratio of the actual occupied volume of the gas molecules to the total volume of the container.

To calculate the fraction of the volume occupied by Ar atoms, we need to use the Van der Waals equation, which accounts for the non-ideal behavior of gases:

\( Z = \frac{{PV}}{{nRT}} = 1 + \frac{{B}}{{V}} - \frac{{A}}{{V^2}} \)

Here, Z is the compressibility factor (which is approximately equal to 1 for an ideal gas), P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Given that the Van der Waals constant b is equal to four times the total volume actually occupied by the molecules, we have:

\( b = 4 \times V_{\text{{actual}}} \)

To find the actual volume, we can rearrange the above equation:

\( V_{\text{{actual}}} = \frac{{b}}{4} \)

Now let's calculate the fraction of the volume occupied by Ar atoms in two different scenarios:

a) At STP (standard temperature and pressure), which is approximately 273 K and 1 atm:

Using the ideal gas equation: PV = nRT, we can rearrange it as:

\( V_{\text{{total}}} = \frac{{nRT}}{{P}} = \frac{{(1 \, \text{{mol}}) \times (0.08206 \, \text{{atm}} \cdot \text{{L/mol/K}}) \times (273 \, \text{{K}})}}{{1 \, \text{{atm}}}} \)

Calculate V_total using the above equation. Then, substitute b = 4 × V_actual and calculate V_actual using \( V_{\text{{actual}}} = \frac{{b}}{4} \).

Finally, calculate the fraction of the volume occupied by Ar atoms by dividing V_actual by V_total.

b) At 100 atm pressure and 0 degrees Celsius:

First, convert 0 degrees Celsius to Kelvin: T = 273 K.

Next, calculate V_total using the ideal gas equation PV = nRT. Rearrange the equation as:

\( V_{\text{{total}}} = \frac{{nRT}}{{P}} = \frac{{(1 \, \text{{mol}}) \times (0.08206 \, \text{{atm}} \cdot \text{{L/mol/K}}) \times (273 \, \text{{K}})}}{{100 \, \text{{atm}}}} \)

Calculate V_total using the above equation. Then, substitute b = 4 × V_actual and calculate V_actual using \( V_{\text{{actual}}} = \frac{{b}}{4} \).

Finally, calculate the fraction of the volume occupied by Ar atoms by dividing V_actual by V_total.

I hope this helps you understand how to approach these calculations! Let me know if you have any further questions.