B Calculate length of side of cube

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To calculate the length of each side of a cubic simulation box containing 500 molecules, first determine the number of moles, which is 8.3e-22 moles based on the fact that 1 mole equals 6.002e+23 molecules. Using the molar volume of 0.6 liters per mole, multiply this by the number of moles to find the volume of the cube in liters. Convert this volume to cubic meters by dividing by 1000, then use the formula for the volume of a cube (V = l^3) to find the length of a side in meters. Finally, convert the length from meters to Ångströms for the desired measurement.
Omsin
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How can I calculate the length L of each side of a cubic simulation box with 500 molecules (in Ångstrøm) when I want molar volumes of V1 = 0.6 l/mol?
 
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How many molecules in 1 mol? What fraction of 1 mol is 500molecules? That gives you the volume of the cube in l.
V=l2 will give you the length of a side of the cube.
All you need to do is the Arithmetic -get the units right, of course,
 
1 mole is 6.002e+23 molecules. 500 molecules have 8.3e-22 moles. I am not really sure what you meant after that. Volume = (liter)^2? Isn't Volume[m^3] = [1000* Liter]
 
Omsin said:
1 mole is 6.002e+23 molecules. 500 molecules have 8.3e-22 moles. I am not really sure what you meant after that. Volume = (liter)^2? Isn't Volume[m^3] = [1000* Liter]
My bad. Of course volume is cubed. But the sums are easy, no?
 
sophiecentaur said:
My bad. Of course volume is cubed. But the sums are easy, no?
Sorry, but I am not really sure how to do this.
 
You have already calculated how many moles will be in your cube (8.3e-22 moles as you previously said).

You want molar volume of 0.6 Liters/mol, so if you multiply this number with the moles, you will get the volume of the cube (V) in Liters. Then you can divide by one thousand to get the Volume in m3.

So, your cube's side will have length ## l = \sqrt[3] V ## meters.

The last step is to change meters to Angstroms.
 
Omsin said:
Sorry, but I am not really sure how to do this.
Sorry. I was busy and couldn't get back to you quickly. But @DoItForYourself has sorted you out I think.
 
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