Calculate N for Thermo Lab Vacuum at 293K

[kopinator]

The best laboratory vacuum has a pressure of about 1.00 x 10-18 atm, or 1.01 x 10-13 Pa. How
many gas molecules are there per cubic centimeter in such a vacuum at 293K?

PV = nkT
P/kT = n/V = N (# of molecules per cm^3)

Since V is typically in units of m^3 or liters, should I make my volume .01V to account for the cm^3?

 

[berkeman]

The best laboratory vacuum has a pressure of about 1.00 x 10-18 atm, or 1.01 x 10-13 Pa. How
many gas molecules are there per cubic centimeter in such a vacuum at 293K?

PV = nkT
P/kT = n/V = N (# of molecules per cm^3)

Since V is typically in units of m^3 or liters, should I make my volume .01V to account for the cm^3?

When making unit conversions, use the trick of multiplying by "1" to help you do the unit conversion.

So to convert from cm^3 to m^3 multiply by "1" like this:

$$1cm^3 * \frac{(1m)^3}{(100cm)^3}$$

The cm^3 unit terms in the numerator and denominator cancel (just like numbers cancel if they are identical in the numerator and denominator of a fraction), and you are left with what in units of m^3? Hint -- It's not 0.01 ...

 

[berkeman]

Oh, and where you have "k" in your Ideal Gas Law equation, you really mean "R", correct?

http://en.wikipedia.org/wiki/Ideal_gas_law

.

 

[BvU]

You are using the right equation. People are conditioned to see n as number of moles and N as number of molecules, so better use N.

Now try to work in a consistent set of units. SI units is what every reasonable person would use.
p in Pascal, Pa with 1Pa = 1 N/m[sup2[/sup]. Conversion: 1 atm = 101325 N/m[sup2[/sup].
k Boltzmann constant, J/K 1.3806488 10^-23 J/K
T 293 K

The equation gives you n/V in molecules/m³.
Since the exercise asks for molecules/cm³, all you have to do is multiply by

( molecules/cm³ ) / (molecules/m³ ) = cm³ / m³ = (cm / m³) = (10^-2)³

[I see berkeman beat me to it, well, good for you!

But I don't agree with him (/her?): pV = NkT is just fine. It's the same as pV = nRT since n = N/N_A and R = k_B*N_A ]

 

[kopinator]

Ok, sweet. Thank you!

 

[berkeman]

Great post BvU. Thank you.