Calculate N for Thermo Lab Vacuum at 293K

AI Thread Summary
To calculate the number of gas molecules per cubic centimeter in a vacuum at 293K with a pressure of 1.00 x 10-18 atm, the Ideal Gas Law (PV = NkT) is used. It's important to ensure consistent unit conversions, particularly from cubic centimeters to cubic meters, which involves multiplying by the appropriate conversion factor. The discussion emphasizes using SI units, where pressure is expressed in Pascals and the Boltzmann constant is applied correctly. The equation can be rearranged to express the number of molecules per cubic centimeter by adjusting the volume unit accordingly. Overall, the conversation clarifies the application of the Ideal Gas Law and the importance of unit consistency in calculations.
kopinator
Messages
41
Reaction score
1
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?
 
Physics news on Phys.org
kopinator said:
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 ... :smile:
 
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/NA and R = kB*NA ]
 
  • Like
Likes 1 person
Ok, sweet. Thank you!
 
Great post BvU. Thank you.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top