# Having trouble using the ideal gas law for this problem.

1. Dec 7, 2009

### lilmul123

1. The problem statement, all variables and given/known data

A pressure as low as 1 * 10^-8 torr can be achieved using an oil diffusion pump. How many molecules are there in 1 cm^3 of a gas at this pressure if its temperature is 371 K?

2. Relevant equations

PV = nRT = NkT

3. The attempt at a solution

I converted 1*10^-8 torr to atm and got 1.31578947 * 10^-11 atm. Then, I converted 1 cm^3 to liters and got .001L. Then, I plugged all known variables into the ideal gas law. When doing PV = nRT, I got n in molecules to be 2568782.388. This was incorrect. I then tried PV = NkT, and N was also incorrect. Where did I go wrong?

2. Dec 7, 2009

### Andrew Mason

Your units are mixed up. Atmospheres are not MKS units. Convert torr to Pascals instead (Newtons/m^2)

AM

3. Dec 7, 2009

### lilmul123

Are liters correct then? Or should that be in m^3? I converted torr to pascals, and found the molecules to be 2.602914125*10^11, which is still incorrect.

4. Dec 7, 2009

### denverdoc

lets see what I get using torrs/atm

volume at stp = 1ml(10^-8/760)*273/371=9.78 x 10-12 ml.

Converting to liters and dividing by 22.4 L/mole I get 4.27 x 10^-16 moles. Multiplying by A's number: 262,837,500 which should be rounded to 2.63 x 10^8. Seems like your answer is off by 1000--maybe liter m^3 conversion?

5. Dec 8, 2009

### Andrew Mason

It is rather difficult to determine where you went wrong if you do not show us your detailed calculations.

n=PV/RT where $P = 10^{-8} Torr = 1.33 \times 10^{-6} Pa$ and $V = 10^{-6} m^3$

$$n = \frac{1.33 \times 10^{-6} \times 10^{-6}}{8.3145 \times 371} = \frac{1.33 \times 10^{-12}}{3.085 \times 10^3} = 4.31 \times 10^{-16} moles$$