How many molecules are in 3.00 cm^3?

Question

Current vacuum technology can achieve a pressure of 1.0 * 10^-10mm of Hg. At this pressure, and at a temperature of 40.0 degrees C.

Answer 1

The ideal gas law states that

PV = nRT where

P is the absolute pressure [Pa],
V is the volume [m3],
n is the amount of substance of gas [mol],
R is the gas constant 8.3143 m^3·Pa/(K·mol), and
T is the temperature in kelvin [K].

so n = PV/RT

P = 1e-10mmHg = 1.333e-8 Pa
V = 3cm^3 = 3e-6m^3
T = 40 degC = 313 K

so

n = 1.333e-8Pa * 3e-6m^3/313K * 8.3143 m^3·Pa/(K·mol)
n = 1.533e-17 mole

1 mole is 6.022 x 10^23 molecule so

1.533e-17 mole * 6.022e23 molecules/mole = 9.25e6 molecules or

9 250 000 molecules

Answer 2

At STP, 1 mole of gas (6.022 * 10^23 molecules) occupies 22.4 liters, or 22,400 cc. The volume you're talking about is (3.00 cc/ 22,400cc) of that, or 1.339*10^-4, so you're reduced to (6.022 * 10^23 * 1.339 * 10^-4) molecules at STP, or 8.065 * 10^19 molecules. STP is 0° C (273.15 K) and 1 bar (750 mm Hg). If the temperature is raised to 40° C, then the pressure is also raised according to the ideal gas law PV = nRT. Keeping V and n the same, P1/T1 = P2/T2; (750 mmHg)/(273.16 K) = (P2)/(273.16 + 40 K), which yields P2 = 859.83 mm Hg. So you're reducing the pressure in your stated problem from 859.83 mm Hg to 1.0 * 10^-10 mm Hg, or a reduction of (1.0 * 10^-10 mm Hg/859.83 mm Hg), or 1.163 * 10^-13. Which leads to the original 3.00 cc volume which, at STP, would have held 8.065 * 10^19 molecules, now holding (8.065 * 10^19 molecules * 1.163 * 10^-13), or 9.38*10^6 molecules.

Answer 3

Gas Equation-:

P V = n R T

First convert pressure to Pascals.....
P = 1.333 x 10^-8 Pascals

(1.333 x 10^-8) x (3 x 10^-2) = n x 8.3 x 313

n = (1.333 x 10^-8) x (3 x 10^-2) / 8.3 x 313

n = 1.5396 x 10^-13 molecules