(a) An ideal gas occupies a volume of 3.0 cm3 at 20°C and atmospheric pressure. Determine the number of molecules of gas in the container.

Question

_____________molecules

(b) If the pressure of the 3.0-cm3 volume is reduced to 2.4 × 10−11 Pa (an extremely good vacuum) while the temperature remains constant, how many moles of gas remain in the container?
______________mol

Answer 1

(a) First, we need to convert the volume from cm3 to m3:

3.0 cm3 × (1 m3 / 10^6 cm3) = 3.0 × 10^-6 m3

We are given the pressure and temperature of the gas as:

P = atmospheric pressure = 1.013 × 10^5 Pa
T = 20°C = (20 + 273.15) K = 293.15 K

We can use the ideal gas law (PV = nRT) to find the number of moles of the gas:

n = PV / RT

The gas constant (R) is given as 8.314 J/(mol K)

n = (1.013 × 10^5 Pa)(3.0 × 10^-6 m3) / (8.314 J/(mol K))(293.15 K)

n ≈ 1.243 × 10^-4 mol

To find the number of molecules, we can use Avogadro's number (N_A = 6.022 × 10^23 molecules/mol):

N = n * N_A

N ≈ (1.243 × 10^-4 mol)(6.022 × 10^23 molecules/mol)

N ≈ 7.48 × 10^19 molecules

So there are approximately 7.48 × 10^19 molecules of gas in the container.

(b) We are given the new pressure as 2.4 × 10^−11 Pa, while the volume and temperature remain constant. We can use the ideal gas law again to find the new number of moles of gas:

n_new = P_new * V / (R * T)

n_new = (2.4 × 10^−11 Pa)(3.0 × 10^-6 m3) / (8.314 J/(mol K))(293.15 K)

n_new ≈ 9.806 × 10^-18 mol

So there are approximately 9.806 × 10^-18 moles of gas remaining in the container under the reduced pressure.