(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.
_____________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
(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.
To determine the number of molecules of gas in the container, we need to use the ideal gas law equation:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin
(a) To find the number of molecules of gas in the container at 20°C and atmospheric pressure, we first need to convert the given temperature from Celsius to Kelvin. The temperature in Kelvin is 273.15 + 20 = 293.15 K.
We also need to convert the given volume from cm³ to m³ because the unit of volume in the ideal gas law equation is in m³. 1 cm³ is equal to 1 x 10⁻⁶ m³.
Volume = 3.0 cm³ = 3.0 x 10⁻⁶ m³
Atmospheric pressure is usually approximately 101325 Pa.
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
(101325 Pa)(3.0 x 10⁻⁶ m³) = n(8.314 J/K/mol)(293.15 K)
Solve for n:
n = (101325 Pa)(3.0 x 10⁻⁶ m³) / (8.314 J/K/mol)(293.15 K)
Calculating the equation will give us the number of moles of gas in the container. To find the number of molecules, we can use Avogadro's number, which is 6.022 x 10²³ molecules/mol. Multiply the number of moles by Avogadro's number to get the number of molecules.
(b) To determine how many moles of gas remain in the container when the pressure is reduced to 2.4 x 10⁻¹¹ Pa, we can use the same equation as above, but this time we need to solve for n:
n = (P)(V) / (RT)
For this scenario, the pressure is 2.4 x 10⁻¹¹ Pa, the volume is still 3.0 x 10⁻⁶ m³, and the temperature remains constant.
Plug the values into the equation:
n = (2.4 x 10⁻¹¹ Pa)(3.0 x 10⁻⁶ m³) / (8.314 J/K/mol)(293.15 K)
Calculate the equation to find the number of moles of gas remaining in the container.
To determine the number of molecules of gas in the container, we can use the ideal gas law:
PV = nRT
where P is the pressure in Pascal (Pa), V is the volume in cubic meters (m^3), n is the number of moles of gas, R is the ideal gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin (K).
(a) To find the number of molecules of gas in the container, we need to find the number of moles of gas first, and then convert it to the number of molecules.
Given:
V = 3.0 cm^3 = 3.0 × 10^(-6) m^3 (since 1 cm^3 = 10^(-6) m^3)
P = atmospheric pressure (which is generally around 101325 Pa at sea level)
T = 20°C = 293 K (since 1°C = 273.15 K)
First, we rearrange the ideal gas law to solve for n:
n = (PV) / (RT)
Substituting the given values:
n = ((101325 Pa) × (3.0 × 10^(-6) m^3)) / ((8.314 J/(mol·K)) × (293 K))
Now we can calculate the number of moles of gas:
n = 0.01289 mol
To convert the number of moles to the number of molecules, we use Avogadro's number:
1 mole = 6.022 × 10^23 molecules
Number of molecules = 0.01289 mol × (6.022 × 10^23 molecules/mol)
Therefore, the number of molecules of gas in the container is:
(a) ________ molecules = 7.77 × 10^21 molecules
(b) In this scenario, the pressure is reduced to an extremely low value of 2.4 × 10^(-11) Pa, and the temperature remains constant at 293 K.
Since the temperature does not change, the number of moles of gas remains the same.
Therefore, the number of moles of gas remaining in the container is:
(b) ________ mol = 0.01289 mol