How many molecules of nitrogen are in 22.4 L of nitrogen at exactly 0 C and 3 atm?
1.81 × 10^24 molec
To determine the number of molecules of nitrogen in 22.4 L at 0 degrees Celsius and 3 atm, we can use the ideal gas law. The ideal gas law is given by:
PV = nRT
where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L·atm / (mol·K))
T = temperature in Kelvin
Step 1: Convert temperature to Kelvin
Given that the temperature is 0 degrees Celsius, we can convert it to Kelvin by adding 273.15:
T = 0 + 273.15 = 273.15 K
Step 2: Plug in the values into the ideal gas law equation and solve for n (number of moles):
(3 atm) * (22.4 L) = n * (0.0821 L·atm / (mol·K)) * (273.15 K)
66.72 = n * 22.41475
n = 66.72 / 22.41475
n = 2.98 moles
Step 3: Convert moles to molecules
One mole of any substance contains 6.022 x 10^23 molecules (Avogadro's number). Therefore, we can calculate the number of molecules of nitrogen in 2.98 moles as follows:
Number of molecules = 2.98 moles * (6.022 x 10^23 molecules / 1 mole)
Number of molecules = 1.794 x 10^24 molecules
Therefore, there are approximately 1.794 x 10^24 molecules of nitrogen in 22.4 L at 0 degrees Celsius and 3 atm.
To determine the number of molecules of nitrogen in a given volume, you need to use the ideal gas law formula. The ideal gas law is given by the equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's convert the given temperature of 0 °C to Kelvin. The formula to convert Celsius to Kelvin is:
T(K) = T(°C) + 273.15
Therefore, the temperature in Kelvin would be:
T(K) = 0 °C + 273.15 = 273.15 K
Now we can rearrange the ideal gas law equation to solve for n (number of moles):
n = PV / RT
n = (3 atm) * (22.4 L) / (0.0821 L·atm/mol·K * 273.15 K)
n ≈ 1.022 moles
To convert moles to molecules, we use Avogadro's number, which states that there are approximately 6.022 x 10^23 molecules in one mole.
Therefore, the number of molecules of nitrogen in 22.4 L at 0 °C and 3 atm would be:
Number of Molecules = n * Avogadro's number
Number of Molecules ≈ 1.022 moles * 6.022 x 10^23 molecules/mole
Number of Molecules ≈ 6.154 x 10^23 molecules
So, there are approximately 6.154 x 10^23 molecules of nitrogen in 22.4 L at exactly 0 °C and 3 atm.
PV = nRT and solve for n = number of mols.
Then 1 mol contains 6.022e 23 molecules.