To solve the problem of finding the number of molecules in one liter of an ideal gas at 300 K and 2 atmospheric pressure with a mean kinetic energy of \(2 \times 10^{-9}\) J per molecule, we can follow these steps:
### Step 1: Understand the Mean Kinetic Energy Formula
The mean kinetic energy (\(K\)) of a molecule in an ideal gas is given by the formula:
\[
K = \frac{3}{2} k_B T
\]
where:
- \(k_B\) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\)),
- \(T\) is the temperature in Kelvin.
### Step 2: Relate Mean Kinetic Energy to Temperature
Given that the mean kinetic energy per molecule is \(2 \times 10^{-9}\) J, we can rearrange the formula to find \(T\):
\[
T = \frac{2K}{3k_B}
\]
Substituting the values:
\[
T = \frac{2 \times (2 \times 10^{-9})}{3 \times (1.38 \times 10^{-23})}
\]
### Step 3: Calculate the Temperature
Calculating \(T\):
\[
T = \frac{4 \times 10^{-9}}{4.14 \times 10^{-23}} \approx 9.66 \times 10^{13} \, \text{K}
\]
However, we already know the temperature is given as 300 K, so we can proceed with the next step.
### Step 4: Use the Ideal Gas Law
The ideal gas law is given by:
\[
PV = nRT
\]
Where:
- \(P\) is the pressure in Pascals,
- \(V\) is the volume in cubic meters,
- \(n\) is the number of moles,
- \(R\) is the universal gas constant (\(8.314 \, \text{J/(mol K)}\)),
- \(T\) is the temperature in Kelvin.
### Step 5: Convert Units
Convert the pressure from atmospheres to Pascals:
\[
P = 2 \, \text{atm} = 2 \times 1.013 \times 10^5 \, \text{Pa} = 2.026 \times 10^5 \, \text{Pa}
\]
Convert the volume from liters to cubic meters:
\[
V = 1 \, \text{L} = 1 \times 10^{-3} \, \text{m}^3
\]
### Step 6: Calculate the Number of Moles
Rearranging the ideal gas law to solve for \(n\):
\[
n = \frac{PV}{RT}
\]
Substituting the values:
\[
n = \frac{(2.026 \times 10^5) \times (1 \times 10^{-3})}{(8.314) \times (300)}
\]
### Step 7: Calculate \(n\)
Calculating \(n\):
\[
n = \frac{2.026 \times 10^2}{2494.2} \approx 0.0812 \, \text{mol}
\]
### Step 8: Calculate the Number of Molecules
To find the number of molecules (\(N\)), we use Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{molecules/mol}\)):
\[
N = n \times N_A = 0.0812 \times 6.022 \times 10^{23} \approx 4.89 \times 10^{22} \, \text{molecules}
\]
### Final Answer
Thus, the number of molecules in one liter of the ideal gas at the given conditions is approximately:
\[
N \approx 4.89 \times 10^{22} \, \text{molecules}
\]
