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Step-by-Step Solution
Step 1: Identify the Given Information
• Pressure, $P = 2\,\text{atm}$
• Temperature, $T = 300\,\text{K}$
• Volume, $V = 1\,\text{L} = 1\times 10^{-3}\,\text{m}^3$
• Mean kinetic energy per molecule, $KE_{\text{mean}} = 2 \times 10^{-9}\,\text{J}$
• We want to find the total number of molecules $N$.
Step 2: Relate Mean Kinetic Energy to $kT$
For an ideal gas molecule, the mean kinetic energy per molecule is given by:
$KE_{\text{mean}} = \frac{3}{2}\,k T,$
where $k$ is the Boltzmann constant. From this,
$\frac{3}{2}\,k T = 2 \times 10^{-9}\,\text{J} \quad \Longrightarrow \quad k T = \frac{2 \times 10^{-9} \times 2}{3} = \frac{4}{3} \times 10^{-9}\,\text{J}.$
Step 3: Convert Pressure to SI Units
1 atm = $1.013 \times 10^5\,\text{Pa}$. Thus,
$P = 2\,\text{atm} = 2 \times 1.013 \times 10^5\,\text{Pa} = 2.026 \times 10^5\,\text{Pa}.$
Step 4: Apply the Ideal Gas Relation in Terms of Molecules
When counting the actual number of molecules (instead of moles), the ideal gas law can be written as:
$P V = N \, k \, T,$
where $N$ is the total number of molecules and $k T$ has been found from the mean kinetic energy.
Step 5: Substitute the Values and Solve for $N$
First, compute $P V$:
$P V = \bigl(2.026 \times 10^5\,\text{Pa}\bigr)\times\bigl(1 \times 10^{-3}\,\text{m}^3\bigr) = 2.026 \times 10^2\,\text{J} = 202.6\,\text{J}.$
Next, we use $k T = \tfrac{4}{3}\times 10^{-9}\,\text{J}$. Therefore,
$N = \frac{P V}{k T}
= \frac{202.6\,\text{J}}{\frac{4}{3}\times 10^{-9}\,\text{J}}
= \frac{202.6}{1.333 \times 10^{-9}}
\approx 1.52 \times 10^{11}.$
Step 6: Conclusion
Hence, the number of molecules in one litre of the gas under the given conditions is approximately $1.5 \times 10^{11}.$
