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Step-by-Step Solution
Step 1: Recall the Expression for Mean Free Path
The mean free path $ \lambda $ of gas molecules is often given by the relation:
$$
\lambda = \frac{k\,T}{\sqrt{2}\,\pi\, d^2 \,P},
$$
where
$k$ is the Boltzmann constant,
$T$ is the absolute temperature,
$d$ is the diameter of the gas molecules,
$P$ is the pressure.
Step 2: Identify the Data at STP
At standard temperature and pressure (STP):
$T_{\text{STP}} = 273\,\mathrm{K}$,
$P_{\text{STP}} = 1\,\mathrm{atm}$,
The mean free path, $ \lambda_{\text{STP}} = 1500\,d $.
Step 3: Mean Free Path at the New Temperature
Temperature is changed to $T_{\text{new}} = 373\,\mathrm{K}$, while the pressure remains $1\,\mathrm{atm}$. The new mean free path, call it $ \lambda_{\text{new}} $, becomes:
$$
\lambda_{\text{new}} = \frac{k \times 373\,\mathrm{K}}{\sqrt{2}\,\pi\, d^2 \times 1\,\mathrm{atm}}.
$$
Step 4: Form the Ratio of Mean Free Paths
To find how $ \lambda_{\text{new}} $ relates to $ \lambda_{\text{STP}} $, consider the ratio:
$$
\frac{\lambda_{\text{new}}}{\lambda_{\text{STP}}}
= \frac{
\tfrac{k \times 373\,\mathrm{K}}{\sqrt{2}\,\pi\, d^2 \times 1\,\mathrm{atm}}
}{
\tfrac{k \times 273\,\mathrm{K}}{\sqrt{2}\,\pi\, d^2 \times 1\,\mathrm{atm}}
}
= \frac{373}{273}.
$$
All factors except temperature cancel out.
Step 5: Calculate the New Mean Free Path
Since $ \lambda_{\text{STP}} = 1500\,d $, we have:
$$
\lambda_{\text{new}} = 1500\,d \times \frac{373}{273} \approx 1500\,d \times 1.366 = 2049\,d.
$$
Final Answer
Therefore, under the new conditions (temperature at $373\,\mathrm{K}$ and the same pressure as STP), the mean free path is approximately $2049\,d$.
