The mean free path of molecules of a certain gas at STP is $1500 \mathrm{~d}$, where $\mathrm{d}$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at $373 \mathrm{~K}$ is approximately:

The mean free path (λ) of molecules in a gas is given by the formula: <br/><br/> $\lambda = \frac{kT}{\sqrt{2}\pi d^2 P}$ <br/><br/> where k is the Boltzmann constant, T is the temperature in Kelvin, d is the diameter of the gas molecules, and P is the pressure. <br/><br/> At STP (standard temperature and pressure), the temperature is $273\mathrm{~K}$ and the pressure is $1\mathrm{~atm}$. We are given that the mean free path at STP is $1500d$. Let's denote the mean free path at $373\mathrm{~K}$ as λ': <br/><br/> $\lambda' = \frac{k(373\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}$ <br/><br/> To find the ratio of the mean free path at $373\mathrm{~K}$ to that at STP, we can divide λ' by λ: <br/><br/> $$\frac{\lambda'}{\lambda} = \frac{\frac{k(373\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}}{\frac{k(273\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}}$$ <br/><br/> The Boltzmann constant, pressure, and molecular diameter cancel out: <br/><br/> $\frac{\lambda'}{\lambda} = \frac{373\mathrm{~K}}{273\mathrm{~K}}$ <br/><br/> Now, we can solve for λ': <br/><br/> $\lambda' = \lambda \cdot \frac{373\mathrm{~K}}{273\mathrm{~K}}$ <br/><br/> Substituting the given value of λ as $1500d$: <br/><br/> $\lambda' = 1500d \cdot \frac{373\mathrm{~K}}{273\mathrm{~K}}$ <br/><br/> $\lambda' \approx 2049d$ <br/><br/> Thus, the mean free path of the molecules at $373\mathrm{~K}$ while maintaining the standard pressure is approximately $2049d$.