Question 264: Thermodynamics & KTG

Question

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:

Accepted Answer

The mean free path (ÃÂ») of molecules in a gas is given by the formula:
$$\lambda = \frac{kT}{\sqrt{2}\pi d^2 P}$$
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.
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 ÃÂ»':
$$\lambda' = \frac{k(373\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}$$
To find the ratio of the mean free path at $$373\mathrm{~K}$$ to that at STP, we can divide ÃÂ»' by ÃÂ»:
$$\frac{\lambda'}{\lambda} = \frac{\frac{k(373\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}}{\frac{k(273\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}}$$
The Boltzmann constant, pressure, and molecular diameter cancel out:
$$\frac{\lambda'}{\lambda} = \frac{373\mathrm{~K}}{273\mathrm{~K}}$$
Now, we can solve for ÃÂ»':
$$\lambda' = \lambda \cdot \frac{373\mathrm{~K}}{273\mathrm{~K}}$$
Substituting the given value of ÃÂ» as $$1500d$$:
$$\lambda' = 1500d \cdot \frac{373\mathrm{~K}}{273\mathrm{~K}}$$
$$\lambda' \approx 2049d$$
Thus, the mean free path of the molecules at $$373\mathrm{~K}$$ while maintaining the standard pressure is approximately $$2049d$$.

Answer

$$750 \mathrm{~d}$$

Answer

$$1500 \mathrm{~d}$$

Answer

$$\mathrm{2049~ d}$$

Answer

$$1098 \mathrm{~d}$$

Thermodynamics & KTG

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:

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