Consider the following statements for air mol | vedclass

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(A) $1$. Average speed $\bar{c}$ is related to $c_{rms}$ as $\bar{c} = \sqrt{\frac{8}{3\pi}} c_{rms} \approx 0.92 c_{rms}$. Since $0.92 < 1$,the avera

Vedclass · Accepted Answer

Only statement $II$ is correct

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(A) $1$. Average speed $\bar{c}$ is related to $c_{rms}$ as $\bar{c} = \sqrt{\frac{8}{3\pi}} c_{rms} \approx 0.92 c_{rms}$. Since $0.92 $2$. In a gas,the mean free path $\lambda$ is typically much larger than the average distance between molecules $d \approx (V/N)^{1/3}$. Thus,statement $(II)$ is correct.$3$. The mean free path is given by $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$. At constant volume,$P \propto T$,so $\lambda$ remains constant with temperature. If pressure is constant,$\lambda \propto T$. However,in an air-tight container,volume is constant,so $\lambda$ does not increase with temperature. Thus,statement $(III)$ is incorrect.$4$. The rms speed is $c_{rms} = \sqrt{\frac{3RT}{M}}$. Since the molar mass of nitrogen $(M_{N_2} = 28 	ext{ g/mol})$ is less than that of oxygen $(M_{O_2} = 32 	ext{ g/mol})$,the rms speed of nitrogen is greater than that of oxygen. Thus,statement $(IV)$ is incorrect.Therefore,only statement $(II)$ is correct.

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