The diffusion coefficient of an ideal gas is | vedclass

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(A) The diffusion coefficient $D$ is proportional to the mean free path $\lambda$ and the mean speed $v$,so $D \propto \lambda \cdot v$.We know that t

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$4$

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(A) The diffusion coefficient $D$ is proportional to the mean free path $\lambda$ and the mean speed $v$,so $D \propto \lambda \cdot v$.We know that the mean free path $\lambda \propto \frac{T}{P}$ and the mean speed $v \propto \sqrt{T}$.Therefore,$D \propto \frac{T}{P} \cdot \sqrt{T} = \frac{T^{3/2}}{P}$.Given that the temperature $T_f = 4T_i$ and the pressure $P_f = 2P_i$,we can write the ratio:$\frac{D_f}{D_i} = \left( \frac{T_f}{T_i} \right)^{3/2} \cdot \left( \frac{P_i}{P_f} \right)$$\frac{D_f}{D_i} = (4)^{3/2} \cdot \left( \frac{1}{2} \right)$$\frac{D_f}{D_i} = 8 \cdot \frac{1}{2} = 4$.Thus,$D_f = 4D_i$,which means $x = 4$.

The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased $4$ times and its pressure is increased $2$ times. As a result,the diffusion coefficient of this gas increases $x$ times. The value of $x$ is

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