JEE Advance - Chemistry (2016 - Paper 1 Offline - No. 5)
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 ___________:
Answer
4
Explanation
The diffusion coefficient (D) is proportional to the mean free path (λ) and the mean speed (Umean), so we can write
$$D ∝ λU_{\text{mean}}$$
The mean free path (λ) is given by
$$λ = \frac{RT}{\sqrt{2} N_0 σP}$$
hence
$$λ ∝ \frac{T}{P}$$
The mean speed (Umean) is given by
$$U_{\text{mean}} = \sqrt{\frac{8RT}{πM}}$$
hence
$$U_{\text{mean}} ∝ \sqrt{T}$$
Therefore,
$$D ∝ \frac{T^{3/2}}{P}$$
The change in the diffusion coefficient would then be given by :
$$\frac{(DC)_2}{(DC)_1} = \frac{P_1}{P_2} \cdot \left(\frac{T_2}{T_1}\right)^{3/2}$$
Substituting $P_2 = 2P_1$ and $T_2 = 4T_1$ into the equation, we get :
$$\frac{(D)_2}{(D)_1} = \frac{1}{2} \cdot (4)^{3/2} = 4$$
So, the diffusion coefficient of the gas increases 4 times. Hence, x = 4.
$$D ∝ λU_{\text{mean}}$$
The mean free path (λ) is given by
$$λ = \frac{RT}{\sqrt{2} N_0 σP}$$
hence
$$λ ∝ \frac{T}{P}$$
The mean speed (Umean) is given by
$$U_{\text{mean}} = \sqrt{\frac{8RT}{πM}}$$
hence
$$U_{\text{mean}} ∝ \sqrt{T}$$
Therefore,
$$D ∝ \frac{T^{3/2}}{P}$$
The change in the diffusion coefficient would then be given by :
$$\frac{(DC)_2}{(DC)_1} = \frac{P_1}{P_2} \cdot \left(\frac{T_2}{T_1}\right)^{3/2}$$
Substituting $P_2 = 2P_1$ and $T_2 = 4T_1$ into the equation, we get :
$$\frac{(D)_2}{(D)_1} = \frac{1}{2} \cdot (4)^{3/2} = 4$$
So, the diffusion coefficient of the gas increases 4 times. Hence, x = 4.
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