Three closed vessels $A, B$ and $C$ are at the same temperature $T$ and contain gases. Vessel $A$ contains only $\mathrm{O}_2, B$ contains only $\mathrm{N}_2$ and $\mathrm{C}$ contains a mixture of equal quantities of $\mathrm{O}_2$ and $\mathrm{N}_2$. If the rms speed of $\mathrm{O}_2$ molecules in vessel $A$ is $v_1$ and that of $\mathrm{N}_2$ molecules in vessel $B$ is $v_2$ then the rms speed of $\mathrm{O}_2$ molecules in vessel $\mathrm{C}$ is

Question

Three closed vessels $A, B$ and $C$ are at the same temperature $T$ and contain gases. Vessel $A$ contains only $\mathrm{O}_2, B$ contains only $\mathrm{N}_2$ and $\mathrm{C}$ contains a mixture of equal quantities of $\mathrm{O}_2$ and $\mathrm{N}_2$. If the rms speed of $\mathrm{O}_2$ molecules in vessel $A$ is $v_1$ and that of $\mathrm{N}_2$ molecules in vessel $B$ is $v_2$ then the rms speed of $\mathrm{O}_2$ molecules in vessel $\mathrm{C}$ is, $\left(\frac{v_1+v_2}{2}\right)$, $v_1$, $\left(v_1 v_2\right)$, $\frac{v_1}{2}$

MARKS App · Accepted Answer

Key Idea Root mean square velocity of any gas is given by $v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ Where, $M=$ molecular weight. As, rms speed of any gas is depends on the temperature of gas and molecular weight of gas. In a mixture of gases (say $\mathrm{N}_2$ and $\mathrm{O}_2$ ) at a constant temperature, rms speed is independent to the quantity (moles) of gas present in the gas mixture. So, in this problem the rms speed of $\mathrm{O}_2$ in vessel $C$ is $v_1$. Hence, the correct option is (b).

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