What is the composition of the gas that diffused out initially from $2^{n d}$ container?

\frac{1}{A} = \frac{3}{7}

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X_{B} = \frac{3}{5}

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X_{A} = \frac{1}{3}

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X_{B} = \frac{1}{4}

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Solution

The correct option is C $X_{A} = \frac{1}{3}$
The expression for the rate of effusion is $\frac{n_{A} / t}{n_{B} / t} = \sqrt{\frac{M_{B}}{M_{A}}}$
The expression for the composition of mixture is $\frac{n_{1} M_{1} + n_{2} M_{2}}{x_{1} + x_{2}} = M_{m i x}$
Thus, $X_{1} M_{1} + (1 - X_{1}) M_{2} = M_{m i x}$
$\Rightarrow X_{1} = \frac{2}{5}, X_{2} = \frac{3}{5}$
The ratio $\Rightarrow \frac{X_{A}}{X_{B}} = \frac{2}{3} \sqrt{\frac{72}{128}} = \frac{1}{2}$
Hence, $X_{A} = \frac{1}{3}, X_{B} = \frac{2}{3}$
Thus, the composition of the gas that diffused out initially from 2nd container is $X_{A} = \frac{1}{3}, X_{B} = \frac{2}{3}$ .

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Similar questions

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1

Prove that:

(i) $\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$

(ii) $\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1$

X_{A}, X_{B}, X_{C}

X_{D}

A, B, C,

D

\frac{1}{1 + x^{(b - a)} + x^{(c - a)}} + \frac{1}{1 + x^{(a - b)} + x^{(c - b)}} + \frac{1}{1 + x^{(b - c)} + x^{(a - c)}} = ?

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