At STP, a closed vessel contains 1 mole each of He and $\mathrm{CH}_4$. Through a small hole, 2L of He and 1…

At STP, a closed vessel contains 1 mole each of He and $\mathrm{CH}_4$. Through a small hole, 2L of He and 1 L of $\mathrm{CH}_4$ was escaped from vessel in ' $t$ ' minutes. What are the mole fractions of He and $\mathrm{CH}_4$ respectively remaining in the vessel? (Assume He and $\mathrm{CH}_4$ as ideal gases. At STP one mole of an ideal gas occupies 22.4 L of volume)
  1. 0.512,0.488
  2. 0.5,0.5
  3. 0.329,0.671
  4. 0.488,0.512

Solution

Initial mole of $\mathrm{He}\left(\mathrm{n}_1\right)=1$
Initial mole of $\mathrm{CH}_4\left(\mathrm{n}_2\right)=1$ No. of mole of He escaped $\left(\mathrm{n}_2^{\prime}\right)=\frac{2}{22.4}=0.089$ No. of mole of $\mathrm{CH}_4$ escaped $\left(\mathrm{n}_2^{\prime}\right)=\frac{2}{22.4}=0.044$ Remaining no. of mole $\left(\mathrm{n}^{\prime \prime}{ }_1\right)=\mathrm{n}_1-\mathrm{n}_1^{\prime}$ $\begin{aligned} & =1-0.089 \\ & =0.90 \end{aligned}$
Remaining no. of mole $\left(\mathrm{n}^{\prime \prime}{ }_2\right)=1-0.044$ $=0.95$
Mole fraction of $\mathrm{He}\left(\chi_{\mathrm{He}}\right)=\frac{0.90}{1.85}=0.488$ Mole fraction of $\mathrm{CH}_4\left(\chi_{\mathrm{CH}_4}\right)=\frac{0.95}{1.85}=0.512$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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