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Question Number 18624 by tawa tawa last updated on 25/Jul/17

$$\mathrm{Argon}\mathrm{diffuses}\mathrm{through}\mathrm{a}\mathrm{hole}\mathrm{under}\mathrm{prescribed}\mathrm{condition}\mathrm{of}\mathrm{temperature}$$$$\mathrm{and}\mathrm{pressure}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}{3\mathrm{cm}}^3\mathrm{per}\mathrm{velocity}.\mathrm{At}\mathrm{what}\mathrm{velocity}\mathrm{will}\mathrm{helium}$$$$\mathrm{diffuse}\mathrm{through}\mathrm{the}\mathrm{same}\mathrm{hole}\mathrm{under}\mathrm{the}\mathrm{same}\mathrm{condition}$$$$(\mathrm{Ar}=29.94\mathrm{g},\mathrm{He}=4\mathrm{g}).$$

Answered by Tinkutara last updated on 26/Jul/17

$$\mathrm{Rate}\mathrm{of}\mathrm{diffusion}\mathrm{of}\mathrm{argon}=r_{\mathrm{Ar}}=\frac{v_{\mathrm{Ar}}}{t}$$$$\mathrm{Rate}\mathrm{of}\mathrm{diffusion}\mathrm{of}\mathrm{helium}=r_{\mathrm{He}}=\frac{v_{\mathrm{He}}}{t}$$$$v_{\mathrm{Ar}}\mathrm{and}{v}_{\mathrm{He}}\mathrm{denote}\mathrm{the}\mathrm{volumes}\mathrm{of}$$$$\mathrm{argon}\mathrm{and}\mathrm{helium}\mathrm{diffused}\mathrm{in}\mathrm{the}\mathrm{same}$$$$\mathrm{time}.$$$$\frac{r_{\mathrm{Ar}}}{r_{\mathrm{He}}}=\sqrt{\frac{{\mathrm{M}}_{\mathrm{He}}}{{\mathrm{M}}_{\mathrm{Ar}}}}=\sqrt{\frac{4}{29.94}}=\frac{v_{\mathrm{Ar}}}{v_{\mathrm{He}}}$$$$\frac{3}{v_{\mathrm{He}}}=\sqrt{\frac{4}{29.94}}$$$$v_{\mathrm{He}}=8.2076{\mathrm{cm}}^3$$

Commented by tawa tawa last updated on 26/Jul/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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