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Argon-diffuses-through-a-hole-under-prescribed-condition-of-temperature-and-pressure-at-the-rate-of-3cm-3-per-velocity-At-what-velocity-will-helium-diffuse-through-the-same-hole-under-the-same-con




Question Number 18624 by tawa tawa last updated on 25/Jul/17
Argon diffuses through a hole under prescribed condition of temperature  and pressure at the rate of  3cm^3  per velocity. At what velocity will helium   diffuse through the same hole under the same condition   (Ar = 29.94 g,  He = 4g).
$$\mathrm{Argon}\:\mathrm{diffuses}\:\mathrm{through}\:\mathrm{a}\:\mathrm{hole}\:\mathrm{under}\:\mathrm{prescribed}\:\mathrm{condition}\:\mathrm{of}\:\mathrm{temperature} \\ $$$$\mathrm{and}\:\mathrm{pressure}\:\mathrm{at}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\:\mathrm{3cm}^{\mathrm{3}} \:\mathrm{per}\:\mathrm{velocity}.\:\mathrm{At}\:\mathrm{what}\:\mathrm{velocity}\:\mathrm{will}\:\mathrm{helium}\: \\ $$$$\mathrm{diffuse}\:\mathrm{through}\:\mathrm{the}\:\mathrm{same}\:\mathrm{hole}\:\mathrm{under}\:\mathrm{the}\:\mathrm{same}\:\mathrm{condition}\: \\ $$$$\left(\mathrm{Ar}\:=\:\mathrm{29}.\mathrm{94}\:\mathrm{g},\:\:\mathrm{He}\:=\:\mathrm{4g}\right). \\ $$
Answered by Tinkutara last updated on 26/Jul/17
Rate of diffusion of argon = r_(Ar)  = (v_(Ar) /t)  Rate of diffusion of helium = r_(He)  = (v_(He) /t)  v_(Ar)  and v_(He)  denote the volumes of  argon and helium diffused in the same  time.  (r_(Ar) /r_(He) ) = (√(M_(He) /M_(Ar) )) = (√(4/(29.94))) = (v_(Ar) /v_(He) )  (3/v_(He) ) = (√(4/(29.94)))  v_(He)  = 8.2076 cm^3
$$\mathrm{Rate}\:\mathrm{of}\:\mathrm{diffusion}\:\mathrm{of}\:\mathrm{argon}\:=\:{r}_{\mathrm{Ar}} \:=\:\frac{{v}_{\mathrm{Ar}} }{{t}} \\ $$$$\mathrm{Rate}\:\mathrm{of}\:\mathrm{diffusion}\:\mathrm{of}\:\mathrm{helium}\:=\:{r}_{\mathrm{He}} \:=\:\frac{{v}_{\mathrm{He}} }{{t}} \\ $$$${v}_{\mathrm{Ar}} \:\mathrm{and}\:{v}_{\mathrm{He}} \:\mathrm{denote}\:\mathrm{the}\:\mathrm{volumes}\:\mathrm{of} \\ $$$$\mathrm{argon}\:\mathrm{and}\:\mathrm{helium}\:\mathrm{diffused}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{time}. \\ $$$$\frac{{r}_{\mathrm{Ar}} }{{r}_{\mathrm{He}} }\:=\:\sqrt{\frac{\mathrm{M}_{\mathrm{He}} }{\mathrm{M}_{\mathrm{Ar}} }}\:=\:\sqrt{\frac{\mathrm{4}}{\mathrm{29}.\mathrm{94}}}\:=\:\frac{{v}_{\mathrm{Ar}} }{{v}_{\mathrm{He}} } \\ $$$$\frac{\mathrm{3}}{{v}_{\mathrm{He}} }\:=\:\sqrt{\frac{\mathrm{4}}{\mathrm{29}.\mathrm{94}}} \\ $$$${v}_{\mathrm{He}} \:=\:\mathrm{8}.\mathrm{2076}\:\mathrm{cm}^{\mathrm{3}} \\ $$
Commented by tawa tawa last updated on 26/Jul/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

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