At what temperature, the rate of effusion of N 2 would be 1.625 times than the rate of SO2 at 500oC ?

rN2rSO2=1.625=M2M1∵ u∝TM and r∝u∴ rN2rSO2=1.625=TN2×MSO2TSO2×MN2=T×64773×28∴ T=893K or 620∘C

At what temperature, the rate of effusion of N 2 would be 1.625 times than the rate of SO2 at 500oC ?

rN2rSO2=1.625=M2M1∵ u∝TM and r∝u∴ rN2rSO2=1.625=TN2×MSO2TSO2×MN2=T×64773×28∴ T=893K or 620∘C

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At what temperature, the rate of effusion of N₂ would be 1.625 times than the rate of SO₂ at 500^oC ?

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373°C

620°C

110°C

173°C

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{_{r_{N_{2}}}}{r_{{SO}_{2}}} = 1 .625 = \sqrt{\frac{M_{2}}{M_{1}}} \\ ∵ u \propto \sqrt{\frac{T}{M}} and r \propto u \\ ∴ \frac{r_{N_{2}}}{r_{{SO}_{2}}} = 1 .625 = \sqrt{\frac{T_{N_{2}} \times M_{{SO}_{2}}}{T_{{SO}_{2}} \times M_{N_{2}}}} = \sqrt{\frac{T \times 64}{773 \times 28}} \\ ∴ T = 893 K or 620^{\circ} C \end{array}$