At what temperature, the rate of effusion of N_2 would be 1.625 times than that of SO_2 at 50^° C ?

Correct answer: (C) 373 ~K. r ∝ U and U=√((3 R T)/M) fracr_1r_2=√fracT_1 M_2~T_2 M_1 or fracr_N_2r_SO_2=√fracT_1 × 64323 × 28=1.625 or T_2=373 ~K

Q25 / 125 · BITSAT 2016 · Chemistry

At what temperature, the rate of effusion of $\mathrm{N}_{2}$ would be 1.625 times than that of $\mathrm{SO}_{2}$ at $50^{\circ} \mathrm{C}$ ?

A
$110 \mathrm{~K}$
B
$173 \mathrm{~K}$
Correct answer
C
$373 \mathrm{~K}$
D
$273 \mathrm{~K}$

Solution

$\quad r \propto U$ and $U=\sqrt{\frac{3 R T}{\text { M }}}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\sqrt{\frac{\mathrm{T}_{1} \mathrm{M}_{2}}{\mathrm{~T}_{2} \mathrm{M}_{1}}}$ or $\frac{\mathrm{r}_{\mathrm{N}_{2}}}{\mathrm{r}_{\mathrm{SO}_{2}}}=\sqrt{\frac{\mathrm{T}_{1} \times 64}{323 \times 28}}=1.625$ or $\mathrm{T}_{2}=373 \mathrm{~K}$

Hence, the correct answer is option: 'C'.

At what temperature, the rate of effusion of $\mathrm{N}_{2}$ would be 1.625 times than that of $\mathrm{SO}_{2}$ at $50^{\circ} \mathrm{C}$ ?

Solution

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