At what temperature, the rate of effusion of N2 would be 1.625 times that of SO2 at 50°C

rN2rSO2=VrmsN2VrmsSO2=TN2TSO2⋅MSO2MN2=TN2323×64281.625=TN2323⋅167TN2=(1.625)2×323×716=373K

At what temperature, the rate of effusion of N2 would be 1.625 times that of SO2 at 50°C

rN2rSO2=VrmsN2VrmsSO2=TN2TSO2⋅MSO2MN2=TN2323×64281.625=TN2323⋅167TN2=(1.625)2×323×716=373K

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

At what temperature, the rate of effusion of N₂ would be 1.625 times that of SO₂ at 50°C

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

110 K

173 K

373 K

273 K

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \frac{r_{N_{2}}}{r_{{SO}_{2}}} = \frac{V_{rms} N_{2}}{V_{rms} {SO}_{2}} = \sqrt{\frac{T_{N_{2}}}{T_{{SO}_{2}}} \cdot \frac{M_{{SO}_{2}}}{M_{N_{2}}}} = \sqrt{\frac{T_{N_{2}}}{323} \times \frac{64}{28}} \\ 1 .625 = \sqrt{\frac{T_{N_{2}}}{323} \cdot \frac{16}{7}} \\ T_{N_{2}} = \frac{(1 .625)^{2} \times 323 \times 7}{16} = 373 K \end{array}$