Question

At what temperature will the rms velocity of so2 be the same as that of o2 at 303 k

Answer 1

Answer:

uRMS=3RT/M‾‾‾‾‾‾‾√

uO2uSO2=TO2MSO2TSO2MO2‾‾‾‾‾‾‾√

uO2=uSO2

∴ TO2MSO2=TSO2MO2

303 × 64 = TSO2×32

TSO2 = 606 K

Explanation:

Answer 2

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GIVEN :

Temperature of O2(T2) = 303 k

TO FIND :

Temperature of SO2.(T1)

PROCESS :

Molar mass of O2(M2) = 16 × 2 = 32

Molar mass of SO2(M1) = 32 + (16 × 2) = 64.

A/Q,

$\sf{V_{r.m.s} \:( SO_{2} )= V_{r.m.s} \:( O_{2} )}$

After putting the formula,

$\sf{ \implies \: \sqrt{ \frac{3 \: R \: T_{1} }{M_{1} } } = \: \sqrt{\frac{3 \: R \: T_{2} }{M_{2} }} }$

After cancelling the common values, we get

$\sf{ \implies \: \: \frac{T_{1}}{M_{1}} = \: \frac{T_{2}}{M_{2}} }$

$\sf {\implies \: \frac{T_{1}}{64} = \frac{303}{32} }$

$\sf{ \implies \: T_{1} \: = \: 303 \: \times \: 2}$

$\sf{ \implies T_{1} = \: 606 \: K}$

Hence our required temperature is 606 k.