Class 12 Chemistry Gases & Solutions States Of Matter

The temperature of the gas is raised from $27_{∘}C$ to $927_{∘}C$, the root mean square velocity is

(a)

$27927 $ times of the earlier value

(b)

same as before

(c)

halved

(d)

doubled

Correct answer: (d)

Solution: Root mean square velocity at $T_{1}$ temperature,

$U_{1}=M3RT_{2} =M3R(27+273) ...(i)$

Root mean square velocity at $T_{2}$ temperature,

$U_{2}=M3RT_{2} =M3R(927+273) ...(ii)$

Eq. (i) divided by Eq. (ii)

$U_{2}U_{1} =927+27327+273 =1200300 =21 $

$U_{2}=2U_{1}$

$U_{1}=M3RT_{2} =M3R(27+273) ...(i)$

Root mean square velocity at $T_{2}$ temperature,

$U_{2}=M3RT_{2} =M3R(927+273) ...(ii)$

Eq. (i) divided by Eq. (ii)

$U_{2}U_{1} =927+27327+273 =1200300 =21 $

$U_{2}=2U_{1}$

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