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When the temperatw-e of a gas is raised ...

When the temperatw-e of a gas is raised from 30 °C to 90 °C, the percentage increase in the R.M.S. velocity of the molecules will be

A

0.6

B

0.1

C

0.15

D

0.3

Text Solution

Verified by Experts

The correct Answer is:
B
` T_1 = 273 +30 = 303 K " " T_2 = 273 +90 = 373 K`
` V_(rms) prop sqrtT `
` (V_1)/(V_2) = sqrt((T_1)/(T_2)) = sqrt((303)/(363))`
`(V_1)/(V_2)=1.1`
`(V_2)/(V_1) -1=1.1 -1 " " [because ` Subtract 1 from both sides]
`(Delta V)/(V) xx 100 `
Percentage incrase ` (Delta V)/(V) = 10%`
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