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When the temperature of a gas is raised ...

When the temperature of a gas is raised from `30^@C` to `90^@C` the percentage increase in the rms 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
`30^@C=303K,90^@C=963K`
rms speed `c prop sqrt T`
hence , `c_2/c_1=sqrt(T_2/T_1)=sqrt(363/303)=sqrt(1.2) approx =1.1`
`therefore` The percentage increase
`=(c_2-c_1)/c_1times100=(c_2/c_1-1)times100`
`=(1.1-1)times100=10%`
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