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Question

The root mean square speed of a gas molecule at 20 C is v. If the temperature of the gas is raised by 10 C, then the root mean square speed of the gas molecule will be

A
2 v
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B
1.08 v
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C
1.02 v
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D
1.5 v
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Solution

The correct option is C 1.02 v
Given:
Initial root mean square speed, v1=v
Initial temperature of the gas, T1=20C=293 K
Change in temperature, ΔT=10 C
So, final temperature of the gas, T2=30C=303 K
To find:
Final root mean square speed, v2=?

We know that, root mean square speed is given by,
vrms=3RTM
Here, R is the gas constant, T is the temperature and M is the molar mass.
So, v1=3R×293M and
v2=3R×303M
[ gas is same, so molar mass is same ]
Now, v2v1=303293
v21.02v11.02 v

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