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Question

When the temperature of a gas is raised from 27 C to 90 C, the percentage increase in the r.m.s. velocity of the molecules will be

A
10%
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B
15%
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C
20%
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D
17.5%
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Solution

The correct option is A. 10%.

The rms velocity of the gas molecules is given by,

vrms = 3RTM

where R = Universal gas constant

T = Temperature of the gas

M = Molar mass of the gas

Thus, for a given gas R and M are constant.

vrmsT

v2rmsv1rms = T2T1 = (273+90)(273+27)==1.1

% increase = (v2v11) ×100 = (1.11)×100=0.1×100= 10%.


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