Q.

Let vAVG , vRMS  and vMP denote mean speed, rms speed and most probable speed, respectively, of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is m. Then,

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a

no molecule can have a speed greater than 2vRMS

b

no molecule can have a speed less than vMP2

c

vMP>vAVG>vRMS

d

the average kinetic energy of a molecule is 34mvMP2

answer is D.

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Detailed Solution

According to the kinetic theory of gases, a molecule of a gas can have any speed between 0 and ¥. Thus, (1) and (2) are obviously wrong.

Also, vRMS=3RTM;  vAVG=8RTπM;   vMP=2RTM .  So, vRMS>vAVG>vMP  
EAVG=12mvRMS2=12m3kTm=32kT     (1)
But  vMP=2kTm
kT=12mvMP2.        (2)
Using (2) in (1),
EAVG=34mvMP2.
 

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