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.
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,