Kinetic Theory

JEE Main level, single correct, medium

39 of 168

Question 39 of 168

Given below are two statements:
Statement I: The temperature of a gas is −73∘C-73^{\circ}C. When the gas is heated to 527∘C527^{\circ}C, the root mean square speed of the molecules is doubled.
Statement II : The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.
In the light of the above statements, choose the correct answer from the options given below:

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Answer: (B) Statement I is true but Statement II is false

Given:
T1=−73∘C=(−73+273)K=200K{T}_{1}=-73^{\circ}C=\left(-73+273\right)K=200K
and T2=527∘C=(527+273)K=800K{T}_{2}=527^{\circ}C=\left(527+273\right)K=800K
Therefore,
Vrms,1Vrms,2=3RT1M3RT2M=T1T2\dfrac{{V}_{rms,1}}{{V}_{rms,2}}=\dfrac{\sqrt{\dfrac{3R{T}_{1}}{M}}}{\sqrt{\dfrac{3R{T}_{2}}{M}}} =\sqrt{\dfrac{{T}_{1}}{{T}_{2}}}
=200800=12=\sqrt{\dfrac{200}{800}} =\dfrac{1}{2}
Vrms,2=2Vrms,1{V}_{rms,2}=2{V}_{rms,1}
Hence,
Statement-I is true.
Now,
PV=nRTPV=nRT
But translational KE=32nRTKE=\dfrac{3}{2}nRT
Hence, Statement-II is false.

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