Kinetic Theory

JEE Main level, single correct, medium

156 of 168

Question 156 of 168

2kg2kg of a monoatomic gas is at a pressure of 4×104Nm−24\times {10}^{4}N{m}^{-2}. The density of the gas is 8kgm−38 kg{m}^{-3}. What is the order of energy of the gas due to its thermal motion?

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Answer: (C) 104J{10}^{4}J

Internal energy is the energy due to thermal motion.
The internal energy of an ideal gas given by

U=f2nRTU=\dfrac{f}{2}nRT (for nmolesnmoles)

U=f2(PV)=f2×P×MρU=\dfrac{f}{2}\left(PV\right)=\dfrac{f}{2}\times P\times \dfrac{M}{\rho } (Since, f=3f=3 for a monoatomic gas.)

U=32×4×104×28U=\dfrac{3}{2}\times 4\times {10}^{4}\times \dfrac{2}{8}

=1.5×104J=1.5\times {10}^{4}J
So, the order of energy due to thermal motion is 104{10}^{4}

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