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Physics Question on kinetic theory

The average energy of molecules in a sample of oxgen gas at 300 K {300 \, K} are 6.21×10−21 J {6.21 \times 10^{-21} \, J}. The corresponding values at 600 K {600\, K} are

A

\ce6.21×10−21 J\ce{6.21 \times 10^{-21} \, J}

B

\ce8.78×10−21 J\ce{8.78 \times 10^{-21} \, J}

C

\ce4.21×10−21 J\ce{4.21 \times 10^{-21} \, J}

D

\ce12.42×10−21 J\ce{12.42 \times 10^{-21} \, J}

Answer

\ce12.42×10−21 J\ce{12.42 \times 10^{-21} \, J}

Explanation

Solution

The average kinetic energy of molecules in a sample of oxygen gas is given by
KEavg=32kT or (KEavg)1(KEavg)2=T1T2{KE_{avg} = \frac{3}{2}kT\, or \, \frac{(KE_{avg})_{1}}{(KE_{avg})_{2}} = \frac{T_{1}}{T_{2}}}
(KEavg)1=6.21×10−21J,T1=300K{(KE_{avg})_{1} =6.21 \times10^{-21} J, T_{1} =300 K}
T2=600 K,(KEavg)2=?{T_2 = 600 \, K , (KE_{avg})_2 = ?}
∴   6.21×10−21(KEavg)2=300600=12\therefore \:\:\: {\frac{6.21 \times10^{-21}}{(KE_{avg})_{2}} =\frac{300}{600} =\frac{1}{2}}
(KEavg)2=12.42×10−21J{(KE_{avg})_{2} = 12.42 \times10^{-21} J}