A) \[\frac{{{E}_{1}}}{\sqrt{2}}\]
B) \[\sqrt{2}{{E}_{1}}\]
C) \[2{{E}_{1}}\]
D) \[\frac{{{E}_{1}}}{2}\]
Correct Answer: C
Solution :
Let the mass of 1 g molecule of a gas be M, and its mean square velocity be \[{{v}^{2}},\]then kinetic energy of 1 g molecule of gas is \[\frac{1}{2}M{{v}^{2}}=\frac{1}{2}M\left( \frac{3RT}{M} \right)=\frac{3}{2}RT\] There are N-molecules in 1 g molecule of gas. Thus, average kinetic energy of one molecule \[=\frac{(3/2)RT}{N}\] \[=\frac{3}{2}\left( \frac{R}{N} \right)T\] where, (R/N) = k is Boltzmann's constant. \[\therefore \] \[KE=\frac{3}{2}kT.\] Given, \[{{T}_{1}}=27{{\,}^{o}}C=273+27=300\,K\] \[{{T}_{2}}=327{{\,}^{o}}C=273+327=600\,K\] \[\therefore \] \[\frac{{{(KE)}_{2}}}{{{(KE)}_{1}}}=\frac{600}{300}=\frac{2}{1}\] \[\Rightarrow \] \[{{(KE)}_{2}}=2{{(KE)}_{1}}\] \[\Rightarrow \] \[{{E}_{2}}=2{{E}_{1}}\]
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