A) \[{{100}^{\circ }}C\]
B) \[{{273}^{\circ }}C\]
C) \[0\,K\]
D) \[{{0}^{\circ }}C\]
Correct Answer: D
Solution :
Square-root of the mean-square velocity of the gas molecules is called the root mean square velocity and is denoted by \[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}\] Where \[T\]is temperature, M is molecular weight, R is gas constant. Let \[{{v}_{H}}\] and \[{{v}_{He}}\] be velocities of hydrogen and Helium respectively. T and T? be the temperature of hydrogen and helium respectively \[{{v}_{He}}=\frac{5}{7}{{v}_{H}}\] \[\sqrt{\frac{3RT'}{{{M}_{He}}}}=\frac{5}{7}\sqrt{\frac{3RT}{{{M}_{H}}}}\] \[\Rightarrow \] \[\sqrt{\frac{T'}{T}}=\frac{5}{7}\sqrt{\frac{{{M}_{He}}}{{{M}_{H}}}}\] \[\Rightarrow \] \[\frac{T'}{T}={{\left( \frac{5}{7} \right)}^{2}}\frac{{{M}_{He}}}{{{M}_{H}}}\] \[\Rightarrow \] \[\frac{T'}{T}=\frac{25\times 2}{49\times 1}=1\] \[\Rightarrow \] \[T'=273\,\,K\] In Centigrade \[T'=273-273={{0}^{\circ }}C\] Note: Faster the motion of molecules of a gas, Higher will be the temperature of the gas.
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