A) A is at temperature T and B at \[\operatorname{T}, T> T\]
B) A is lowered to a temperature \[{{\operatorname{T}}_{2}}, {{T}_{2}}< T\] while B is at T
C) Both A and B are raised to a higher temperature
D) Both A and B are placed at lower temperature
Correct Answer: B
Solution :
\[{{\left( {{U}_{AV}} \right)}_{A}}=\sqrt{\frac{8\,RT}{\pi {{M}_{A}}}}\,\,and\,\,{{\left( {{U}_{rms}} \right)}_{B}}=\sqrt{\frac{3RT}{{{M}_{B}}}}\] \[\therefore \,\,\,\,\,\,\,\,\frac{8}{3\pi }=\frac{{{M}_{A}}}{{{M}_{B}}}\] \[A\left( {{U}_{AV}} \right)=\sqrt{\frac{8\,R{{T}_{2}}}{\pi {{M}_{A}}}}\,\,for\,\,B\,\,{{V}_{AV}}=\sqrt{\frac{8RT}{\pi \,{{M}_{B}}}}\] \[\frac{{{T}_{2}}}{T}=\frac{{{M}_{A}}}{{{M}_{B}}}=\frac{8}{3\pi }\] \[\therefore \,\,\,\,\,\,{{T}_{2}}=\frac{8}{3\pi }T\,\,\,\,or\,\,\,{{T}_{2}}<T\,\,\,\]
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