Question
A gas mixture consists of molecules of type 1, 2 and 3, with molar masses $${m_1} > {m_2} > {m_3}.{v_{rms}}$$ and $$\overline K $$ are the $$r.m.s.$$ speed and average kinetic energy of the gases. Which of the following is true?
A.
$${\left( {{v_{rms}}} \right)_1} < {\left( {{v_{rms}}} \right)_2} < {\left( {{v_{rms}}} \right)_3}\,{\text{and}}\,{\left( {\overline K } \right)_1} = {\left( {\overline K } \right)_2} = {\left( {\overline K } \right)_3}$$
B.
$${\left( {{v_{rms}}} \right)_1} = {\left( {{v_{rms}}} \right)_2} = {\left( {{v_{rms}}} \right)_3}\,{\text{and}}\,{\left( {\overline K } \right)_1} = {\left( {\overline K } \right)_2} > {\left( {\overline K } \right)_3}$$
C.
$${\left( {{v_{rms}}} \right)_1} > {\left( {{v_{rms}}} \right)_2} > {\left( {{v_{rms}}} \right)_3}\,{\text{and}}\,{\left( {\overline K } \right)_1} < {\left( {\overline K } \right)_2} > {\left( {\overline K } \right)_3}$$
D.
$${\left( {{v_{rms}}} \right)_1} > {\left( {{v_{rms}}} \right)_2} > {\left( {{v_{rms}}} \right)_3}\,{\text{and}}\,{\left( {\overline K } \right)_1} < {\left( {\overline K } \right)_2} < {\left( {\overline K } \right)_3}$$
Answer :
$${\left( {{v_{rms}}} \right)_1} < {\left( {{v_{rms}}} \right)_2} < {\left( {{v_{rms}}} \right)_3}\,{\text{and}}\,{\left( {\overline K } \right)_1} = {\left( {\overline K } \right)_2} = {\left( {\overline K } \right)_3}$$
Solution :
$${v_{rms}} \propto \frac{1}{{\sqrt M }} \Rightarrow {\left( {{v_{rms}}} \right)_1} < {\left( {{v_{rms}}} \right)_2} < {\left( {{v_{rms}}} \right)_3}$$
also in mixture temperature of each gas will be same, hence kinetic energy also remains same.