Answer
a) Greater than.
b) Less than.
Work Step by Step
a) Since the two gases are in contact for a long time, they have the same temperature.
Now we need to figure out whether their molecules have the same RMS speed or not.
We know that the RMS speed is given by
$$v_{\rm rms} =\sqrt{\dfrac{3k_BT}{m}}$$
Hence,
$$\dfrac{v_{\rm rms,He} }{v_{\rm rms,Ar} }=\dfrac{\sqrt{\dfrac{3k_BT}{m_{\rm He}}}}{\sqrt{\dfrac{3k_BT}{m_{\rm Ar}}}}$$
$$\dfrac{v_{\rm rms,He} }{v_{\rm rms,Ar} }=\sqrt{\dfrac{m_{\rm Ar}}{m_{\rm He}}}\tag 1$$
Now we need to find the mass of helium and argon.
$$n_{He}=\dfrac{m_{\rm He}}{M_{\rm He}}$$
Hence,
$$m_{\rm He}=n_{He}M_{\rm He}=(0.1)(4)=0.4\;\rm g$$
By the same approach,
$$m_{\rm Ar}=n_{Ar}M_{\rm Ar}=(0.2)(39.9)=7.98\;\rm g$$
Plugging into (1);
$$\dfrac{v_{\rm rms,He} }{v_{\rm rms,Ar} }=\sqrt{\dfrac{7.98 }{0.4 }}=\bf 4.5 $$
Therefore,
$$v_{\rm rms,He} =4.5v_{\rm rms,Ar} $$
And hence,
$$\boxed{v_{\rm rms,He}\gt v_{\rm rms,Ar} }$$
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b) The thermal energy for monatomic gases, like helium and argon, is given by
$$E_{th}=\dfrac{3}{2}nRT$$
So,
$$\dfrac{(E_{th})_{He}}{(E_{th})_{Ar}}=\dfrac{\frac{3}{2}n_{He}RT}{\frac{3}{2}n_{Ar}RT}$$
We know that both gases are at the same temperature.
So,
$$\dfrac{(E_{th})_{He}}{(E_{th})_{Ar}}=\dfrac{ n_{He} }{n_{Ar} }=\dfrac{0.1}{0.2}=\frac{1}{2}$$
Therefore,
$${(E_{th})_{Ar}}=2{(E_{th})_{He}}$$
And hence,
$$\boxed{{(E_{th})_{Ar}}\gt {(E_{th})_{He}}}$$
