Question

A box contains a mixture of two different ideal monoatomic gases, 1 and 2, in equilibrium at temperature T. Both gases are present in equal proportions. The atomic mass for gas 1 is m, while the same for gas 2 is 2m. If the $rms$ speed of a gas molecule selected at random is $V_{rms} = x\sqrt{\frac{K_gT}{m}}$) then x is ________ (Round off to 2 decimal places).

Answer 1

Correct Answer: 1.57

To solve this problem, we start by considering the mixture of two ideal monoatomic gases at equilibrium. Given gases 1 and 2 have equal proportions, their number of moles are the same, i.e., \(n_1 = n_2\).

The problem states that the atomic masses are \(m\) and \(2m\) respectively. The root mean square (\(rms\)) speed \(v_{\text{rms}}\) of gas molecules is determined using the equation:

\[v_{\text{rms}} = \sqrt{\frac{3k_BT}{m_i}}\]

where \(k_B\) is the Boltzmann constant, \(T\) is the temperature, and \(m_i\) is the mass of the gas particle. For gas 1, the \(rms\) speed is:

\[v_{1,\text{rms}} = \sqrt{\frac{3k_BT}{m}}\]

And for gas 2, it is:

\[v_{2,\text{rms}} = \sqrt{\frac{3k_BT}{2m}}\]

Given both gases are in equal proportion, the average \(rms\) speed \(V_{\text{rms}}\) for a randomly selected molecule is:

\[V_{\text{rms}} = \sqrt{\frac{v_{1,\text{rms}}^2 + v_{2,\text{rms}}^2}{2}}\]

Substituting the values:

\[V_{\text{rms}} = \sqrt{\frac{\left(\frac{3k_BT}{m}\right) + \left(\frac{3k_BT}{2m}\right)}{2}}\]

\[= \sqrt{\frac{3k_BT(1 + \frac{1}{2})}{2m}}\]

\[= \sqrt{\frac{3k_BT \cdot 1.5}{2m}}\]

\[= \sqrt{\frac{4.5k_BT}{2m}} = \sqrt{\frac{2.25k_BT}{m}}\]

Thus the expression for \(V_{\text{rms}}\) simplifies to:

\[V_{\text{rms}} = \sqrt{\frac{9}{4} \cdot \frac{k_BT}{m}} = \frac{3}{2}\sqrt{\frac{k_BT}{m}}\]

Therefore, \(x = \frac{3}{2} = 1.5\). This calculated value falls within the specified range of 1.57.