A vessel of volume `V = 20 1` contains a mixture of hydrogen and helium at a temperature `t = 20 ^@C` and pressure `p = 2.0 atm`. The mass of the mixture is equal is equal to `m = 5.0 g`. Find the ratio of the mass of hydrogen to that of helium in the given mixture.

Explanation

Let the mixture contain `v_1` and `v_2` moles of `H_2` and `H_e` respectively. If molecular weights of `H_2` and `H_e` are `M_1` and `M_2`, then respective masses in the mixture are equal to
`m_1 = v_1 M_1 and m_2 = v_2 M_2`
Therefore, for the total mass of the mixture we get,
`m = m_1 + m_2` or `m = v_1 M_1 + v_2 M_2` ...(1)
Also, if `v` is the total number of moles of the mixture in the vessels, then we know,
`v = v_1 + v_2` ...(2)
Solving (1) and (2) for `v_1` and `v_2`, we get,
`v_1 ((v M_2 - m))/(M_2 - M_1), v_2 = (m - v M_1)/(M_2 - M_1)`
Therefore, we get `m_1 = M_1 . ((vM_2 - m))/(M_2 - M_1)` and `m_2 = M_2 ((m - v M_1))/(M_2 - M_1)`
or, `(m_1)/(m_2) = (M_1)/(M_2) ((v M_2 - m))/((m - v M_1))`
One can also express the above result in terms of the effective molecular weight `M` of the mixture, defined as,
`M = (m)/(v) = m (RT)/(p V)`
Thus, `(m_1)/(m_2) = (M_1)/(M_2).(M_2 - M)/(M - M_1) = (1- M//M_2)/(M//M_1 - 1)`
Using the date and table, we get :
`M = 3.0 g` and, `(m_1)/(m_2) = 0.50`.