QUESTION:
A vessel of volume V = 20\rm \ L contains a mixture of hydrogen and helium at a temperature T = 20\rm \ ^0C and pressure P = 2.0\rm \ atm. The mass of the mixture is equal to m = 5.0\rm \ g. Find the ratio of the mass of hydrogen to that of helium in the given mixture.
To solve this problem, you must have an idea of the ideal gas law.
SOLUTION:
Let:
- n_1 and n_2 are the number of moles of hydrogen and helium in the mixture,
- m is the mass of the mixture, and
- P, V and T are the pressure, volume and temperature of the mixture,
Given:
- V = 20\rm \ L
- T = 20\rm \ ^0C = 293\rm \ K
- P = 2.0\rm \ atm
- m = 5.0\rm \ g
(We have Gas constant, R = 0.0821\rm \ L.atm.mol^{-1}.K^{-1})
From the ideal gas equation
\begin{aligned}PV &= (n_1+n_2)RT\\\Rightarrow (2.0\rm \ atm) \times (20\rm \ L) & = (n_1+n_2)( 0.0821\rm \ L•atm•mol^{-1}K^{-1})(293\rm \ K) \\\Rightarrow n_1+n_2&= 1.663 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\\end{aligned}
The mass of the mixture is (in grams) is given by
\begin{aligned}n_1\times 2+n_2\times 4 &= 5\\ \Rightarrow 2n_1+4n_2 &= 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\\\end{aligned}
From equations 1 and 2, we get
n_1= 0.826 and n_2=0.837
Therefore, the ratio of the mass of hydrogen to that of helium in the given mixture is
\dfrac{m_{H}}{m_{He}}=\dfrac{2n_1}{4n_2} = \dfrac{2\times 0.826}{4\times 0.837 } = 0.493
