Question

Pressure of 1 g of an ideal gas A at ${27}^{\circ }C$ is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar.
Find a relationship between their molecular masses.

Answer 1

For ideal gas A, the ideal gas equation is given by,
$p_{A}V=n_{A}RT\dots \left(i\right)$
Where, $p_A$ and $n_A$ represent the pressure and number of moles of gas A.
For ideal gas B, the ideal gas equation is given by,
$p_{B}V=n_{B}RT\dots \left(ii\right)$
Where, $p_B$ and $n_B$ represent the pressure and number of moles of gas B.
[V and T are constants for gases A and B]
From equation (i), we have
$p_{A}V=\frac{m_A}{M_A}RT\Rightarrow \frac{p_A{M}_A}{m_A}=\frac{RT}{V}\dots \left(iii\right)$
From equation (ii), we have
$p_{B}V=\frac{m_B}{M_B}RT\Rightarrow \frac{p_B{M}_B}{m_B}=\frac{RT}{V}\dots \left(iv\right)$
Where, $M_A$ and $M_B$ are the molecular masses of gases A and B respectively.
Now, from equations (iii) and (iv), we have
$\frac{p_A{M}_A}{m_A}=\frac{p_B{M}_B}{m_b}\dots \left(v\right)$
Given,
$m_A=1g{p}_A=2bar{m}_B=2g{p}_B=(3-2)=1bar$
(Since total pressure is 3 bar)
Substituting these values in equation (v), we have
$\frac{2\times M_A}{1}=\frac{1\times M_B}{2}\Rightarrow 4M_A=M_B$
Thus, a relationship between the molecular masses of A and B is given by
$4M_A=M_B$

The molecular mass of B is four times of molecular mass of A.