Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar.
Question:

Pressure of 1 g of an ideal gas A at 27 °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.

Solution:

For ideal gas A, the ideal gas equation is given by,

$p_{\mathrm{A}} V=n_{\mathrm{A}} \mathrm{R} T \ldots \ldots .(\mathrm{i})$

Where, pA and nA represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by,

$p_{B} V=n_{B} R T$

Where, pand nB 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

$p_{A} V=\frac{m_{A}}{\mathrm{M}_{A}} \mathrm{R} T \Rightarrow \frac{p_{A} \mathrm{M}_{A}}{m_{A}}=\frac{\mathrm{R} T}{V}$…….(iii)

From equation (ii), we have

$p_{\mathrm{B}} V=\frac{m_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \mathrm{R} T \Rightarrow \frac{p_{\mathrm{B}} \mathrm{M}_{\mathrm{B}}}{m_{\mathrm{B}}}=\frac{\mathrm{R} T}{V}$ …….(iv)

Where, Mand MB are the molecular masses of gases A and B respectively.

Now, from equations (iii) and (iv), we have

$\frac{p_{A} \mathrm{M}_{A}}{m_{A}}=\frac{p_{B} \mathrm{M}_{B}}{m_{B}} \ldots \ldots \ldots(v)$

Given,

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

$\frac{2 \times \mathrm{M}_{A}}{1}=\frac{1 \times \mathrm{M}_{B}}{2}$

$\Rightarrow 4 \mathrm{M}_{A}=\mathrm{M}_{B}$

Thus, a relationship between the molecular masses of A and B is given by

$4 \mathrm{M}_{A}=\mathrm{M}_{B}$

 

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