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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.

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

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Given Pressure of 1g of an ideal gas A at $27^{\circ}C$ is $P_{A} = 2\ bar$.

When 2g of another ideal gas B is introduced in the same flask at the same temperature,

The pressure becomes $P_{A}+P_{B}= 3\ bar$.

$\implies P_{B }= 1\ bar$.

We can assume the molecular masses of A and B be $M_{A}\ and\ M_{B}$ respectively.

The ideal gas equation, $PV=nRT$

So, we have $P_{A}V=n_{A}RT$   and   $P_{B}V=n_{B}RT$

Therefore, $\frac{P_{A}}{P_{B}} = \frac{n_{A}}{n_{B}} = \frac{1M_{B}}{2M_{A}} = \frac{M_{B}}{2M_{A}}$

or, $\frac{M_{B}}{M_{A}} = 2\times\frac{P_{A}}{P_{B}} = 2\times \frac{2}{1} =4$

Hence the relation between the two gases is $M_{B} =4M_{A}$.

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