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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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