Q. 53.8( 91 Votes )

# 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.

Answer :

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

⇒ P_{a}V = n_{a}RT

Where, Pa 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

Where,P_{a} and n_{b} represent the pressure and number of moles of gas B.

[V and T are constants for gases A and B]

Therefore, we have,

⇒

Where, M_{a} and M_{b} are the molecular masses of gases A and B respectively.

Now, we have

Substituting the values

m_{a} = 1 g

P_{a} = 2 bar

m_{b} = 2 g

P_{b} = 1 bar.

Given,

(Since total pressure is 3 bar)

Substituting these values in equation we have

⇒

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

⇒ 4M_{a} = M_{b}

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