When 2g of a gas A is introduced into an evacuated flask kept at $25^{0} C$ , the pressure is found to be 1 atmosphere. If 3g of another gas is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal behaviour, the ratio of their molecular weights $M_{A} : M_{B}$ is

3 : 1

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1 : 3

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2 : 3

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3 : 2

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Solution

The correct option is B
1 : 3

$P_{1} V_{1} = \frac{W_{1}}{M_{1}} R T_{1}$

$\frac{P_{A} V_{A}}{P_{B} V_{B}} = \frac{W_{A}}{W_{B}} \frac{M_{B}}{M_{A}} \frac{T_{A}}{T_{B}}$

$\frac{M_{A}}{M_{B}} = 1 : 3$

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Similar questions

Q. When 2g of a gas A are introduced into an evacuated flask kept at

25^{0} C

, the pressure is found to be one atmosphere, if 3g of another gas B are then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, Calculate the ratio of molecular weights

M_{A} : M_{B}

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