Pressure of 1 g of an ideal gas ‘X’ at $27^{0} C$ is 2P bar. When 2 g of another ideal gas ‘Y’ is introduced into the same flask at same temperature, the pressure becomes 1.5 times. Find out $\frac{M o l a r m a s s o f Y}{M o l a r m a s s o f X}$

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answer is 4.

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Detailed Solution

Since, V and T are constant, we have,
$\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}$

$\frac{P_{1} M_{1}}{w_{1}} = \frac{P_{2} M_{2}}{w_{2}}$
$\frac{2 P \times M_{1}}{1} = \frac{P \times M_{2}}{2}$ $\Rightarrow \frac{M_{2}}{M_{1}} = 4$

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