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

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Solution

The correct option is D

$ρ_{A} V = \frac{w_{A}}{M_{A}} R T a n d ρ_{t o t a l} V = ⟮ \frac{w_{A}}{M_{A}} + \frac{w_{B}}{M_{B}} ⟯ R T$

Substituting the given values , we get

$2 b a r \times V = ⟮ \frac{1 g}{M_{A}} ⟯ R T a n d 3 b a r \times V = ⟮ \frac{1 g}{M_{A}} + \frac{2 g}{M_{B}} ⟯ R T$

From the two equations, we get

$\frac{3}{2} = \frac{\frac{1 g}{M_{A}} + \frac{2 g}{M_{B}}}{\frac{1 g}{M_{A}}} = \frac{\frac{M_{B} + 2 M_{A}}{M_{A} M_{B}}}{\frac{1}{M_{A}}}$

$\frac{M_{B} + 2 M_{A}}{M_{B}} = 1 + \frac{2 M_{A}}{M_{B}}$

This gives , $\frac{2 M_{A}}{M_{B}} = \frac{3}{2} - 1 = \frac{1}{2} \Rightarrow \frac{M_{A}}{M_{B}} = \frac{1}{4} \Rightarrow M_{B} = 4 M_{A}$

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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 the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

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