Pressure of $1 g$ of an ideal gas $A$ at $27^{\circ} C$ is found to be $2 b a r$ . When $2 g$ of another ideal gas $B$ is introduced in the same flask at same temperature the pressure becomes $3 b a r$ . What would be the ratio of their molecular masses?

4 : 1

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1 : 4

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1 : 8

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2 : 8

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Solution

The correct option is A $1 : 4$
For gas $A, P_{A} V = \frac{m_{A}}{M_{A}} R T . . . . (1)$
For gas $B, P_{B} V = \frac{m_{B}}{M_{B}} R T . . . . (2)$
Dividing equation $(1)$ by equation $(2)$ gives
$\frac{P_{A}}{P_{B}} = \frac{m_{A}}{m_{B}} \frac{M_{B}}{M_{A}}$
$P_{A} + P_{B} = 3 \Rightarrow P_{B} = 3 - 2 = 1 b a r$
$\frac{M_{A}}{M_{B}} = (\frac{m_{A}}{m_{B}}) (\frac{P_{B}}{P_{A}})$

$= (\frac{1 g}{2 g}) (\frac{1 b a r}{2 b a r})$

$\Rightarrow \frac{M_{A}}{M_{B}} = 1 : 4$ .

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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. What would be the ratio of their molecular masses?

The correct option is A 1:4For gas A,PAV=mAMART....(1)For gas B,PBV=mBMBRT....(2)Dividing equation (1) by equation (2) givesPAPB=mAmBMBMAPA+PB=3⇒PB=3−2=1 barMAMB=(mAmB)(PBPA)=(1 g2 g)(1 bar2 bar)⇒MAMB=1:4.

Pressure of $1 g$ of an ideal gas $A$ at $27^{\circ} C$ is found to be $2 b a r$ . When $2 g$ of another ideal gas $B$ is introduced in the same flask at same temperature the pressure becomes $3 b a r$ . What would be the ratio of their molecular masses?