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The presence of 1 g of an ideal gas A at...

The presence of `1 g` of an ideal gas `A` at `27^(@)C is 2 bar`. When `2 g` of another ideal gas `B` is added to the same flask at the same temperature, the pressure becomes `3 bar`. The relationship between their molar masses is

A

`M_(A) = (1)/(4)M_(B)`

B

`M_(A) = M_(B)`

C

`M_(A) = (1)/(2) M_(B)`

D

`M_(A) = 4 M_(B)`

Text Solution

Verified by Experts

The correct Answer is:
A
According to data ,
`{:(p_(A) = 2 ba r,,,p_(B) = 1 ba r,),(m_(A) = 1 g,,,m_(B) = 2g,):}`
According to the ideal gas equation `(pV = nRT)`, we have
`p_(A)V = n_(A)RT`
`p_(B)V = n_(B)RT`.
Dividing the two equation , we get
`(p_(A))/(p_(B)) = (n_(A))/(n_(B)) = (m_(A))/(M_(A)) xx (M_(B))/(m_(B)) = (m_(A))/(m_(B))(M_(B))/(M_(A))`
`(2 ba r)/(1 ba r) = (1 g)/(2 g) xx (M_(B))/(M_(A))`
Thus, `M_(A) = (1)/(4)M_(B)`
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