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Pressure of 1g of an ideal gas A at 27^(...

Pressure of `1g` of an ideal gas `A` at `27^(@)C` is found to be 2 bar when `2g` of another ideal gas `B` is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship thieir molecular masses .

A

`4:1`

B

`1:4`

C

`1:8`

D

`2:8`

Text Solution

Verified by Experts

The correct Answer is:
B
For gas A, `P_(A)V=(m_(A))/(M_(A))RT` …..(1)
For gas B, `P_(B)V=(m_(B))/(M_(B))RT` …..(2)
Dividing equation (1) by equation (2) gives
`(P_(A))/(P_(B))=(m_(A))/(m_(B))(M_(B))/(M_(A))`
`P_(A)+P_(B)=3 rArr P_(B)=3-2=1` bar
`(M_(A))/(M_(B))=((m_(A))/(m_(B)))((P_(B))/(P_(A)))=((1g)/(2g))(("1 bar")/("2 bar")) rArr (M_(A))/(M_(B))= 1 : 4`
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