If the ration of the masses of `SO_(3)` and `O_(2)` gases confined in a vessel is `1 : 1` , then the ratio of their partial pressure would be

Options

A. `5 : 2`
B. `2 : 5`
C. `2 : 1`
D. `1 : 2`

Answer

B

Explanation

Let m be the mass of `SO_(3)` and `O_(2)` enclosed in the vessel.
Number of moles of `SO_(3)=(m)/(80)`
Number of moles of `O_(2)=(m)/(32)`
Partial pressure of `SO_(3), P_(A)=(m)/(80)xx(R xx T)/(V)`
Partial pressure of `O_(2), P_(B)=(m)/(32)xx(R xx T)/(V)`
Now, `(P_(A))/(P_(B))=(m)/(80)xx(32)/(m)=(2)/(5)`
Hence, ratio of partial pressure of m g of `SO_(3)` and `O_(2)` is 2 : 5.