A certain quantiy of gas occupies a volume of `0.8 L` collected over water at `300 K` and a pressure `0.92 "bar"`. The same gas occupies a volume of `0.08 L` at `STP` in dry conditions. Calculate the aqueous tension at `300 K`.

Explanation

Let the aqueous tension at `300 K` be `P"bar"`.
The pressure of dry gas at `300 K=(0.92-P) "bar"`
Now, `P_(1)=(0.92-P)"bar"`, `P_(2)= 1"bar"`
`V_(1)=0.8 L`, `V_(2)=0.08 L`
`T_(1)=300 K`, `T_(2)=273.15 K`
Using gas equation, we get
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
or `P_(1)=(P_(2)V_(2)T_(1))/(V_(1)T_(2))`
Subsituting the values, we get
`0.92-P=(1xx0.08xx300)/(0.8xx273.15)=0.109 `
or aqueous tension `=0.92-0.109=0.811 "bar"`