At a particular temperature and pressure, the volume of 12 g of `H_2` gas is 134.5 L. what will be the final volume of the gas if 4 g of `H_(2)` gas is added to it at the same temperature and pressure?

Explanation

12 g `H_(2)=(12)/(2)=6" mol of "H_(2)`.
similarly, 4 g `H_(2)=2` mol of `H_(2)`.
We know, at constant temperature and pressure, `(V_(1))/(V_(2))=(n_(1))/(n_(2))`
given that, `n_(1)=6,n_(2)=6+2=8, V_(1)=134.5L and V_(2)=?`
`therefore V_(2)=(n_(2))/(n_(1))xxV_(1)=(8)/(6)xx134.5=179.3L`
Therefore, at the same temperature and pressure, if 4g of `H_(2)` gas is added to 12 g of `H_(2)` gas, then the final volume of the gas will be 179.3 L.