If `200 mL " of " N_(2) " at " 25^(@)C` and a pressure of 250 mm are mixed with `350 mL " of " O_(2) " at " 25^(@)C` and a pressure of 300 mm so that, the volume of resulting mixture is `300 mL`, what would be the final pressure of the mixture at `25^(@)C`?

Explanation

Since temperature is constant, Boyle's Law is applicable
Step I. Calculation of partical pressure of `N_(2)` after mixing.
`V_(1) = 200 mL` , `V_(1) = 300` mL
`P_(1) = 250` mm and `P_(2) = ?`
According to Boyle's Law, `P_(1)V_(1) = P_(2)V_(2)`
or `P_(2) = (P_(1)V_(1))/(V_(2)) = ((250 mm) xx (200 mL))/((300 mL)) = 166.6 mm`
Step II. Calculation of partial pressure of `O_(2)` after mixing
`V_(1) = 350 mL` , `V_(2) = 300 mL`
`P_(1) = 300 mm` , `P_(2) = ?`
According to Boyle's Law, `P_(1)V_(1) = P_(2)V_(2)`
`P_(2) = (P_(1)V_(1))/(V_(2)) = ((300 mm) xx (350 mL))/((300 mL)) = 350 mm`
Step III. Calculation of total pressure of the gaseous mixture
`P = P_(1) + P_(2) = 166.6 + 350 = 516.6 mm`.