A 4.0 dm^(3) flask containing N_(2) at 4.0 bar was connected to a 6.0 dm^(3) flask containing helium at 6.0 bar, and the gases were allowed to mix isothermally, then the total pressure of the resulting mixture will be

A 4.0 dm^(3) flask containing N_(2) at 4.0 bar was connected to a 6.0

1.	A 4.0 dm^(3) flask containing N_(2) at 4.0 bar was connected to a 6.0 dm^(3) flask containing helium at 6.0 bar, and the gases were allowed to mix isothermally, then the total pressure of the resulting mixture will be
Answer» 10.0 bar 5.2 bar 1.6 bar 5.0 bar Solution :At constant temperature, `P_(1)V_(1)+P_(2)V_(2)=P_(3)(V_(1)+V_(2))` `(4.0" bar")(4.0" dm"^(3))+(6.0" bar")(6.0" dm"^(3))` `=P_(3)(4.0+6.0" dm"^(3))` or `P_(3)=(16+36)/(10)=(52)/(10)=5.2" bar"`