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A 4.0 dm^(3) flask containing N(2) at 4...

A `4.0 dm^(3)` flask containing `N_(2) at 4` bar was connected to a `6.0 dm^(3)` flask containing helium at `6` bar , and the gases were allowed to mix isothermally. The total pressure of the resulting mixture will be

A

10.0 bar

B

5.2 bar

C

3.6 bar

D

1.6 bar

Text Solution

Verified by Experts

The correct Answer is:
B
According to Boyle.s law, at constant temperature,
`p_(1)V_(1)+p_(2)V_(2)=p_("mix")xxV_("mix")`
`4xx4+6xx6=p_("mix")xx(6+4)`
`thereforep_("mix")=(52)/(10)="5.2 bar"`
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