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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

1.6 bar

D

5.0 bar

Text Solution

Verified by Experts

The correct Answer is:
B
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"`
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