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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 beA. 10.0 barB. 5.2 barC. 1.6 barD. 5.0 bar |
Answer» Correct Answer - 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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