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» <html><body><p>10.0 bar <br/>5.2 bar <br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.6 bar <br/>5.0 bar</p>Solution :At constant temperature, <br/> `P_(1)V_(1)+P_(2)V_(2)=P_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)(V_(1)+V_(2))`<br/> `(4.0" bar")(4.0" dm"^(3))+(6.0" bar")(6.0" dm"^(3))` <br/> `=P_(3)(4.0+6.0" dm"^(3))` <br/> or `P_(3)=(16+36)/(10)=(<a href="https://interviewquestions.tuteehub.com/tag/52-324553" style="font-weight:bold;" target="_blank" title="Click to know more about 52">52</a>)/(10)=5.2" bar"`</body></html>