You have helium gas in the right side of a two-bulbed container connected by a valve. Initially the valve is closed.?

The left side of the container is evacuated (it is a vacuum; the volume is 9.43L). The helium in the right side of the container is at 4.40 atm (the volume of the right side of the container is 3.23L). After the valve is opened, what is the pressure of the helium gas in the entire container? Assume constant temperature, ideal behavior, and that the volume of the valve is too small to worry about.

3 Answers

  • 2 months ago
    Favorite Answer

    (4.40 atm) x (3.23 L) / (3.23 L + 9.43 L) = 1.12 atm

  • 2 months ago

    Boyle's Law.

    P1V1 = P2V2

    (4.40 atm) (3.23 L) = (x) (12.66 L

    Solve for x

    The 12.66 comes from 3.23 + 9.43

  • 2 months ago

    The bridge would have to be a long longer to reach the other side of the bay?

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