A 285-mL flask contains pure helium at a pressure of 750 torr . A second flask with a volume of 455 mL contains pure argon at a pressure of

Question

A 285-mL flask contains pure helium at a pressure of 750 torr . A second flask with a volume of 455 mL contains pure argon at a pressure of 732 torr .If we connect the two flasks through a stopcock and we open the stopcock, what is the partial pressure of helium

in progress 0
Bình An 5 months 2021-08-29T12:59:22+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-29T13:00:51+00:00

    Answer:

    P_{He}^|=288.85torr

    Explanation:

    Given data

    P_{He}=750 torr\\V_{He}=285mL\\P_{Ar}=732 torr\\V_{Ar}=455 mL

    Required

    Partial pressure of helium

    Solution

    First calculate the total volume of gas mixture once the stopcock is opened

    So

    V_{total}=V_{He}+V_{Ar}\\V_{total}=285mL+455mL\\V_{total}=740mL

    As temperature remains constant,by Boyle’s Law we calculate the partial pressure of the helium gas

    So

    P_{He}V_{He}=P_{He}^|V_{total}\\P_{He}^|=\frac{P_{He}V_{He}}{V_{total}}\\P_{He}^|=\frac{750torr*285mL}{740mL}\\  P_{He}^|=288.85torr

    0
    2021-08-29T13:01:16+00:00

    Explanation:

    Below is an attachment containing the solution.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )