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7 July, 02:56

a sealed container filled with argon gas at 35 c has a pressure of 832 torr. if the volume of the container is decreased by a factor of 2 what will happen to the pressure? you may assume the temperature remains at 35 c

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Answers (2)
  1. 7 July, 02:58
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    If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.

    Explanation:

    Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure

    V ∝ 1/P or V₁·P₁ = V₂·P₂

    Where:

    V₁ = Initial volume

    V₂ = Final volume = V₁/2

    P₁ = Initial pressure = 832 torr

    P₂ = Final pressure = Required

    From V₁·P₁ = V₂·P₂ we have,

    P₂ = V₁·P₁/V₂ = V₁·P₁ / (V₁/2)

    P₂ = 2·V₁·P₁/V₁ = 2·P₁ = 2 * 832 torr = 1664 torr
  2. 7 July, 03:25
    0
    The pressure will increase by a factor of 2 and is now 1664 torr

    Explanation:

    The question says for us to assume that temperature is constant. Now, since we are given pressure and volumw, we will use Boyle's law which is a law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature i. e V ∝ 1/P

    Thus, PV = K

    Where K is a constant.

    So,

    P₁•V₁ = P₂•V₂ = k

    Where:

    P₁ = Initial pressure

    V₁ = Initial volume

    P₂ = Final pressure

    V₂ = Final volume = V₁/2

    From the question, P₁ = 832 torr; we are told that volume is decreased by 2, thus, V₂ = V₁/2

    Now, we want to find out what will happen to the pressure P₂.

    Let's make P₂ the subject;

    P₁•V₁ = P₂•V₂

    Thus, P₂ = (P₁•V₁) / V₂

    Plugging in the relevant values to obtain;

    P₂ = (P₁•2V₁) / V₁

    V₁ will cancel out and we have;

    P₂ = 2P₁

    Now, we are given that P₁ = 832 torr, Thus,

    P₂ = 2P₁ = 2 * 832 torr = 1664 torr
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