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  1. #1
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    word problem

    The volume V of a gas varies inversely as the pressure P upon it. The volume of a gas is 200 cu cm under a pressure of 32kg/sq. cm
    a. find the variation constant
    b. what will be its volume under a presure of 40 kg/sq. cm
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by imbadatmath View Post
    The volume V of a gas varies inversely as the pressure P upon it. The volume of a gas is 200 cu cm under a pressure of 32kg/sq. cm
    a. find the variation constant
    b. what will be its volume under a presure of 40 kg/sq. cm
    Hint: to vary inversly means they have this relationship.

    $\displaystyle V = \frac kT$ .........for $\displaystyle k$ some constant (call the proportionality constant)
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    im sorry that made no sense to me lol i am so lost
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by imbadatmath View Post
    The volume V of a gas varies inversely as the pressure P upon it. The volume of a gas is 200 cu cm under a pressure of 32kg/sq. cm
    a. find the variation constant
    b. what will be its volume under a presure of 40 kg/sq. cm
    I think Jhevon should have wrote it as:

    $\displaystyle V = \frac{k}{P}$

    Where $\displaystyle V$ is the volume and $\displaystyle P$ the pressure.

    In the first question, to find the constant $\displaystyle k$ you need to put the given values for $\displaystyle V$ and $\displaystyle P$ in the equation, and solve for $\displaystyle k$.

    In the second question you now make use of that value for $\displaystyle k$ when solving for the volume, $\displaystyle V$.
    Last edited by janvdl; Apr 1st 2008 at 01:02 PM. Reason: Typo
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    I think Jhevon should have wrote it as:

    $\displaystyle V = \frac{k}{P}$

    Where $\displaystyle V$ is the volume and $\displaystyle P$ the pressure.

    In the first question, to find the constant $\displaystyle k$ you need to put the given values for $\displaystyle V$ and $\displaystyle P$ in the equation, and solve for $\displaystyle k$.

    In the second question you know make use of that value for $\displaystyle k$ when solving for the volume, $\displaystyle V$.
    yes, i meant P not T
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  6. #6
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    i am confused how to solve for k
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  7. #7
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    As janvdl said, substitute in the given values of V and P and solve for k.

    $\displaystyle
    200 = \frac{k}{32}
    $

    Solve for k, multiply both sides by 32.
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  8. #8
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    32*200=6400 is that the value of k
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  9. #9
    Bar0n janvdl's Avatar
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    Quote Originally Posted by imbadatmath View Post
    32*200=6400 is that the value of k
    Yes. Now use that value for $\displaystyle k$ and the given value for $\displaystyle P$, to solve for $\displaystyle V$. (The second question.)
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  10. #10
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    Yes so you have the proportionality formula:

    $\displaystyle
    V = \frac{6400}{P}
    $

    To find V when P = 40, substutute P = 40 into the formula.
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  11. #11
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    so the volume in the second one owuld equal 160
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