what is the integral of the equation: (1-k)dV/V = dT/T [k=Cp/Cv and it is constant]?

how do you get this: (V2/V1)^(1-k)=T2/T1 [k=Cp/Cv]

2 Answers

  • 10 years ago
    Favorite Answer

    Its straight forward integration. Integrate the LHS between V1 and V2 (with respect to V) .. and at the same time integrate the RHS (with respect to T of course) between the limits T1 and T2

    what you will get is

    (1-k) INT [ dV/V ] = INT [ dT/T ] ..... ( 1-k is constant so you can take it out of the integral) ( I am using INT for integral...)

    which if course gives

    (1-k) ln [V] = ln [T] .... of course you must put in your limits... i cant draw that on here

    you will get

    (1-k) ln [V2/V1] = ln [T2/T1]

    use the property of logs that ..... a ln[x] =ln[x^a] to get

    ln [ (V2/V1)^(1-k) ] = ln [T2/T1]

    now take exponents

    (V2/V1)^(1-k) = T2/T1

    and there is your answer!

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    4 years ago

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