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# 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 agoFavorite 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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