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How to rearrange and get T2 from this equation?

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    P_1 \left( \frac{P_2 V_2 T_1}{T_2 P_1} \right)^n = P_2V_2^n

    I know that V2^n is on both sides so cancels out, but not sure what to do next.
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    (Original post by mh1985)
    P_1 \left( \frac{P_2 V_2 T_1}{T_2 P_1} \right)^n = P_2V_2^n

    I know that V2^n is on both sides so cancels out, but not sure what to do next.
    So once you expand and cancel V2^n you'll have:

    \frac{P_1 P_2^n T_1^n}{T_2^n P_1^n}\ = P_2

    Now you need to have a go at factorising T2^n out of the equation to get T2^n on its own.
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    P1(P2V2T1/T2P1) power n = P2V2 POWER N
    I will suggest that you have to take an 1/n to both sides because form this there will no effect on your equation.
    after that you will say that at the left side you will be get rid of power n. and your equation will be re-arrange like
    P1 Power 1/n (P2V2T1/T2P1)= P2 Power 1/n x V2

    1/T2= P2 power 1/n x v2 x p1/p1 Power 1/'n x P2V2T1
    t2 = p1power 1/nxp2v2t1 / p2 power 1/n x v2 x p1

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    if i am wrong then please tell me the rigth one method to re-arrange that equation.


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