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A ideal gas whose adiabatic exponent equals `gamma` is expanded so that the amount of heat transferred to the gas is equal to twice of decrease of its internal energy. The equation of the process is `TV^((gamma-1)/k)=` constant (where `T` and `V` are absolute temeprature and volume respectively.

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The correct Answer is:
3

`dQ=dU+dW`
`dQ=-2dU`
`-3dU=dW`
`-3nCvdT=PdV`
`-3nCvdT=PdV`
`(-3nRdT)/(gamma-1)=((nRT)/V)dV`
From this `TV^((gamma-1)/3)=` constant
So, `K=3`
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