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As a result of the isobaric heating by D...

As a result of the isobaric heating by `DeltaT=72K`, one mole of a certain ideal gas obtain an amount of heat `Q=1.6kJ`. Find the work performed by the gas, the increment of its internal energy and `gamma`.

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The correct Answer is:
A
`W=nRDeltaT`
`=(1)(8.31)(72)
`=600J=0.6kJ`
`DeltaU=Q-W=1.0kJ`
`gamma=C_(p)/C_(V)=(nC_pDeltaT)/(nC_VDeltaT)=(Q)/(DeltaU)`
`=1.6/1.0=1.6`
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