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An ideal monoatomic gas initially in sta...

An ideal monoatomic gas initially in state 1 with pressure `P_(1)=20` atm and volume `V_(1)1500cm^(3)` it is then taken to state 2 with pressure `P_(2)=1.5P_(1)` and volume `V_(2)=2V_(1)` find the change in internal energy in this process in KJ. (take `1atm` lit `=100J`)

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The correct Answer is:
9
`DeltaE=nCv(T_(2)-T_(1))`
`DeltaE=nxx(3)/(2)R((P_(2)V_(2))/(nR)-(P_(1)V_(1))/(nR))`
`=(3)/(2)(1.5xx20xx2xx1.5-20xx1.5)`
`=90lit-atm`
`=9000J`
`=9KJ`
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