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`)

An ideal monoatomic gas initially in state 1 with pressure `P_(1)=20`

1.	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`)
Answer» Correct Answer - 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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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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