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An ideal monoatomic gas is initially in ...

An ideal monoatomic gas is initially in state `1` with pressure `p_(1) = 20 atm` and volume `v_(1) = 1500 cm^(3)`. If is then taken to state `2` with pressure `p_(2) = 1.5 p_(1)` and volume `v_(2) = 2v_(1)`. The change in internal energy from state `1` to state `2` is equal to

A

`2000J`

B

`3000 J`

C

`6000 J`

D

`9000 J`

Text Solution

Verified by Experts

The correct Answer is:
D
`p_(1) = 20 xx 10^(5) N//m^(2), v_(1) = 1500 xx 10^(-6) m^(3)`
`p_(2) = 30 xx 10^(5) N//m^(2) , v_(2) = 300 xx 10^(-6) m^(3)`
`T_(1) = (p_(1)v_(1))/(nR)` and `T_(2) = (p_(2)v_(2))/(nR)`
`dU = nC_(v)dT = n. (3)/(2)R. (T_(2)-T_(1))`
`=n.(3)/(2)R ((p_(2)v_(2)-p_(1)v_(1)))/(nR)`
`= (3)/(2) (p_(2)v_(2)-p_(1)v_(1)) = 9000 J`
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